invalid for tag treeselect according to TLD

将jeesite的一个oa项目从myeclipse 换到eclipse上面,点击修改用户的信息就出现这个问题,之前一直好好的。
org.apache.jasper.JasperException: /WEB-INF/views/modules/sys/userForm.jsp (line: 177, column: 16) Attribute okFunction invalid for tag treeselect according to TLD
tld相关的jar包jstl和standard我都有 ,但是tomcat启动时显示TLD skipped. URI: http://java.sun.com/jstl/xml_rt is already defined ,之后登陆系统其它功能都没问题 就是涉及用户表单这块不行 ,查看可以,修改一点就报错

1个回答

Csdn user default icon
上传中...
上传图片
插入图片
抄袭、复制答案,以达到刷声望分或其他目的的行为,在CSDN问答是严格禁止的,一经发现立刻封号。是时候展现真正的技术了!
其他相关推荐
struts2.3升级到2.5的问题
在官网上下载了2.5-min-lib把原来的2.3版本替换之后,修改本部分代码后成功启动,结果报了: org.apache.jasper.JasperException: /index.jsp (line: 400, column: 11) Attribute id invalid for tag iterator according to TLD 迭代器标签没有id这个属性。有没有大神知道该怎么解决。
Programming the EDSAC
Problem Description The world's first full-scale, stored-program, electronic, digital computer was the EDSAC (Electronic Delay Storage Automatic Calculator). The EDSAC had an accumulator-based instruction set, operating on 17-bit words(and 35-bit double words), and used a 5-bit teletypewriter code for input and output. The EDSAC was programmed using a very simple assembly language: a single letter opcode followed by an unsigned decimal address, followed by the the letter "F"(for full word) or "D"(for double word). For example, the instruction "A 128 F" would mean "add the full word at location 128 to the accumulator", and would be assembled into the 17-bit binary value, 11100000100000000, consisting of a 5-bit opcode(11100="add"), an 11-bit operand (00010000000 = 128), and a single 0 bit denoting a full word operation(a 1 bit would indicate a double word operation). Although arithmetic on the EDSAC was fixed point two's complement binary, it was not mere intger arithmetic (as is common with modern machines). The EDSAC hardware assumed a binary point between the lrftmost bit and its immediate successor. Thus the hardware could handle only values in the range -1.0 <= x < 1.0 . For example: As you can see, the largest possible positive value was: and the smallest possible positive value was: (This also happens to be the increment between successive values on the EDSAC). By a curious coincidence(or an elegant design decision), the opcode for the add operation(11100) was the same as the teleprinter code for the letter"A". The opcode for subtract was the same as the teleprinter code for "S"(01100), and so on. This simplified the programming for the assembler (which, incidentally, was a mere 31 instructions long). The EDSAC teleprinter alphabet was "PQWERTYUIOJ#SZK*?F@D!HNM&LXGABCV"(with "P"=00000, "Q"=00001, and so on, up to "V"=11111) UNfortunately, the EDSAC assembler had no special directives for data values. On the other hand, there was no reason that ordinary instructions couldn't be used for this , thus, an EDSAC programmer desiring to reserve space for the constant 3/4(represented as 01100000000000000) would use the instruction "S O F" and for 1/3(which is approximately represented as 00101010101010101)"T 682 D", and so on. Your job is to write a program that will translate decimal input value into the appropriate EDSAC instructions. Input The frist line of input contains a single integer P,(1<=P<=1000),which is the number of data sets that follow. Each data set is a single line that contains of two space separated values N and D. N is the data set number. D is the decimal number of the from sd.ddd...., where s is an optional minus sign, and d is any decimal digit(0,-9). There will be at least 1 and at most 16 digits after the decimal point. Output For each data set there is one line of output. It contains the data set number(N)followed by a single space, followed by the EDSAC instruction necessary to specify the given constant. The instruction should be printed as follows:the"opcode"character followed by a space followed by the operand(as a non-negative decimal integer)followed by a space followed by an'F'or'D'(as appropriate). If the constant cannot be represented exactly in 17 bits,the value is to be rounded toward zero(up for negative,down for positive numbers). If the input value D is not in the range -1.0 <= D <= 1.0, the string"INVALID VALUE"should be printed instead of an EDSAC instruction. Sample Input 16 1 0.5 2 -0.5 3 -1.0000000 4 0.1 5 0.0000152587890625 6 0.0000152587890624 7 0.0000152587890626 8 -0.0000152587890625 9 -0.0000152587890624 10 -0.0000152587890626 11 0.9999999999999999 12 -0.9999999999999999 13 -5.3 14 9.1 15 -1.0000000000000001 16 0.31415926 Sample Output 1 I 0 F 2 & 0 F 3 ? 0 F 4 Q 1228 D 5 P 0 D 6 P 0 F 7 P 0 D 8 V 2047 D 9 P 0 F 10 V 2047 D 11 * 2047 D 12 ? 0 D 13 INVALID VALUE 14 INVALID VALUE 15 INVALID VALUE 16 T 54 F
Judges's Final Problem C语言怎么做
Problem Description For any ACM/ICPC participant, it is always discouraging to discover that the real problems they face in the programming world are seldom as interesting as the tasks they face in the algorithm-related contests. While the efficiency of your algorithm is certainly vital to the overall system, still it is seldom valued so much as other factors such as readability, and the ability of coding accurately often takes precedence in many projects. Still, we judges believe that a balance between these two can be achieved - a good programmer should be able to come up with code that is fast, reliable and readable in a reasonable amount of time - and that is why you, the contestant, is asked to solve the problem of building a ranking system for us. You're probably familiar with the following paragraph taken from Rules of ACM/ICPC: A problem is solved when it is accepted by the judges. Teams are ranked according to the most problems solved. Teams who solve the same number of problems are ranked by least total time. The total time is the sum of the time consumed for each problem solved. The time consumed for a solved problem is the time elapsed from the beginning of the contest to the submittal of the accepted run plus 20 penalty minutes for every rejected run for that problem regardless of submittal time. There is no time consumed for a problem that is not solved. During a contest, the following kinds of requests might be submitted by participants: Submission (Format: S [minute]:[Team No]:[Problem ID]:[Result]) If a team submits a problem which they have already solved, then this submission should be considered invalid and ignored by your system; otherwise, the submission is valid and saved for further processing - if the result is 1 (Correct), then this submission should be considered accepted by the system. Query a team's place on the rank list. (Format: R [Team No]) Sort the teams according to the contest rules mentioned above. A tie may occur between two teams who solved equal number of problems and same penalty. Two teams who solved same number of problems and have same penalty should always have same rank; however, when querying the Kth team on the rank list, even if many teams share the same rank, only one team should be printed (please refer to the rule given below). Query the Kth team on the rank list. (Format: T [k]) If a tie occurs, return the team with the minimum last accepted submission time. Note that the submission time of two teams might be different (i.e., in seconds) even though it may appear otherwise in the input. Initially every team has no submissions, and the contest will last no more than 5 hours. Input There are multiple test cases in the input. Each test case starts with two integers, N and M, (1 <= N <= 10000, 1 <= M <= 10), the number of teams participating in the competition, and the total number of problems in the contest. Teams are numbered from 0 to N - 1. Each of the following lines is either in one of the three formats listed above, or a single line "Contest Ends" followed by an empty line indicating the end of current test case. It is guaranteed that there are no more than 100000 requests in each test case. Input ends with End-of-File. Output For every submission request, if the submitted program is accepted, you need to output a line whose format is [Team No] [Problem ID]; for every query about team's place on the rank list, output the result in one single line; for every query about the team with the Kth highest score, if a team is found, output a line with one integer, the number of the team, or -1 if you are unable to find such a team. Please print a blank line after each test case. Sample Input 5 8 T 1 T 2 S 5:0:A:0 S 8:0:A:1 S 9:1:B:1 S 15:0:A:1 T 1 T 2 T 3 R 0 R 1 R 2 R 3 Contest Ends Sample Output 0 -1 [0][A] [1][B] 1 0 2 1 0 2 2
求助!谁知道C++UDP打洞?
