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- r语言
- spring
- maven
- java
- tomcat
- intellij-idea
个人按照实例编写Spring Boot Upload File页面，很简单的操作
1、上传页面
<body>
<h1>File Upload</h1>
<form action="/upload" method="post" enctype="multipart/form-data">
<input type="file" name="file">
<br><br>
<input type="submit" value="Submit">
</form>
</body>
2、处理方法
@RequestMapping("/upload")
public String singleFileUpload(@RequestParam("file")MultipartFile file, RedirectAttributes redirectAttributes) {
if (file.isEmpty()) {
redirectAttributes.addFlashAttribute("message", "Please select a file to upload");
return "redirect:/uploadStatus";
}
try {
byte[] bytes = file.getBytes();
Path path = Paths.get(UPLOADED_FOLDER + file.getOriginalFilename());
Files.write(path, bytes);
redirectAttributes.addFlashAttribute("message", "You successfully uploaded " + file.getOriginalFilename());
} catch (IOException e) {
e.printStackTrace();
}
return "redirect:/uploadStatus";
}
@RequestMapping("/uploadStatus")
public String uploadStatus() {
return "uploadStatus";
}
3、GlobalExceptionHandler
@ControllerAdvice
public class GlobalExceptionHandler {
@ExceptionHandler(MultipartException.class)
public String handleError1(MultipartException e, RedirectAttributes redirectAttributes) {
redirectAttributes.addFlashAttribute("message", e.getCause().getMessage());
return "redirect:/uploadStatus";
}
}
实例运行中，正常上传文件没有问题，在超过预定设置大小后，控制台报错
2021-01-18 10:24:27.730 WARN 2248 --- [nio-8080-exec-2] .m.m.a.ExceptionHandlerExceptionResolver : Resolved [org.springframework.web.multipart.MaxUploadSizeExceededException: Maximum upload size exceeded; nested exception is java.lang.IllegalStateException: org.apache.tomcat.util.http.fileupload.impl.SizeLimitExceededException: the request was rejected because its size (77299500) exceeds the configured maximum (52428800)]
2021-01-18 10:24:27.771 WARN 2248 --- [nio-8080-exec-3] .m.m.a.ExceptionHandlerExceptionResolver : Resolved [org.springframework.web.multipart.MaxUploadSizeExceededException: Maximum upload size exceeded; nested exception is java.lang.IllegalStateException: org.apache.tomcat.util.http.fileupload.impl.SizeLimitExceededException: the request was rejected because its size (77299500) exceeds the configured maximum (52428800)]
2021-01-18 10:24:27.797 WARN 2248 --- [nio-8080-exec-4] .m.m.a.ExceptionHandlerExceptionResolver : Resolved [org.springframework.web.multipart.MaxUploadSizeExceededException: Maximum upload size exceeded; nested exception is java.lang.IllegalStateException: org.apache.tomcat.util.http.fileupload.impl.SizeLimitExceededException: the request was rejected because its size (77299500) exceeds the configured maximum (52428800)]
不知道为什么是3条？
全局异常GlobalExceptionHandler中的方法也捕捉到异常，但是不会执行重定向return "redirect:/uploadStatus";
- html5
- vue.js
- css
- node.js
- javascript
- 深度学习
- java
- eclipse
/*首先D盘根目录下创建students.txt文件（没有D盘或其它特殊情况的同学可以选择在其它盘符下创建）。 接收控制台输入的学生信息实例化学生对象Student，将对象装入ArrayList<Student>集合中。使用对象流能够将集合写入students.txt文件，也要能从students.txt文件中反序列化得到集合，进而获得学生的信息。 完成复制students.txt功能。复制后的文件与students.txt放在相同目录下，名为students拷贝.txt。（二）首先D盘根目录下创建students.txt文件（没有D盘或其它特殊情况的同学可以选择在其它盘符下创建）。 （三）接收控制台输入的学生信息实例化学生对象Student，将对象装入ArrayList<Student>集合中。使用对象流能够将集合写入students.txt文件，也要能从students.txt文件中反序列化得到集合，进而获得学生的信息。（四） 完成复制students.txt功能。复制后的文件与students.txt放在相同目录下，名为students拷贝.txt。我写的代码 为什么读取时只读取了第一次输入的数据啊 以后输入的读取不出来 拜托*/
