1、 地址：http://125.88.70.16:8082/EPG/jsp/defaulthdcctv/gx2/page/toidc/getRtspToIDC.jsp
2、 第一点中的地址是16服务器上的测试地址，如要项目上线，需将更换为现网地址

1、 http://125.88.70.16:8082/EPG/jsp/defaulthdcctv/gx2/page/toidc/getRtspToIDC.jsp?foreignId=4244&contentType=4243&callback=getRtspURL
2、 回调函数：
function getRtspURL (rtsp){
//函数体…
}

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2个回答

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VC从ftp服务器上下载文件问题，VC怎么实现显示ftp服务器上指定路径下的所有文件呢，望高手指点指点！rn之前在网上看到的例子都是，静态路径给定一个文件然后实现下载到本地。。。怎么动态由自己选择一个文件然后下载到本地上

6、A real polynomial P(x) of degreen or less is given byrnP(x) = a0 a1x a2x平方.. anxn次方 rnWith coefficients a0,a1,a2,...,an representing real numbers.rnYou are part of a design team and your task is to define a class to represent P(x) and to evaluate the value of the polynomial P(x) at x.rnrna) Providing that the value of n is always less than 100, what is the most suitable data structure to store coefficients a0,a1,a2,...,an ? Define this data structure and name it a.rnb) Define the class ‘Polynomial’ that contains the following three data items(private data):rn• a Answer to part a.rn• n Polynomial degree.rn• xrnThe class ‘Polynomial’ also contains three member functions:rn• setn sets the value of n ( passed as a parameter) and returns nothing.rn• SetCoefficients sets a0,a1,a2,...,an by asking the user to enter them. This function has no parameter and returns nothing;rn• Evaluate evaluates the value of polynomial p at x (passed as a parameter) and returns the result as an integer value.rnNote that you just need to write the definition of the class. Do not write body of the functions in this part.rnc) Give the implementation of the function ‘Evaluate’.rnAssume that the function ‘setCoefficient’ is always called before calling the function ‘Evaluate’ so you are sure that a0,a1,a2,...,an were set before the function ‘Evaluate’ and you can use them in your function.rn

http://zhidao.baidu.com/question/351697095.htmlrnrn问题描述在这个网址上，csdn不知道要怎么发图，所以就把这个网址粘过来了，求高手给点思路啊。rn如果方便，可以加我QQ：87574619rn谢谢了

50分~~~~高手给点思路吧~~~~~

java web 高手给点思路？？

【求助】请高手给点思路

rn 1 问题我有一个工程组现在我要在一个工程组中的一个工程中使用另一个工程中的模块，界面，控件。注意不是activex工程，只是标准的exe工程，有没有办法。rnrn 2如何在一个执行的EXE文件界面中注册控件，并在EXE中记录，下次在调用此EXE能的自动加入。还有怎样做到这样的功能（我在界面上用代码创建了一个控件注意只有操作时才会创建如点某一按钮时等。我怎样才能在EXE文件打开的界面中保存它的位置，类型，大小；下次使用EXE时它将存在。就有点像windows在桌面创建了快捷图标，下一次启动WINDOWS时快捷图标就显示在哪儿）能给思路就行 并不局限于vb

#include rn#include rnusing namespace std;rnrntemplate rnint indexOfKey(T list[], int low, int high, T key);rnrntemplaternint indexOfKey(T list[], T key, int size);rnrnint main()rnrn int list1[10] = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9;rn int index1 = indexOfKey(list1, 2, 10);rn cout << "The index of the key is " << index1 << endl;rnrn double list2[10] = 1.2, 2.5, 3.1, 4, 5, 6, 7, 8.1, 9.5;rn int index2 = indexOfKey(list2, 8.1, 10);rn cout << "The index of the key is " << index2 << endl;rnrn string list3[4] = "BeiJing", "ShangHai", "HuNan", "ChenZhou";rn int index3 = indexOfKey(list3, "BeiJing", 4);rn cout << "The index of the key is " << index3 << endl;rnrn return 0;rnrnrnrntemplate rnint indexOfKey(T list[], int low, int high, T key)rnrn int low = 0;rn int high = size - 1;rnrn if (low > high)rn return -1 - low;rn elsern rn int mid = (low + high) / 2;rn if (key > list[mid])rn return indexOfKey(list, mid + 1, high, key);rn elsern return indexOfKey(list, low, mid + 1, key);rn rnrnrntemplate rnint indexOfKey(T list[], T key, int size)rnrn int low = 0;rn int high = size - 1;rn return indexOfKey(list, low, mid + 1, key);rn