# 为什么我的代码不在一个局域网内无法连接,也无反应? 服务器(UDP打洞): ``` #include <Winsock2.h> #include <stdio.h> #include <stdlib.h> #define DEFAULT_PORT 4768 #define BUFFER_SIZE 100 int main() { int serverPort = DEFAULT_PORT; WSADATA wsaData; SOCKET serverListen; struct sockaddr_in serverAddr; if (WSAStartup(MAKEWORD(2,2),&wsaData) != 0 ) { printf("Failed to load Winsock.\n"); return -1; } serverListen = socket(AF_INET,SOCK_DGRAM,0); if (serverListen == INVALID_SOCKET) { printf("socket() failed:%d\n",WSAGetLastError()); return -1; } serverAddr.sin_family = AF_INET; serverAddr.sin_port = htons(serverPort); serverAddr.sin_addr.s_addr = htonl(INADDR_ANY); if (bind(serverListen,(LPSOCKADDR)&serverAddr,sizeof(serverAddr)) == SOCKET_ERROR) { printf("bind() failed:%d\n",WSAGetLastError()); return -1; } struct sockaddr_in sourceAddr1; int sourceAddrLen1 = sizeof(sourceAddr1); SOCKET sockC1 = socket(AF_INET,SOCK_DGRAM,0); char bufRecv1[BUFFER_SIZE]; int len; len = recvfrom(serverListen, bufRecv1, sizeof(bufRecv1), 0,(struct sockaddr*)&sourceAddr1,&sourceAddrLen1); if (len == SOCKET_ERROR) { printf("recv() failed:%d\n", WSAGetLastError()); return -1; } printf("C1 IP:[%s],PORT:[%d]\n",inet_ntoa(sourceAddr1.sin_addr) ,ntohs(sourceAddr1.sin_port)); struct sockaddr_in sourceAddr2; int sourceAddrLen2 = sizeof(sourceAddr2); SOCKET sockC2 = socket(AF_INET,SOCK_DGRAM,0); char bufRecv2[BUFFER_SIZE]; len = recvfrom(serverListen, bufRecv2, sizeof(bufRecv2), 0,(struct sockaddr*)&sourceAddr2,&sourceAddrLen2); if (len == SOCKET_ERROR) { printf("recv() failed:%d\n", WSAGetLastError()); return -1; } printf("C2 IP:[%s],PORT:[%d]\n",inet_ntoa(sourceAddr2.sin_addr) ,ntohs(sourceAddr2.sin_port)); char bufSend1[BUFFER_SIZE]; memset(bufSend1,'\0',sizeof(bufSend1)); char* ip2 = inet_ntoa(sourceAddr2.sin_addr); char port2[10]; itoa(ntohs(sourceAddr2.sin_port),port2,10); for (int i=0;i<strlen(ip2);i++) { bufSend1[i] = ip2[i]; } bufSend1[strlen(ip2)] = '^'; for (int i=0;i<strlen(port2);i++) { bufSend1[strlen(ip2) + 1 + i] = port2[i]; } len = sendto(sockC1,bufSend1,sizeof(bufSend1),0,(struct sockaddr*)&sourceAddr1,sizeof(sourceAddr1)); if (len == SOCKET_ERROR) { printf("send() failed:%d\n",WSAGetLastError()); return -1; } else if (len == 0) { return -1; } else { printf("send() byte:%d\n",len); } char bufSend2[BUFFER_SIZE]; memset(bufSend2,'\0',sizeof(bufSend2)); char* ip1 = inet_ntoa(sourceAddr1.sin_addr); char port1[10];//C1的port itoa(ntohs(sourceAddr1.sin_port),port1,10); for (int i=0;i<strlen(ip1);i++) { bufSend2[i] = ip1[i]; } bufSend2[strlen(ip1)] = '^'; for (int i=0;i<strlen(port1);i++) { bufSend2[strlen(ip1) + 1 + i] = port1[i]; } len = sendto(sockC2,bufSend2,sizeof(bufSend2),0,(struct sockaddr*)&sourceAddr2,sizeof(sourceAddr2)); if (len == SOCKET_ERROR) { printf("send() failed:%d\n",WSAGetLastError()); return -1; } else if (len == 0) { return -1; } else { printf("send() byte:%d\n",len); } closesocket(serverListen); closesocket(sockC1); closesocket(sockC2); WSACleanup(); return 0; } ``` 用户C1: ``` #include<Winsock2.h> #include<stdio.h> #include<stdlib.h> #include <thread> #define PORT 7777 #define BUFFER_SIZE 100 int main(int argc,char* argv[]) { char APTTTTT[]="127.0.0.1"; scanf("%s",APTTTTT); int HYH=4768; WSADATA wsaData; struct sockaddr_in serverAddr; struct sockaddr_in thisAddr; thisAddr.sin_family = AF_INET; thisAddr.sin_port = htons(PORT); thisAddr.sin_addr.s_addr = htonl(INADDR_ANY); if (WSAStartup(MAKEWORD(2,2),&wsaData) != 0) { printf("Failed to load Winsock.\n"); return -1; } serverAddr.sin_family = AF_INET; serverAddr.sin_port = htons(HYH); serverAddr.sin_addr.s_addr = inet_addr(APTTTTT); SOCKET sockS = socket(AF_INET,SOCK_DGRAM,0); if (sockS == INVALID_SOCKET) { printf("socket() failed:%d\n",WSAGetLastError()); return -1; } if (bind(sockS,(LPSOCKADDR)&thisAddr,sizeof(thisAddr)) == SOCKET_ERROR) { printf("bind() failed:%d\n",WSAGetLastError()); return -1; } SOCKET sockC = socket(AF_INET,SOCK_DGRAM,0); if (sockC == INVALID_SOCKET) { printf("socket() failed:%d\n",WSAGetLastError()); return -1; } char bufSend[] = "I am C1"; char bufRecv[BUFFER_SIZE]; memset(bufRecv,'\0',sizeof(bufRecv)); struct sockaddr_in sourceAddr; int sourceAddrLen = sizeof(sourceAddr); struct sockaddr_in oppositeSideAddr; int len; len = sendto(sockS,bufSend,sizeof(bufSend),0,(struct sockaddr*)&serverAddr,sizeof(serverAddr)); if (len == SOCKET_ERROR) { printf("sendto() failed:%d\n", WSAGetLastError()); return -1; } len = recvfrom(sockS, bufRecv, sizeof(bufRecv), 0,(struct sockaddr*)&sourceAddr,&sourceAddrLen); if (len == SOCKET_ERROR) { int TRYY=WSAGetLastError(); if(TRYY!= WSAECONNRESET) { printf("recvfrom() failed:%d\n", WSAGetLastError()); return -1; } } closesocket(sockS); if (bind(sockC,(LPSOCKADDR)&thisAddr,sizeof(thisAddr)) == SOCKET_ERROR) { printf("bind() failed:%d\n",WSAGetLastError()); return -1; } char ip[20]; char port[10]; int i; for (i=0;i<strlen(bufRecv);i++) if (bufRecv[i] != '^') ip[i] = bufRecv[i]; else break; ip[i] = '\0'; int j; for (j=i+1;j<strlen(bufRecv);j++) port[j - i - 1] = bufRecv[j]; port[j - i - 1] = '\0'; oppositeSideAddr.sin_family = AF_INET; oppositeSideAddr.sin_port = htons(atoi(port)); oppositeSideAddr.sin_addr.s_addr = inet_addr(ip); unsigned long ul = 1; ioctlsocket(sockC, FIONBIO, (unsigned long*)&ul); while (1) { Sleep(1000); len = sendto(sockC,bufSend,sizeof(bufSend),0,(struct sockaddr*)&oppositeSideAddr,sizeof(oppositeSideAddr)); if (len == SOCKET_ERROR) { printf("while sending package to C2 , sendto() failed:%d\n", WSAGetLastError()); return -1; }else { printf("successfully send package to C2\n"); } len = recvfrom(sockC, bufRecv, sizeof(bufRecv), 0,(struct sockaddr*)&sourceAddr,&sourceAddrLen); if (len == WSAEWOULDBLOCK) { continue; }else { printf("C2 IP:[%s],PORT:[%d]\n",inet_ntoa(sourceAddr.sin_addr) ,ntohs(sourceAddr.sin_port)); printf("C2 says:%s\n",bufRecv); break; } } char buff[1024]={}; while(1) { memset(buff, 0, 1024); scanf("%s",buff); sendto(sockC,buff,sizeof(buff),0,(struct sockaddr*)&oppositeSideAddr,sizeof(oppositeSideAddr)); recvfrom(sockC, bufRecv, sizeof(bufRecv), 0,(struct sockaddr*)&sourceAddr,&sourceAddrLen); printf("%s\n\n",bufRecv); } closesocket(sockC); } ``` 用户C2: ``` #include<Winsock2.h> #include<stdio.h> #include<stdlib.h> #include <thread> #define PORT 8888 #define BUFFER_SIZE 100 int main(int argc,char* argv[]) { char APTTTTT[]="127.0.0.1"; scanf("%s",APTTTTT); int HYH=4768; WSADATA wsaData; struct sockaddr_in serverAddr; struct sockaddr_in thisAddr; thisAddr.sin_family = AF_INET; thisAddr.sin_port = htons(PORT); thisAddr.sin_addr.s_addr = htonl(INADDR_ANY); if (WSAStartup(MAKEWORD(2,2),&wsaData) != 0) { printf("Failed to load Winsock.