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
public class Test1 {
public static void main(String[] args) throws FileNotFoundException, IOException, ClassNotFoundException {
Scanner sc=new Scanner(System.in);
FileOutputStream file=new FileOutputStream("students.txt");
while(true) {
System.out.println("请选择序号：1.向文件写入学生信息 2.读文件显示学生信息 3.退出");
int n=sc.nextInt();
List<Student> list=new ArrayList<>();
switch (n) {
case 1:
ObjectOutputStream oos=new ObjectOutputStream(file);
System.out.println("请输入学生的姓名，年龄和性别逗号分隔（例如：张三，12，男）");
String all=sc.next();
String arr[]=all.split(",");
String name=arr[0];
int age=Integer.parseInt(arr[1]);
String gender=arr[2];
list.add(new Student(name,age,gender));
oos.writeObject(list);
oos.flush();
if(oos!=null) {
oos.close();
}
break;
case 2:
ObjectInputStream ois=new ObjectInputStream(new FileInputStream("students.txt"));
List<Student> list2=(List<Student>)ois.readObject();
for (Student student : list2) {
System.out.println(student);
}
break;
case 3:
ObjectInputStream newois=new ObjectInputStream(new FileInputStream("students.txt"));
ObjectOutputStream newoos=new ObjectOutputStream(new FileOutputStream("students拷贝.txt"));
newoos.writeObject(newois);
System.exit(0);
break;
default:
System.out.println("输入错误");
}
}
}
}- jquery
- c++
- java
如图所示，nacos在Linux机器上启动成功了，最上面是启动日志，查看端口也是监听状态，并且是被nacos的进程占用的，下面用curl访问路径，没有展示页面- c语言
- java
- mysql
- javascript
- css3
- css
- c++
这几天学习VC++ 用的VS2019 ，利用剪贴板向QQ聊天窗口 粘贴内容，但是第2次调用时就出错，代码如下：
void CSendQQmsg::CopyToClipboard(CString str)
{
if (str.GetLength() >= 40)
{
return;
}
TCHAR a[40];
memcpy(a, str.GetBuffer(0), (str.GetLength() + 1) * sizeof(TCHAR));
HGLOBAL hMem;
LPTSTR pStr;
// 分配内存空间
hMem = GlobalAlloc(GHND | GMEM_SHARE, (str.GetLength() + 1) *sizeof(TCHAR));
pStr = (LPTSTR)GlobalLock(hMem);
wcsncpy_s(pStr, 40, a, 40);
GlobalUnlock(hMem);
if (OpenClipboard(NULL)) //二次调用在这次出错
{
TRACE("剪贴板打开成功\n");
EmptyClipboard(); // 清空
SetClipboardData(CF_UNICODETEXT, hMem);// 设置剪贴板文本
CloseClipboard(); //关闭剪贴板
}
else
{
TRACE("剪贴板打开失败 %x\n",GetLastError());
GlobalFree(hMem);
}
}
C不多，求师傅们指教，感激不尽！
- python
import matplotlib
import numpy as np
import matplotlib as mpl
import matplotlib.pyplot as plt
import matplotlib.animation as animation
#坐标轴显示成中文
mpl.rcParams['font.sans-serif'] = ['SimHei'] # 指定默认字体
mpl.rcParams['axes.unicode_minus'] = False # 解决保存图像是负号'-'显示为方块的问题
#提取你想要的文本数据
filename = 'disp_6.out'
X, Y = [], []
with open(filename, 'r') as f:
lines = f.readlines()
for line in lines:
value = [float(s) for s in line.split()]
#选取你想要的文档列数，代码中0表示文本里的第一列
X.append(value[0])
Y.append(value[1])
这是我读取我想要文档数据的代码段落
FuncAnimation只能画已经固定好了的方程动态图
import numpy as np
import matplotlib.pyplot as plt
from matplotlib.animation import FuncAnimation #导入负责绘制动画的接口
fig, ax = plt.subplots()
x, y= [], []
line, = plt.plot([], [], '.-',color='red')
nums = 100 #需要的帧数
filename = 'disp_6.out'
with open(filename, 'r') as f:
lines = f.readlines()
for line in lines:
value = [float(s) for s in line.split()]
# 选取你想要的文档列数，代码中0表示文本里的第一列
x.append(value[0])
y.append(value[1])
# 用matplotlib绘制一个图形
def init():
ax.set_xlim(-5, 50)
ax.set_ylim(-10, 10)
return line
def update(i):
if len(x)>=nums: #通过控制帧数来避免不断的绘图
return line
line.set_data(x[i],y[i])
return line
ani = FuncAnimation(fig, update, frames=nums, #nums输入到frames后会使用range(nums)得到一系列step输入到update中去
init_func=init)
plt.show()
这样没办法运行....
