Silverlight的HTTP POST几个变量，最简单的例子（最少的代码） [英] Silverlight HTTP POST few variables, SIMPLEST example (least code)

问题描述

您好我想从Silverlight的一些数据发布到一个网站。结果
我发现下面的链接和它的作品...结果
但是.... 这个例子是如此拟订该让我的眼睛受伤了。结果
还..柔性例子是干净多了/少代码..结果
结果
我说必须有一个更好的解决方案..

有关参考。我们发布2个变量（串），并读出结果（字符串）。

从链路解决方案：

  1 // C＃
 2. //创建一个请求对象
 3. HttpWebRequest的要求=（HttpWebRequest的）WebRequest.Create（新的URI（POST_ADDRESS，UriKind.Absolute））; 
 4. request.Method =POST; 
 5 //千万不要错过了这个
 6. request.ContentType =应用/的X WWW的形式，进行了urlencoded 
 7. request.BeginGetRequestStream（新的AsyncCallback（RequestReady），请求）; 
 8 
 9. //透过POST数据
 10.无效RequestReady（IAsyncResult的asyncResult）
 11. {
 12. HttpWebRequest的要求= asyncResult.AsyncState作为HttpWebRequest的; 
 13.流流= request.EndGetRequestStream（asyncResult）; 
 14，
 15. //哈克解决多线程问题
 16. //我觉得这是一个错误
 17. this.Dispatcher.BeginInvoke（委托（） 
 18. {
 19. //发送POST变量
 20. StreamWriter的作家=新的StreamWriter（流）; 
 21. writer.WriteLine（键1 =值） ; 
 22. writer.WriteLine（键2 =值）; 
 23 writer.Flush（）; 
 24 writer.Close（）; 
 25 
 26. request.BeginGetResponse（新的AsyncCallback（ResponseReady），请求）; 
 27.}）; 
 28} 
 29 
 30. //得到的结果
 31.无效ResponseReady（IAsyncResult的asyncResult）
 32. {
 33的HttpWebRequest要求= asyncResult.AsyncState为HttpWebRequest的; 
 34. HttpWebResponse响应=（HttpWebResponse）request.EndGetResponse（asyncResult）; 
 35 
 36. this.Dispatcher.BeginInvoke（委托（）
 37. {
 38.流responseStream = response.GetResponseStream（）; 
 39的StreamReader读者=新的StreamReader（responseStream）; 
 40. //得到的结果文本
 41.字符串结果= reader.ReadToEnd（）; 
 42.}）; 
 43} 
  

解决方案

您可以使用Web客户端发送表单数据。如果你不关心的成功确认这将是非常短暂：

  WebClient的WC =新的WebClient（）; 
 wc.Headers [内容类型] =应用/的X WWW的形式，进行了urlencoded 
 wc.UploadStringAsync（新的URI（postUrl），POST，VAL1 =参数1&放大器;将val2 =参数2）; 
  

Hello I want to post some data from silverlight to a website.
I found the following link and it works...
However.... This example was so elaborate it made my eyes hurt.
Also.. the flex example was much cleaner/less code..

I'd say there must be a better solution...

For reference.. We post 2 variables (strings) and read out the result (string).

The solution from the link :

   1. // C#  
   2. // Create a request object  
   3. HttpWebRequest request = (HttpWebRequest)WebRequest.Create(new Uri(POST_ADDRESS, UriKind.Absolute));  
   4. request.Method = "POST";  
   5. // don't miss out this  
   6. request.ContentType = "application/x-www-form-urlencoded";  
   7. request.BeginGetRequestStream(new AsyncCallback(RequestReady), request);  
   8.   
   9. // Sumbit the Post Data  
  10. void RequestReady(IAsyncResult asyncResult)  
  11. {  
  12.     HttpWebRequest request = asyncResult.AsyncState as HttpWebRequest;  
  13.     Stream stream = request.EndGetRequestStream(asyncResult);  
  14.   
  15.     // Hack for solving multi-threading problem  
  16.     // I think this is a bug  
  17.     this.Dispatcher.BeginInvoke(delegate()  
  18.     {  
  19.         // Send the post variables  
  20.         StreamWriter writer = new StreamWriter(stream);  
  21.         writer.WriteLine("key1=value1");  
  22.         writer.WriteLine("key2=value2");  
  23.         writer.Flush();  
  24.         writer.Close();  
  25.   
  26.         request.BeginGetResponse(new AsyncCallback(ResponseReady), request);  
  27.     });  
  28. }  
  29.   
  30. // Get the Result  
  31. void ResponseReady(IAsyncResult asyncResult)  
  32. {  
  33.     HttpWebRequest request = asyncResult.AsyncState as HttpWebRequest;  
  34.     HttpWebResponse response = (HttpWebResponse)request.EndGetResponse(asyncResult);  
  35.   
  36.     this.Dispatcher.BeginInvoke(delegate()  
  37.     {  
  38.         Stream responseStream = response.GetResponseStream();  
  39.         StreamReader reader = new StreamReader(responseStream);  
  40.     // get the result text  
  41.         string result = reader.ReadToEnd();  
  42.     });  
  43. }  

解决方案

You can use WebClient to send form data. If you don't care about confirmation of success it will be very short:

WebClient wc = new WebClient();
wc.Headers["Content-type"] = "application/x-www-form-urlencoded";
wc.UploadStringAsync(new Uri(postUrl), "POST", "val1=param1&val2=param2");

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