\n"); return -1; } serverAddr.sin_family = AF_INET; serverAddr.sin_port = htons(HYH); serverAddr.sin_addr.s_addr = inet_addr(APTTTTT); SOCKET sockS = socket(AF_INET,SOCK_DGRAM,0); if (sockS == INVALID_SOCKET) { printf("socket() failed:%d\n",WSAGetLastError()); return -1; } if (bind(sockS,(LPSOCKADDR)&thisAddr,sizeof(thisAddr)) == SOCKET_ERROR) { printf("bind() failed:%d\n",WSAGetLastError()); return -1; } SOCKET sockC = socket(AF_INET,SOCK_DGRAM,0); if (sockC == INVALID_SOCKET) { printf("socket() failed:%d\n",WSAGetLastError()); return -1; } char bufSend[] = "I am C2"; char bufRecv[BUFFER_SIZE]; memset(bufRecv,'\0',sizeof(bufRecv)); struct sockaddr_in sourceAddr; int sourceAddrLen = sizeof(sourceAddr); struct sockaddr_in oppositeSideAddr; int len; len = sendto(sockS,bufSend,sizeof(bufSend),0,(struct sockaddr*)&serverAddr,sizeof(serverAddr)); if (len == SOCKET_ERROR) { printf("sendto() failed:%d\n", WSAGetLastError()); return -1; } len = recvfrom(sockS, bufRecv, sizeof(bufRecv), 0,(struct sockaddr*)&sourceAddr,&sourceAddrLen); if (len == SOCKET_ERROR) { int TRYY=WSAGetLastError(); if(TRYY!= WSAECONNRESET) { printf("recvfrom() failed:%d\n", WSAGetLastError()); return -1; } } closesocket(sockS); if (bind(sockC,(LPSOCKADDR)&thisAddr,sizeof(thisAddr)) == SOCKET_ERROR) { printf("bind() failed:%d\n",WSAGetLastError()); return -1; } char ip[20]; char port[10]; int i; for (i=0;i<strlen(bufRecv);i++) if (bufRecv[i] != '^') ip[i] = bufRecv[i]; else break; ip[i] = '\0'; int j; for (j=i+1;j<strlen(bufRecv);j++) port[j - i - 1] = bufRecv[j]; port[j - i - 1] = '\0'; oppositeSideAddr.sin_family = AF_INET; oppositeSideAddr.sin_port = htons(atoi(port)); oppositeSideAddr.sin_addr.s_addr = inet_addr(ip); unsigned long ul = 1; ioctlsocket(sockC, FIONBIO, (unsigned long*)&ul); while (1) { Sleep(1000); len = sendto(sockC,bufSend,sizeof(bufSend),0,(struct sockaddr*)&oppositeSideAddr,sizeof(oppositeSideAddr)); if (len == SOCKET_ERROR) { printf("while sending package to C1 , sendto() failed:%d\n", WSAGetLastError()); return -1; }else { printf("successfully send package to C1\n"); } len = recvfrom(sockC, bufRecv, sizeof(bufRecv), 0,(struct sockaddr*)&sourceAddr,&sourceAddrLen); if (len == WSAEWOULDBLOCK) { continue; }else { printf("C1 IP:[%s],PORT:[%d]\n",inet_ntoa(sourceAddr.sin_addr) ,ntohs(sourceAddr.sin_port)); printf("C1 says:%s\n",bufRecv); break; } } char buff[1024]={}; while(1) { memset(buff, 0, 1024); scanf("%s",buff); sendto(sockC,buff,sizeof(buff),0,(struct sockaddr*)&oppositeSideAddr,sizeof(oppositeSideAddr)); recvfrom(sockC, bufRecv, sizeof(bufRecv), 0,(struct sockaddr*)&sourceAddr,&sourceAddrLen); printf("%s\n\n",bufRecv); } closesocket(sockC); } ```
IDLE import torch报错:SyntaxError: invalid syntax
>>> import torch Traceback (most recent call last): File "<pyshell#0>", line 1, in <module> import torch File "C:\Python\Python37\torch.py", line 2 t.abs? ^ SyntaxError: invalid syntax >>> ``` ```
ElasticSearch 6.3.2版本整合springBoot 报错 java.io.StreamCorruptedException: invalid internal transport message format, got (48,54,54,50)
在服务器上搭建了一个单机的ElasticSearch 版本6.3.2,因为本地访问服务器的9200端口访问不到(network.host:0.0.0.0启动报错),所以用nginx做了代理9002端口对应9200,9001端口对应9300,直接访问访问成功如下:![图片说明](https://img-ask.csdn.net/upload/202002/08/1581096007_348190.jpg)![图片说明](https://img-ask.csdn.net/upload/202002/08/1581096016_248026.png) 然后我用写了个Demo但是一直报错 java.io.StreamCorruptedException: invalid internal transport message format, got (48,54,54,50); 配置文件如下: ![图片说明](https://img-ask.csdn.net/upload/202002/08/1581096398_340973.jpg) ![图片说明](https://img-ask.csdn.net/upload/202002/08/1581096609_898535.jpg) 错误日志和pom.xml文件如下: ``` exception caught on transport layer [NettyTcpChannel{localAddress=/192.168.1.103:61678, remoteAddress=/xxx.xxx.xxx.xxx:9001}], closing connection io.netty.handler.codec.DecoderException: java.io.StreamCorruptedException: invalid internal transport message format, got (48,54,54,50) at io.netty.handler.codec.ByteToMessageDecoder.callDecode(ByteToMessageDecoder.java:472) ~[netty-codec-4.1.31.Final.jar:4.1.31.Final] at io.netty.handler.codec.ByteToMessageDecoder.channelRead(ByteToMessageDecoder.java:278) ~[netty-codec-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.invokeChannelRead(AbstractChannelHandlerContext.java:362) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.invokeChannelRead(AbstractChannelHandlerContext.java:348) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.fireChannelRead(AbstractChannelHandlerContext.java:340) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.handler.logging.LoggingHandler.channelRead(LoggingHandler.java:241) [netty-handler-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.invokeChannelRead(AbstractChannelHandlerContext.java:362) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.invokeChannelRead(AbstractChannelHandlerContext.java:348) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.fireChannelRead(AbstractChannelHandlerContext.java:340) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.DefaultChannelPipeline$HeadContext.channelRead(DefaultChannelPipeline.java:1434) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.invokeChannelRead(AbstractChannelHandlerContext.java:362) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.AbstractChannelHandlerContext.invokeChannelRead(AbstractChannelHandlerContext.java:348) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.DefaultChannelPipeline.fireChannelRead(DefaultChannelPipeline.java:965) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.nio.AbstractNioByteChannel$NioByteUnsafe.read(AbstractNioByteChannel.java:163) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.nio.NioEventLoop.processSelectedKey(NioEventLoop.java:648) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.nio.NioEventLoop.processSelectedKeysPlain(NioEventLoop.java:548) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.nio.NioEventLoop.processSelectedKeys(NioEventLoop.java:502) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.channel.nio.NioEventLoop.run(NioEventLoop.java:462) [netty-transport-4.1.31.Final.jar:4.1.31.Final] at