- c语言
- css3
class Solution {
public:
static bool cmp(int a, int b) {
return a >= b;
}
int breakfastNumber(vector<int>& staple, vector<int>& drinks, int x) {
int ans = 0;
int stapleLen = staple.size();
cout << stapleLen << endl;
int drinksLen = drinks.size();
cout << drinksLen << endl;
sort(staple.begin(), staple.end(), cmp);
sort(drinks.begin(), drinks.end());
int j = 0;
for (int i = 0; i < stapleLen; i++) {
ans += j;
ans %= 1000000007;
for (; j < drinksLen; j++) {
if (staple[i] + drinks[j] <= x) {
ans++;
ans %= 1000000007;
} else {
break;
}
}
}
return ans;
}
};
提示信息：
=================================================================
==42==ERROR: AddressSanitizer: heap-buffer-overflow on address 0x7f22768b87fc at pc 0x00000034d6ad bp 0x7ffc746101b0 sp 0x7ffc746101a8
READ of size 4 at 0x7f22768b87fc thread T0
#6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
0x7f22768b87fc is located 4 bytes to the left of 524288-byte region [0x7f22768b8800,0x7f2276938800)
allocated by thread T0 here:
#6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6+0x270b2)
Shadow bytes around the buggy address:
0x0fe4ced0f0a0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0fe4ced0f0b0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0fe4ced0f0c0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0fe4ced0f0d0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
0x0fe4ced0f0e0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa
=>0x0fe4ced0f0f0: fa fa fa fa fa fa fa fa fa fa fa fa fa fa fa[fa]
0x0fe4ced0f100: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0fe4ced0f110: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0fe4ced0f120: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0fe4ced0f130: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
0x0fe4ced0f140: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
Shadow byte legend (one shadow byte represents 8 application bytes):
Addressable: 00
Partially addressable: 01 02 03 04 05 06 07
Heap left redzone: fa
Freed heap region: fd
Stack left redzone: f1
Stack mid redzone: f2
Stack right redzone: f3
Stack after return: f5
Stack use after scope: f8
Global redzone: f9
Global init order: f6
Poisoned by user: f7
Container overflow: fc
Array cookie: ac
Intra object redzone: bb
ASan internal: fe
Left alloca redzone: ca
Right alloca redzone: cb
Shadow gap: cc
==42==ABORTING
类解决方案 { 公共: 静态bool cmp(int a，int b) { 返回a >= b; } int breakfastNumber(vector<int>& stub，vector<int>& 饮料，int x) { int = 0; int stapleLen = 订书钉.尺寸 (); cout <stapleLen <endl; int drinksLen = 饮料尺寸 (); cout << drinksLen <endl; sort (studin.begin()，stapin.end()，cmp); 排序 (饮料.开始 () 、饮料.结束 ()); int j = 0; for (int i = 0; i < stapleLen; i ++) { ans + = j; ans % = 1000000007; 用于 (; j < drinksLen; j ++) { 如果 (主食 [i] + 饮料 [j] <= x) { ans ++; ans % = 1000000007; } else { 打破; } } } 退货; } }; 提示信息: ================================================================= = = 42 = ERROR: AddressSanitizer: 堆-缓冲区-在pc 0x00000034d6ad bp处的地址0x7f22768b87fc 746101b0 sp 0x7ffc746101a8 在0x7f22768b87fc线程T0读取大小为4 #6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6 + 0x270b2) 0x7f22768b87fc位于524288字节区域左侧4字节 [0x7f22768b8800，0x7f2276938800) 由线程T0在这里分配: #6 0x7f2279d720b2 (/lib/x86_64-linux-gnu/libc.so.6 + 0x270b2) 围绕越野车地址的阴影字节: 0x0fe4ced0f0a0: fa 0x0fe4ced0f0b0: fa 0x0fe4ced0f0c0: fa 0x0fe4ced0f0d0: fa 0x0fe4ced0f0e0: fa =>0x0fe4ced0f0f0: fa[fa] 0x0fe4ced0f100: 00 0x0fe4ced0f110: 00 0x0fe4ced0f120: 00 0x0fe4ced0f130: 00 0x0fe4ced0f140: 00 阴影字节图例 (一个阴影字节代表8个应用程序字节): 可寻址: 00 部分可寻址: 01 02 03 04 05 06 07 堆左redzone: fa 自由堆区域: fd 堆栈左redzone: f1 堆栈中redzone: f2 堆栈右红区: f3 返回后堆栈: f5 作用域后堆栈使用: f8 全球redzone: f9 全球init订单: f6 被用户中毒: f7 容器溢出: fc 数组cookie: ac Intra对象redzone: bb ASan内部: fe 左alloca redzone: ca 右alloca redzone: cb 阴影间隙: cc = = 42 = = 引产
- 职场和发展
- java
将number1.txt中的数据读取到内存,然后对数据进行排序后,再写入到number2.txt文件中.