[img=http://hi.csdn.net/attachment/201106/23/1950766_1308794125O593.jpg][/img]rn我这有如上两个表，我想找出下表中district中的一个或多个区域被包含在上表中的district的下表记录，该怎么查？rn注：上表中的区域个数为1到10个，下表中的区域个数为1到3个。rn现在知道上表中一条记录的ID，求下表符合条件的记录。rn请大侠给出思路，当然最好有简单代码。可以是sql，也可C#或Linq。谢谢诶rn

DescriptionrnIt is often helpful for computer users to see a visual representation of the file structure on their computers. The "explorer" in Microsoft Windows is an example of such a system. Before the days of graphical user interfaces, however, such visual representations were not possible. The best that could be done was to show a static "map"of directories and files, using indentation as a guide to directory contents. For example: rnROOTrnrn| DIR1rnrn| File1rnrn| File2rnrn| File3rnrn| DIR2rnrn| DIR3rnrn| File1rnrnFile1rnrnFile2rnrnrnThis shows that the root directory contains two files and three subdirectories. The first subdirectory contains 3 files, the second is empty and the third contains one file.rnInputrnWrite a program that reads a series of data sets representing a computer file structure. A data set ends with a line containing a single *, and the end of valid data is denoted by a line containing a single #. The data set contains a series of file and directory names. (The root directory is assumed to be the starting point.) The end of a directory is denoted by a ']' Directory names begin with a lower case 'd' and file names begin with a lower case 'f' File names may or may not have an extension (such as fmyfile.dat or fmyfile). File and directory names may not contain spaces. rnOutputrnNote that the contents of any directory should list any subdirectories first, followed by files, if any. All files should be in alphabetical order within each directory. Note that each data set output is marked by the label "DATA SET x:" where x denotes the number of the set, beginning at 1. Note also the blank line between the output data sets. Each level of indentation should show a '|' followed by 5 spaces. rnSample Inputrnfile1rnfile2rndir3rndir2rnfile1rnfile2rn]rn]rnfile4rndir1rn]rnfile3rn*rnfile2rnfile1rn*rn#rnSample OutputrnDATA SET 1:rnROOTrn| dir3rn| | dir2rn| | file1rn| | file2rn| dir1rnfile1rnfile2rnfile3rnfile4rnrnDATA SET 2:rnROOTrnfile1rnfile2rn

buffer[0] ^= buffer[1];rnbuffer[2] ^= buffer[3];rn rnbuffer[0] ^= buffer[2];rnbuffer[0] ^= ( buffer[0] >> 16 );rnbuffer[0] ^= ( buffer[0] >> 8);rnbuffer里储存的是一个128位的二进制数，请问如何将以上程序用公式表达出来呢?请高手不吝赐教！rn 我试着这样表达rn buffer1=buffer ( buffer*2（-64次方） ) rn buffer2=buffer1 ( buffer1*2（-32次方） ) rn buffer3=buffer2 ( buffer2*2（-16次方） ) rn buffer4=buffer3 ( buffer3*2（-8 次方）) rn但是，这样，2（-64次方）数太小，无法表示，请问有什么办法表示出来呢?

[code=sql] select rn A.MO_NUMBER,rn (decode(A.GROUP_NAME , 'PACKING', count(*))) PACKING,rn (decode(A.GROUP_NAME, 'MAC INPUT', count(*))) MAC_INPUT,rn (decode(A.GROUP_NAME, 'FT2', count(*))) FT2,rn (decode(A.GROUP_NAME, 'AOI II', count(*))) AOI_II,rn (decode(A.GROUP_NAME, 'MAC COMPARE', count(*))) MAC_COMPARE,rn (decode(A.GROUP_NAME, 'ASY-1', count(*))) ASY_1,rn (decode(A.GROUP_NAME, 'VISUAL INSPECT', count(*))) VISUAL_INSPECTrnfrom rn SFISM4.R_WIP_TRACKING_T a rn where A.MO_NUMBER='005700000405-1'rngroup by A.MO_NUMBER, A.GROUP_NAME[/code]rnrn輸出如下：rn005700000405-1 20 rn005700000405-1 4 rn005700000405-1 1 rn005700000405-1 9 rn005700000405-1 4 rn005700000405-1 rn005700000405-1 2 rn005700000405-1 18

rn rn 贵美商城rn rn rn rn rn rn rn rn rn rn 手机充值-IP卡-电话卡rn 网游-点卡-金币-代练rn rn rn 移动rn 联通rn 魔兽rn 跑跑卡丁车rn rn rn rn rn

[code=C#]rnstring countryDHLId = "Area" + countryInfo.DHLArea.ToString();rnrnvar info = from df in db.DHLFreightsrn where df.WeightMax >= weight && df.WeightMin <= weightrn select newrn rn countryDHLIdrn ;rnrndouble dhlFreight = double.Parse(info.First().countryDHLId);rnrnrn[/code]rnrn我想查DHLFreights 的某一个字段 countryDHLId 可能是Area1~~~Area9，可是为什么不能把这个字段的数据查出来呢