io.netty.util.concurrent.SingleThreadEventExecutor$5.run(SingleThreadEventExecutor.java:897) [netty-common-4.1.31.Final.jar:4.1.31.Final] at java.lang.Thread.run(Thread.java:748) [na:1.8.0_191] Caused by: java.io.StreamCorruptedException: invalid internal transport message format, got (48,54,54,50) at org.elasticsearch.transport.TcpTransport.validateMessageHeader(TcpTransport.java:1315) ~[elasticsearch-6.3.2.jar:6.4.3] at org.elasticsearch.transport.netty4.Netty4SizeHeaderFrameDecoder.decode(Netty4SizeHeaderFrameDecoder.java:36) ~[transport-netty4-client-6.4.3.jar:6.4.3] at io.netty.handler.codec.ByteToMessageDecoder.decodeRemovalReentryProtection(ByteToMessageDecoder.java:502) ~[netty-codec-4.1.31.Final.jar:4.1.31.Final] at io.netty.handler.codec.ByteToMessageDecoder.callDecode(ByteToMessageDecoder.java:441) ~[netty-codec-4.1.31.Final.jar:4.1.31.Final] ... 19 common frames omitted ``` pom.xml文件如下 ``` <?xml version="1.0" encoding="UTF-8"?> <project xmlns="http://maven.apache.org/POM/4.0.0" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://maven.apache.org/POM/4.0.0 https://maven.apache.org/xsd/maven-4.0.0.xsd"> <modelVersion>4.0.0</modelVersion> <parent> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-parent</artifactId> <version>2.1.1.RELEASE</version> <relativePath/> <!-- lookup parent from repository --> </parent> <groupId>com</groupId> <artifactId>elasticsearchtest</artifactId> <version>0.0.1-SNAPSHOT</version> <name>elasticsearchtest</name> <description>Demo project for Spring Boot</description> <properties> <java.version>1.8</java.version> </properties> <dependencies> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter</artifactId> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-test</artifactId> <scope>test</scope> <exclusions> <exclusion> <groupId>org.junit.vintage</groupId> <artifactId>junit-vintage-engine</artifactId> </exclusion> </exclusions> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-web</artifactId> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-thymeleaf</artifactId> </dependency> <dependency> <groupId>org.elasticsearch</groupId> <artifactId>elasticsearch</artifactId> <version>6.3.2</version> </dependency> <dependency> <groupId>org.elasticsearch.client</groupId> <artifactId>transport</artifactId> <version>6.3.2</version> </dependency> <dependency> <groupId>org.elasticsearch.client</groupId> <artifactId>elasticsearch-rest-high-level-client</artifactId> <version>6.3.2</version> </dependency> <dependency> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-starter-test</artifactId> <scope>test</scope> </dependency> <dependency> <groupId>org.projectlombok</groupId> <artifactId>lombok</artifactId> <version>1.16.10</version> </dependency> </dependencies> <build> <plugins> <plugin> <groupId>org.springframework.boot</groupId> <artifactId>spring-boot-maven-plugin</artifactId> </plugin> </plugins> </build> </project> ``` 请大神帮忙看看,谢谢啦!
nginx websocket配置 Handshake failed due to invalid Upgrade header: null
Handshake failed due to invalid Upgrade header: null 我已经按要求 nginx配置了.还是显示NULL, 应该怎么配置呢?奇怪. 我们服务有开发一个web-socket :rent-socket2 ![图片说明](https://img-ask.csdn.net/upload/201912/29/1577616262_624494.jpg) ![图片说明](https://img-ask.csdn.net/upload/201912/29/1577616150_528675.png) 页面去打开的时候: ![图片说明](https://img-ask.csdn.net/upload/201912/29/1577616201_859674.jpg) 去后台看错误: ![图片说明](https://img-ask.csdn.net/upload/201912/29/1577616234_392531.jpg) 我要怎么样才能连这个WEB-SOCKET呢
swift invalid number of items in section 0
collection view报错 代码如下: let selectIndex = self.photoData.seletedAssetArray.index(of: self.previewPhotoArray[currentIndex]) if selectIndex != nil{ self.rightButton.asyncSetImage(UIImage.zyCreateImageWithView(view: ZYPhotoNavigationViewController.zyGetSelectNuberView(index: "\(selectIndex! + 1)")), for: .selected) } if selectIndex != nil{ if self.rightButton.isSelected == false { let indexPath = IndexPath.init(row: selectIndex!, section: 0) self.photoData.seletedAssetArray.remove(at: self.photoData.seletedAssetArray.index(of: self.previewPhotoArray[currentIndex])!) thumbnailCollectionView.deleteItems(at: [indexPath]) }else{ let indexPath = IndexPath.init(row: selectIndex!, section: 0) thumbnailCollectionView.insertItems(at: [indexPath]) thumbnailCollectionView.reloadItems(at: [indexPath]) thumbnailCollectionViewCellToggeleSelect(indexPath: indexPath) } } 报错信息: 'Invalid update: invalid number of items in section 0. The number of items contained in an existing section after the update (1) must be equal to the number of items contained in that section before the update (1), plus or minus the number of items inserted or deleted from that section (1 inserted, 0 deleted) and plus or minus the number of items moved into or out of that section (0 moved in, 0 moved out) 希望指教,谢谢
为什么cookie中带有空格的字符串会添加失败