实现步骤:
1.在当前项目根目录下，创建一个number1.txt文件。
2.定义一个测试类：Test
2.1创建一个File对象，关联number1.txt文件
2.2通过File对象进行判断，如果当前对象是一个文件的话，则继续执行下面步骤
2.3创建一个BufferedReader对象，关键已有的File对象
2.4创建TreeSet集合对象，泛型为Integer(由于题目要求数字按从小到大进行排序,选择TreeSet数据类型较为合适)
2.5通过输入流对象读取文件中内容，将每个数字都保存到集合中(注意读取出来的数据是字符串需要将字符串转成Integer类型)
2.6创建一个BufferedWriter对象，关联到当前项目下的number2.txt文件
2.7通过输出流对象将TreeSet集合中的所有元素全部写出
源文件数字：
转换后：
- python
书上的举例，假设你想要根据字符串中不同的字母数量对一个字符串集合进行排序：
In [177]: strings = ['foo', 'card', 'bar', 'aaaa', 'abab']
这里我们可以将一个匿名函数传给列表的sort方法：
In [178]: strings.sort(key=lambda x: len(set(list(x))))
In [179]: strings
Out[179]: ['aaaa', 'foo', 'abab', 'bar', 'card']
178行里面的x是什么，我有点看不懂，请问能有老师解释一下178行的意思吗？
我是初学者，请老师解惑，谢谢！！！非常感谢！！！
- android-studio
- android
- 问答团队
- html5
- css3
- css
- firefox
- javascript
- python
import pygame
class moving():
def __init__(self):
pygame.init()
self.screen = pygame.display.set_mode((500, 300))
self.bg_color = self.screen.fill((65, 222, 65))
self.font = pygame.font.Font(None, 36)
self.text = self.font.render('Hello There', 1, (10, 10, 10))
self.rect_width = 50
self.rect_height = 50
self.rect_moving = 10
self.rect1 = pygame.Surface((60, 60), pygame.SRCALPHA)
def text1(self, screen, text):
textpos = text.get_rect()
textpos.centerx = screen.get_rect().centerx
textpos.bottom = screen.get_rect().bottom / 2
screen.blit(text, textpos)
def moving_rect(self,ai_setting):
pygame.draw.rect(self.rect1, (255, 255, 255), [0, 0, self.rect_width, self.rect_height])
x = 0
for i in range(20):
x += self.rect_moving
self.screen.blit(self.rect1, [x, 0])
def run():
pygame.init()
ai_setting = moving()
ai_setting.text1(ai_setting.screen, ai_setting.text)
ai_setting.moving_rect(ai_setting)
while True:
for event in pygame.event.get():
if event.type == pygame.QUIT:
return
pygame.display.update()
pygame.display.flip()
run()
代码如上，我想要让其矩形进行移动，但是移动却出现问题。。是建立了多个矩形然后X轴进行变化。。如何建立单个矩形进行移动
- 开发语言
Sub test()
Sheets("打印发货单").Cells.Clear
Set dNum = CreateObject("scripting.dictionary")
With Sheets("发货单明细")
LastRowA = .Cells(Rows.Count, "a").End(xlUp).Row
For i = 2 To LastRowA
Key = .Cells(i, 8).Value
dNum(Key) = ""
Next
arr = dNum.keys
'.Range("a12").Resize(1, UBound(arr) + 1) = arr
'.Range("a12").Resize(UBound(arr) + 1) = Application.Transpose(arr)
For i = 0 To UBound(arr)
Sheets("模板").[A3] = "客户:"
Sheets("模板").[c3] = ""
Sheets("模板").[f2] = ""
Sheets("模板").Range("A5:F14").ClearContents
k = 0
For j = 2 To LastRowA
If .Cells(j, 8) = arr(i) Then
Sheets("模板").[A3] = "客户:" & .Cells(j, 2)
Sheets("模板").[c3] = .Cells(j, 1)
Sheets("模板").[f2] = arr(i)
Sheets("模板").Cells(5 + k, 1) = .Cells(j, 3)
Sheets("模板").Cells(5 + k, 2) = .Cells(j, 4)
Sheets("模板").Cells(5 + k, 3) = .Cells(j, 5)
Sheets("模板").Cells(5 + k, 4) = .Cells(j, 6)
Sheets("模板").Cells(5 + k, 5) = .Cells(j, 7)
k = k + 1
End If
Next
Sheets("模板").Range("A1:g19").Copy
Sheets("打印发货单").Activate
Sheets("打印发货单").Cells(i * 19 + 1, 1).Select
Selection.PasteSpecial Paste:=xlPasteColumnWidths, Operation:=xlNone, _
SkipBlanks:=False, Transpose:=False
ActiveSheet.Paste
Next
End With
End Sub
'1 ,18,35
'1 2 3