我在时间的dd日和HH添加一个空格 reponse.addCookie时就会报错, ``` java.lang.IllegalArgumentException: An invalid character [32] was present in the Cookie value at org.apache.tomcat.util.http.Rfc6265CookieProcessor.validateCookieValue(Rfc6265CookieProcessor.java:162) at org.apache.tomcat.util.http.Rfc6265CookieProcessor.generateHeader(Rfc6265CookieProcessor.java:111) at org.apache.catalina.connector.Response.generateCookieString(Response.java:989) at org.apache.catalina.connector.Response.addCookie(Response.java:937) at org.apache.catalina.connector.ResponseFacade.addCookie(ResponseFacade.java:386) at com.zero.demo.Demo04Cookie.doPost(Demo04Cookie.java:61) ``` 而如果没有空格就可以添加成功 ```java Date date=new Date(); SimpleDateFormat simpleDateFormat = new SimpleDateFormat("yyyy年MM月dd日 HH:mm:ss"); String loginTime = simpleDateFormat.format(date); System.out.println(loginTime); System.out.println("编码前"+loginTime); String encode = URLEncoder.encode(loginTime, "UTF-8"); System.out.println("编码后"+encode); String decode = URLDecoder.decode(encode, "UTF-8"); System.out.println("解码后"+decode); try{ Cookie loginTime1 = new Cookie("loginTime",decode); System.out.println( loginTime1.getName()); System.out.println(loginTime1.getValue()); response.addCookie(loginTime1); }catch (Exception e) { e.printStackTrace(); System.out.println("未能添加cookie"); } ```
Unexpected CUDA error: invalid argument
我在使用tensorflow-gpu时,运行如下的代码: ``` import tensorflow as tf a = tf.constant([1.0,2.0,3.0],shape = [3],name='a') print(a) ``` 能正确显示 ``tf.Tensor([1. 2. 3.], shape=(3,), dtype=float32)`` 但当我加入以下两行代码 ``` b = tf.constant([1.0,2.0,3.0], shape = [3],name='b') a+b ``` 会报如下的错误 ``` F tensorflow/stream_executor/cuda/cuda_driver.cc:209] Check failed: err == cudaSuccess || err == cudaErrorInvalidValue Unexpected CUDA error: invalid argument ``` 请大神帮忙解答 ================================================ 参照这篇博客,重新安装了一下,问题已经解决了 [https://blog.csdn.net/weixin_43411129/article/details/101390407](https://blog.csdn.net/weixin_43411129/article/details/101390407)
Triangle 三角的问题
Problem Description A triangle is one of the basic shapes of geometry: a polygon with three corners or vertices and three sides or edges which are line segments. Triangles can be classified according to the relative lengths of their sides: · In an equilateral triangle all sides have the same length. An equilateral triangle is also a regular polygon with all angles measuring 60°. · In an isosceles triangle, two sides are equal in length. An isosceles triangle also has two angles of the same measure; namely, the angles opposite to the two sides of the same length; this fact is the content of the Isosceles triangle theorem. · In a scalene triangle, all sides are unequal. The three angles are also all different in measure. Some (but not all) scalene triangles are also right triangles. Input The first line of input contains an integer (1 <= T <= 100), the number of test cases. T test data sets follow, each data set consists of 3 integers A, B and C, where (1 <= A,B,C <= 1,000,000) the triangle side lengths. Output For each test case, print “equilateral”, “isosceles” or “scalene” describing the triangle type. If the input doesn’t create a valid triangle output “invalid!”. Follow the output format below. Sample Input 2 3 3 4 6 4 2 Sample Output Case #1: isosceles Case #2: invalid!
C++嵌入Assembler(汇编)汇编内出错?
<!--blog--> <h4><b>我在C++中插入了汇编代码,但汇编代码老报错,不知什么原因。求各位大神帮助!</b></h4> 代码: ```C++ #include <iostream> using namespace std; int main(int argc,char *argv[]) { volatile int a[3] = {0,0,0}; for (int i = 0;i < 3;i ++) cout << a[i] << ends; cout << endl; asm ("inc [esp + 2]");//Assembler:inc [esp + 2] (++ a[1];) for (int i = 0;i < 3;i ++) cout << a[i] << ends; cout << endl; return 0; } ``` 报错信息: ```Message (File(C:\Users\ADMINI~1\AppData\Local\Temp\ccR9LmJd.s)) Assembler messages: (Line(42)) (File(C:\Users\ADMINI~1\AppData\Local\Temp\ccR9LmJd.s)) Error: invalid char '[' beginning operand 1 `[esp+2]' ``` 之后我想是不是这个汇编器不支持<c style="color:red;">Intel</c>,就把那一行汇编改成了<c style="color:red;">AT&T</c>格式的: ``` asm ("inc [esp + 2]");//Assembler:inc [esp + 2] (++ a[1];) ``` | | \/ ``` asm ("inc 2(%esp)");//Assembler:inc 2(%esp) (++ a[1];) ``` 但还是报错了: ``` (File(C:\Users\ADMINI~1\AppData\Local\Temp\ccEzueAV.s)) Assembler messages: (Line(42)) (File(C:\Users\ADMINI~1\AppData\Local\Temp\ccEzueAV.s)) Error: no instruction mnemonic suffix given and no register operands; can't size instruction ``` </h6><b><c style="color:green">本人对<cc style="color:red;">AT&T</cc>不熟悉,麻烦大家看看哪错了,致以感谢!</c></b></h6>
Network 是怎么编写的
Problem Description The ALPC company is now working on his own network system, which is connecting all N ALPC department. To economize on spending, the backbone network has only one router for each department, and N-1 optical fiber in total to connect all routers. The usual way to measure connecting speed is lag, or network latency, referring the time taken for a sent packet of data to be received at the other end. Now the network is on trial, and new photonic crystal fibers designed by ALPC42 is trying out, the lag on fibers can be ignored. That means, lag happened when message transport through the router. ALPC42 is trying to change routers to make the network faster, now he want to know that, which router, in any exactly time, between any pair of nodes, the K-th high latency is. He needs your help. Input There are only one test case in input file. Your program is able to get the information of N routers and N-1 fiber connections from input, and Q questions for two condition: 1. For some reason, the latency of one router changed. 2. Querying the K-th longest lag router between two routers. For each data case, two integers N and Q for first line. 0<=N<=80000, 0<=Q<=30000. Then n integers in second line refer to the latency of each router in the very beginning. Then N-1 lines followed, contains two integers x and y for each, telling there is a fiber connect router x and router y. Then q lines followed to describe questions, three numbers k, a, b for each line. If k=0, Telling the latency of router a, Ta changed to b; if k>0, asking the latency of the k-th longest lag router between a and b (include router a and b). 0<=b<100000000. A blank line follows after each case. Output For each question k>0, print a line to answer the latency time. Once there are less than k routers in the way, print "invalid request!" instead. Sample Input 5 5 5 1 2 3 4 3 1 2 1 4 3 5 3 2 4 5 0 1 2 2 2 3 2 1 4 3 3 5 Sample Output 3 2 2 invalid request!
已安装的Nginx追加自带模块始终报错
* Centos7系统下已安装的nginx需要追加http_limit_req_module模块。按照网上的方法,下载了对应版本的源码, 1. 在配置参数后追加--with-http_limit_req_module,出现以下报错 ``` ./configure: error: invalid option "--with-http_limit_req_module" ``` 2. 使用--add-module=http_limit_req_module报找不到config文件 ``` adding module in http_limit_req_module ./configure: error: no http_limit_req_module/config was found ``` 3. 将模块名改成ngx_http_limit_req_module报的也是类似的错误 # 求大神看看应该怎么添加 下面是完整配置参数,nginx版本是1.17.8,--with-http_limit_req_module前的参数是用nginx -V查出来的 ``` ./configure --prefix=/etc/nginx --sbin-path=/usr/sbin/nginx --modules-path=/usr/lib64/nginx/modules --conf-path=/etc/nginx/nginx.conf --error-log-path=/var/log/nginx/error.log --http-log-path=/var/log/nginx/access.log --pid-path=/var/run/nginx.pid --lock-path=/var/run/nginx.lock --http-client-body-temp-path=/var/cache/nginx/client_temp --http-proxy-temp-path=/var/cache/nginx/proxy_temp --http-fastcgi-temp-path=/var/cache/nginx/fastcgi_temp --http-uwsgi-temp-path=/var/cache/nginx/uwsgi_temp --http-scgi-temp-path=/var/cache/nginx/scgi_temp --user=nginx --group=nginx --with-compat --with-file-aio --with-threads --with-http_addition_module --with-http_auth_request_module --with-http_dav_module --with-http_flv_module --with-http_gunzip_module --with-http_gzip_static_module --with-http_mp4_module --with-http_random_index_module --with-http_realip_module --with-http_secure_link_module --with-http_slice_module --with-http_ssl_module --with-http_stub_status_module --with-http_sub_module --with-http_v2_module --with-mail --with-mail_ssl_module --with-stream --with-stream_realip_module --with-stream_ssl_module --with-stream_ssl_preread_module --with-cc-opt='-O2 -g -pipe -Wall -Wp,-D_FORTIFY_SOURCE=2 -fexceptions -fstack-protector-strong --param=ssp-buffer-size=4 -grecord-gcc-switches -m64 -mtune=generic -fPIC' --with-ld-opt='-Wl,-z,relro -Wl,-z,now -pie' --with-http_limit_req_module ```
Judges's Final Problem 怎么编写呢
Problem Description For any ACM/ICPC participant, it is always discouraging to discover that the real problems they face in the programming world are seldom as interesting as the tasks they face in the algorithm-related contests. While the efficiency of your algorithm is certainly vital to the overall system, still it is seldom valued so much as other factors such as readability, and the ability of coding accurately often takes precedence in many projects. Still, we judges believe that a balance between these two can be achieved - a good programmer should be able to come up with code that is fast, reliable and readable in a reasonable amount of time - and that is why you, the contestant, is asked to solve the problem of building a ranking system for us. You're probably familiar with the following paragraph taken from Rules of ACM/ICPC: A problem is solved when it is accepted by the judges. Teams are ranked according to the most problems solved. Teams who solve the same number of problems are ranked by least total time. The total time is the sum of the time consumed for each problem solved. The time consumed for a solved problem is the time elapsed from the beginning of the contest to the submittal of the accepted run plus 20 penalty minutes for every rejected run for that problem regardless of submittal time. There is no time consumed for a problem that is not solved. During a contest, the following kinds of requests might be submitted by participants: Submission (Format: S [minute]:[Team No]:[Problem ID]:[Result]) If a team submits a problem which they have already solved, then this submission should be considered invalid and ignored by your system; otherwise, the submission is valid and saved for further processing - if the result is 1 (Correct), then this submission should be considered accepted by the system. Query a team's place on the rank list. (Format: R [Team No]) Sort the teams according to the contest rules mentioned above. A tie may occur between two teams who solved equal number of problems and same penalty. Two teams who solved same number of problems and have same penalty should always have same rank; however, when querying the Kth team on the rank list, even if many teams share the same rank, only one team should be printed (please refer to the rule given below). Query the Kth team on the rank list. (Format: T [k]) If a tie occurs, return the team with the minimum last accepted submission time. Note that the submission time of two teams might be different (i.e., in seconds) even though it may appear otherwise in the input. Initially every team has no submissions, and the contest will last no more than 5 hours. Input There are multiple test cases in the input. Each test case starts with two integers, N and M, (1 <= N <= 10000, 1 <= M <= 10), the number of teams participating in the competition, and the total number of problems in the contest. Teams are numbered from 0 to N - 1. Each of the following lines is either in one of the three formats listed above, or a single line "Contest Ends" followed by an empty line indicating the end of current test case. It is guaranteed that there are no more than 100000 requests in each test case. Input ends with End-of-File. Output For every submission request, if the submitted program is accepted, you need to output a line whose format is [Team No] [Problem ID]; for every query about team's place on the rank list, output the result in one single line; for every query about the team with the Kth highest score, if a team is found, output a line with one integer, the number of the team, or -1 if you are unable to find such a team. Please print a blank line after each test case. Sample Input 5 8 T 1 T 2 S 5:0:A:0 S 8:0:A:1 S 9:1:B:1 S 15:0:A:1 T 1 T 2 T 3 R 0 R 1 R 2 R 3 Contest Ends Sample Output 0 -1 [0][A] [1][B] 1 0 2 1 0 2 2
神奇的语句报错:return HttpResponse(post_lists)
django的部分代码如下 ``` from django.shortcuts import render from django.http import HttpResponse from .models import Post # Create your views here. def homepage(request): posts = Post.objects.all() post_lists = list() for count,post in enumerate(posts): post_lists.append("No.{}:".format(str(count)+str(post)+"<br>") return HttpResponse(post_lists) ``` 报错信息如下: python -m py_compile "views.py" (在目录 C:\venv\mblog\mainsite 中) File "views.py", line 10 :return HttpResponse(post_lists) SyntaxError: invalid syntax 编译失败。 return语句还能报错,小白一脸懵逼。
Judges's Final Problem 怎么写啊
Problem Description For any ACM/ICPC participant, it is always discouraging to discover that the real problems they face in the programming world are seldom as interesting as the tasks they face in the algorithm-related contests. While the efficiency of your algorithm is certainly vital to the overall system, still it is seldom valued so much as other factors such as readability, and the ability of coding accurately often takes precedence in many projects. Still, we judges believe that a balance between these two can be achieved - a good programmer should be able to come up with code that is fast, reliable and readable in a reasonable amount of time - and that is why you, the contestant, is asked to solve the problem of building a ranking system for us. You're probably familiar with the following paragraph taken from Rules of ACM/ICPC: A problem is solved when it is accepted by the judges. Teams are ranked according to the most problems solved. Teams who solve the same number of problems are ranked by least total time. The total time is the sum of the time consumed for each problem solved. The time consumed for a solved problem is the time elapsed from the beginning of the contest to the submittal of the accepted run plus 20 penalty minutes for every rejected run for that problem regardless of submittal time. There is no time consumed for a problem that is not solved. During a contest, the following kinds of requests might be submitted by participants: Submission (Format: S [minute]:[Team No]:[Problem ID]:[Result]) If a team submits a problem which they have already solved, then this submission should be considered invalid and ignored by your system; otherwise, the submission is valid and saved for further processing - if the result is 1 (Correct), then this submission should be considered accepted by the system. Query a team's place on the rank list. (Format: R [Team No]) Sort the teams according to the contest rules mentioned above. A tie may occur between two teams who solved equal number of problems and same penalty. Two teams who solved same number of problems and have same penalty should always have same rank; however, when querying the Kth team on the rank list, even if many teams share the same rank, only one team should be printed (please refer to the rule given below). Query the Kth team on the rank list. (Format: T [k]) If a tie occurs, return the team with the minimum last accepted submission time. Note that the submission time of two teams might be different (i.e., in seconds) even though it may appear otherwise in the input. Initially every team has no submissions, and the contest will last no more than 5 hours. Input There are multiple test cases in the input. Each test case starts with two integers, N and M, (1 <= N <= 10000, 1 <= M <= 10), the number of teams participating in the competition, and the total number of problems in the contest. Teams are numbered from 0 to N - 1. Each of the following lines is either in one of the three formats listed above, or a single line "Contest Ends" followed by an empty line indicating the end of current test case. It is guaranteed that there are no more than 100000 requests in each test case. Input ends with End-of-File. Output For every submission request, if the submitted program is accepted, you need to output a line whose format is [Team No] [Problem ID]; for every query about team's place on the rank list, output the result in one single line; for every query about the team with the Kth highest score, if a team is found, output a line with one integer, the number of the team, or -1 if you are unable to find such a team. Please print a blank line after each test case. Sample Input 5 8 T 1 T 2 S 5:0:A:0 S 8:0:A:1 S 9:1:B:1 S 15:0:A:1 T 1 T 2 T 3 R 0 R 1 R 2 R 3 Contest Ends Sample Output 0 -1 [0][A] [1][B] 1 0 2 1 0 2 2
请问关于JPanel 中paint 方法重写的问题,void is an invalid type for the variable paint
import java.awt.Graphics; import javax.swing.JPanel; public class ipanel extends JPanel { public ipanel () { public void paint(Graphics g) { g.drawLine(0, 0, 10, 10); } } } public class Test { public static void main(String[] args) { new ipanel(); } } 编译会出现void is an invalid type for the variable paint的错误,即使在paint加入父类方法也没用。 可是在ipanel的构造器中应该是可以出现函数的吧? 请教大家,谢谢!
终于明白阿里百度这样的大公司,为什么面试经常拿ThreadLocal考验求职者了
点击上面↑「爱开发」关注我们每晚10点,捕获技术思考和创业资源洞察什么是ThreadLocalThreadLocal是一个本地线程副本变量工具类,各个线程都拥有一份线程私...
程序员必须掌握的核心算法有哪些?
由于我之前一直强调数据结构以及算法学习的重要性,所以就有一些读者经常问我,数据结构与算法应该要学习到哪个程度呢?,说实话,这个问题我不知道要怎么回答你,主要取决于你想学习到哪些程度,不过针对这个问题,我稍微总结一下我学过的算法知识点,以及我觉得值得学习的算法。这些算法与数据结构的学习大多数是零散的,并没有一本把他们全部覆盖的书籍。下面是我觉得值得学习的一些算法以及数据结构,当然,我也会整理一些看过...
Linux(服务器编程):15---两种高效的事件处理模式(reactor模式、proactor模式)
前言 同步I/O模型通常用于实现Reactor模式 异步I/O模型则用于实现Proactor模式 最后我们会使用同步I/O方式模拟出Proactor模式 一、Reactor模式 Reactor模式特点 它要求主线程(I/O处理单元)只负责监听文件描述符上是否有事件发生,有的话就立即将时间通知工作线程(逻辑单元)。除此之外,主线程不做任何其他实质性的工作 读写数据,接受新的连接,以及处...
阿里面试官问我:如何设计秒杀系统?我的回答让他比起大拇指
你知道的越多,你不知道的越多 点赞再看,养成习惯 GitHub上已经开源 https://github.com/JavaFamily 有一线大厂面试点脑图和个人联系方式,欢迎Star和指教 前言 Redis在互联网技术存储方面使用如此广泛,几乎所有的后端技术面试官都要在Redis的使用和原理方面对小伙伴们进行360°的刁难。 作为一个在互联网公司面一次拿一次Offer的面霸,打败了...
五年程序员记流水账式的自白。
不知觉已中码龄已突破五年,一路走来从起初铁憨憨到现在的十九线程序员,一路成长,虽然不能成为高工,但是也能挡下一面,从15年很火的android开始入坑,走过java、.Net、QT,目前仍处于android和.net交替开发中。 毕业到现在一共就职过两家公司,目前是第二家,公司算是半个创业公司,所以基本上都会身兼多职。比如不光要写代码,还要写软著、软著评测、线上线下客户对接需求收集...
C语言魔塔游戏
很早就很想写这个,今天终于写完了。 游戏截图: 编译环境: VS2017 游戏需要一些图片,如果有想要的或者对游戏有什么看法的可以加我的QQ 2985486630 讨论,如果暂时没有回应,可以在博客下方留言,到时候我会看到。 下面我来介绍一下游戏的主要功能和实现方式 首先是玩家的定义,使用结构体,这个名字是可以自己改变的 struct gamerole { char n...
一文详尽系列之模型评估指标
点击上方“Datawhale”,选择“星标”公众号第一时间获取价值内容在机器学习领域通常会根据实际的业务场景拟定相应的不同的业务指标,针对不同机器学习问题如回归、分类、排...
究竟你适不适合买Mac?
我清晰的记得,刚买的macbook pro回到家,开机后第一件事情,就是上了淘宝网,花了500元钱,找了一个上门维修电脑的师傅,上门给我装了一个windows系统。。。。。。 表砍我。。。 当时买mac的初衷,只是想要个固态硬盘的笔记本,用来运行一些复杂的扑克软件。而看了当时所有的SSD笔记本后,最终决定,还是买个好(xiong)看(da)的。 已经有好几个朋友问我mba怎么样了,所以今天尽量客观...
程序员一般通过什么途径接私活?
二哥,你好,我想知道一般程序猿都如何接私活,我也想接,能告诉我一些方法吗? 上面是一个读者“烦不烦”问我的一个问题。其实不止是“烦不烦”,还有很多读者问过我类似这样的问题。 我接的私活不算多,挣到的钱也没有多少,加起来不到 20W。说实话,这个数目说出来我是有点心虚的,毕竟太少了,大家轻喷。但我想,恰好配得上“一般程序员”这个称号啊。毕竟苍蝇再小也是肉,我也算是有经验的人了。 唾弃接私活、做外...
压测学习总结(1)——高并发性能指标:QPS、TPS、RT、吞吐量详解
一、QPS,每秒查询 QPS:Queries Per Second意思是“每秒查询率”,是一台服务器每秒能够相应的查询次数,是对一个特定的查询服务器在规定时间内所处理流量多少的衡量标准。互联网中,作为域名系统服务器的机器的性能经常用每秒查询率来衡量。 二、TPS,每秒事务 TPS:是TransactionsPerSecond的缩写,也就是事务数/秒。它是软件测试结果的测量单位。一个事务是指一...
Python爬虫爬取淘宝,京东商品信息
小编是一个理科生,不善长说一些废话。简单介绍下原理然后直接上代码。 使用的工具(Python+pycharm2019.3+selenium+xpath+chromedriver)其中要使用pycharm也可以私聊我selenium是一个框架可以通过pip下载 pip installselenium -ihttps://pypi.tuna.tsinghua.edu.cn/simple/ ...
阿里程序员写了一个新手都写不出的低级bug,被骂惨了。
这种新手都不会范的错,居然被一个工作好几年的小伙子写出来,差点被当场开除了。
Java工作4年来应聘要16K最后没要,细节如下。。。
前奏: 今天2B哥和大家分享一位前几天面试的一位应聘者,工作4年26岁,统招本科。 以下就是他的简历和面试情况。 基本情况: 专业技能: 1、&nbsp;熟悉Sping了解SpringMVC、SpringBoot、Mybatis等框架、了解SpringCloud微服务 2、&nbsp;熟悉常用项目管理工具:SVN、GIT、MAVEN、Jenkins 3、&nbsp;熟悉Nginx、tomca...
2020年,冯唐49岁:我给20、30岁IT职场年轻人的建议
点击“技术领导力”关注∆每天早上8:30推送 作者|Mr.K 编辑| Emma 来源|技术领导力(ID:jishulingdaoli) 前天的推文《冯唐:职场人35岁以后,方法论比经验重要》,收到了不少读者的反馈,觉得挺受启发。其实,冯唐写了不少关于职场方面的文章,都挺不错的。可惜大家只记住了“春风十里不如你”、“如何避免成为油腻腻的中年人”等不那么正经的文章。 本文整理了冯...
程序员该看的几部电影
1、骇客帝国(1999) 概念:在线/离线,递归,循环,矩阵等 剧情简介: 不久的将来,网络黑客尼奥对这个看似正常的现实世界产生了怀疑。 他结识了黑客崔妮蒂,并见到了黑客组织的首领墨菲斯。 墨菲斯告诉他,现实世界其实是由一个名叫“母体”的计算机人工智能系统控制,人们就像他们饲养的动物,没有自由和思想,而尼奥就是能够拯救人类的救世主。 可是,救赎之路从来都不会一帆风顺,到底哪里才是真实的世界?如何...
Python绘图,圣诞树,花,爱心 | Turtle篇
每周每日,分享Python实战代码,入门资料,进阶资料,基础语法,爬虫,数据分析,web网站,机器学习,深度学习等等。 公众号回复【进群】沟通交流吧,QQ扫码进群学习吧 微信群 QQ群 1.画圣诞树 import turtle screen = turtle.Screen() screen.setup(800,600) circle = turtle.Turtle()...
作为一个程序员,CPU的这些硬核知识你必须会!
CPU对每个程序员来说,是个既熟悉又陌生的东西? 如果你只知道CPU是中央处理器的话,那可能对你并没有什么用,那么作为程序员的我们,必须要搞懂的就是CPU这家伙是如何运行的,尤其要搞懂它里面的寄存器是怎么一回事,因为这将让你从底层明白程序的运行机制。 随我一起,来好好认识下CPU这货吧 把CPU掰开来看 对于CPU来说,我们首先就要搞明白它是怎么回事,也就是它的内部构造,当然,CPU那么牛的一个东...
还记得那个提速8倍的IDEA插件吗?VS Code版本也发布啦!!
去年,阿里云发布了本地 IDE 插件 Cloud Toolkit,仅 IntelliJ IDEA 一个平台,就有 15 万以上的开发者进行了下载,体验了一键部署带来的开发便利。时隔一年的今天,阿里云正式发布了 Visual Studio Code 版本,全面覆盖前端开发者,帮助前端实现一键打包部署,让开发提速 8 倍。 VSCode 版本的插件,目前能做到什么? 安装插件之后,开发者可以立即体验...
破14亿,Python分析我国存在哪些人口危机!
一、背景 二、爬取数据 三、数据分析 1、总人口 2、男女人口比例 3、人口城镇化 4、人口增长率 5、人口老化(抚养比) 6、各省人口 7、世界人口 四、遇到的问题 遇到的问题 1、数据分页,需要获取从1949-2018年数据,观察到有近20年参数:LAST20,由此推测获取近70年的参数可设置为:LAST70 2、2019年数据没有放上去,可以手动添加上去 3、将数据进行 行列转换 4、列名...
2019年除夕夜的有感而发
天气:小雨(加小雪) 温度:3摄氏度 空气:严重污染(399) 风向:北风 风力:微风 现在是除夕夜晚上十点钟,再有两个小时就要新的一年了; 首先要说的是我没患病,至少现在是没有患病;但是心情确像患了病一样沉重; 现在这个时刻应该大部分家庭都在看春晚吧,或许一家人团团圆圆的坐在一起,或许因为某些特殊原因而不能团圆;但不管是身在何处,身处什么境地,我都想对每一个人说一句:新年快乐! 不知道csdn这...
听说想当黑客的都玩过这个Monyer游戏(1~14攻略)
第零关 进入传送门开始第0关(游戏链接) 请点击链接进入第1关: 连接在左边→ ←连接在右边 看不到啊。。。。(只能看到一堆大佬做完的留名,也能看到菜鸡的我,在后面~~) 直接fn+f12吧 &lt;span&gt;连接在左边→&lt;/span&gt; &lt;a href="first.php"&gt;&lt;/a&gt; &lt;span&gt;←连接在右边&lt;/span&gt; o...
在家远程办公效率低?那你一定要收好这个「在家办公」神器!
相信大家都已经收到国务院延长春节假期的消息,接下来,在家远程办公可能将会持续一段时间。 但是问题来了。远程办公不是人在电脑前就当坐班了,相反,对于沟通效率,文件协作,以及信息安全都有着极高的要求。有着非常多的挑战,比如: 1在异地互相不见面的会议上,如何提高沟通效率? 2文件之间的来往反馈如何做到及时性?如何保证信息安全? 3如何规划安排每天工作,以及如何进行成果验收? ...... ...
作为一个程序员,内存和磁盘的这些事情,你不得不知道啊!!!
截止目前,我已经分享了如下几篇文章: 一个程序在计算机中是如何运行的?超级干货!!! 作为一个程序员,CPU的这些硬核知识你必须会! 作为一个程序员,内存的这些硬核知识你必须懂! 这些知识可以说是我们之前都不太重视的基础知识,可能大家在上大学的时候都学习过了,但是嘞,当时由于老师讲解的没那么有趣,又加上这些知识本身就比较枯燥,所以嘞,大家当初几乎等于没学。 再说啦,学习这些,也看不出来有什么用啊!...
2020年的1月,我辞掉了我的第一份工作
其实,这篇文章,我应该早点写的,毕竟现在已经2月份了。不过一些其它原因,或者是我的惰性、还有一些迷茫的念头,让自己迟迟没有试着写一点东西,记录下,或者说是总结下自己前3年的工作上的经历、学习的过程。 我自己知道的,在写自己的博客方面,我的文笔很一般,非技术类的文章不想去写;另外我又是一个还比较热衷于技术的人,而平常复杂一点的东西,如果想写文章写的清楚点,是需要足够...
别低估自己的直觉,也别高估自己的智商
所有群全部吵翻天,朋友圈全部沦陷,公众号疯狂转发。这两周没怎么发原创,只发新闻,可能有人注意到了。我不是懒,是文章写了却没发,因为大家的关注力始终在这次的疫情上面,发了也没人看。当然,我...
这个世界上人真的分三六九等,你信吗?
偶然间,在知乎上看到一个问题 一时间,勾起了我深深的回忆。 以前在厂里打过两次工,做过家教,干过辅导班,做过中介。零下几度的晚上,贴过广告,满脸、满手地长冻疮。 再回首那段岁月,虽然苦,但让我学会了坚持和忍耐。让我明白了,在这个世界上,无论环境多么的恶劣,只要心存希望,星星之火,亦可燎原。 下文是原回答,希望能对你能有所启发。 如果我说,这个世界上人真的分三六九等,...
节后首个工作日,企业们集体开晨会让钉钉挂了
By 超神经场景描述:昨天 2 月 3 日,是大部分城市号召远程工作的第一天,全国有接近 2 亿人在家开始远程办公,钉钉上也有超过 1000 万家企业活跃起来。关键词:十一出行 人脸...
Java基础知识点梳理
虽然已经在实际工作中经常与java打交道,但是一直没系统地对java这门语言进行梳理和总结,掌握的知识也比较零散。恰好利用这段时间重新认识下java,并对一些常见的语法和知识点做个总结与回顾,一方面为了加深印象,方便后面查阅,一方面为了掌握好Android打下基础。
2020年全新Java学习路线图,含配套视频,学完即为中级Java程序员!!
新的一年来临,突如其来的疫情打破了平静的生活! 在家的你是否很无聊,如果无聊就来学习吧! 世上只有一种投资只赚不赔,那就是学习!!! 传智播客于2020年升级了Java学习线路图,硬核升级,免费放送! 学完你就是中级程序员,能更快一步找到工作! 一、Java基础 JavaSE基础是Java中级程序员的起点,是帮助你从小白到懂得编程的必经之路。 在Java基础板块中有6个子模块的学...
B 站上有哪些很好的学习资源?
哇说起B站,在小九眼里就是宝藏般的存在,放年假宅在家时一天刷6、7个小时不在话下,更别提今年的跨年晚会,我简直是跪着看完的!! 最早大家聚在在B站是为了追番,再后来我在上面刷欧美新歌和漂亮小姐姐的舞蹈视频,最近两年我和周围的朋友们已经把B站当作学习教室了,而且学习成本还免费,真是个励志的好平台ヽ(.◕ฺˇд ˇ◕ฺ;)ノ 下面我们就来盘点一下B站上优质的学习资源: 综合类 Oeasy: 综合...
如何优雅地打印一个Java对象?
你好呀,我是沉默王二,一个和黄家驹一样身高,和刘德华一样颜值的程序员。虽然已经写了十多年的 Java 代码,但仍然觉得自己是个菜鸟(请允许我惭愧一下)。 在一个月黑风高的夜晚,我思前想后,觉得再也不能这么蹉跎下去了。于是痛下决心,准备通过输出的方式倒逼输入,以此来修炼自己的内功,从而进阶成为一名真正意义上的大神。与此同时,希望这些文章能够帮助到更多的读者,让大家在学习的路上不再寂寞、空虚和冷。 ...
相关热词 c#导入fbx c#中屏蔽键盘某个键 c#正态概率密度 c#和数据库登陆界面设计 c# 高斯消去法 c# codedom c#读取cad文件文本 c# 控制全局鼠标移动 c# temp 目录 bytes初始化 c#
立即提问