后台编写HttpWebRequest的POST请求,必须注意的一个小细节
通过HttpWebRequest来进行POST请求的时候,如果发现在连续操作的情况下,会导致程序反应迟钝或被卡死,可以查看一下,是否是因为缺少了获取服务器状态这一关键步骤,所以才导致本地程序因为等待服务器的响应而暂停等待,正常完整的发送请求代码示例如下:
1 HttpWebRequest myReq = (HttpWebRequest)WebRequest.Create(UrlText.Text + FileUpload1.FileName);
2
3 myReq.Method = "POST";
4 myReq.ContentType = "application/x-www-form-urlencoded";
5 myReq.ContentLength = FileUpload1.FileBytes.Length;
6 Stream outStream = myReq.GetRequestStream();
7 outStream.Write(FileUpload1.FileBytes, 0, FileUpload1.FileBytes.Length);
8 outStream.Close();
9
10 // 获取服务器反馈结果
11 using (HttpWebResponse response = (HttpWebResponse)myReq.GetResponse())
12 {
13 if (response.StatusCode != HttpStatusCode.OK)
14 {
15 throw new Exception("上传文件返回结果错误!");
16 }
2
3 myReq.Method = "POST";
4 myReq.ContentType = "application/x-www-form-urlencoded";
5 myReq.ContentLength = FileUpload1.FileBytes.Length;
6 Stream outStream = myReq.GetRequestStream();
7 outStream.Write(FileUpload1.FileBytes, 0, FileUpload1.FileBytes.Length);
8 outStream.Close();
9
10 // 获取服务器反馈结果
11 using (HttpWebResponse response = (HttpWebResponse)myReq.GetResponse())
12 {
13 if (response.StatusCode != HttpStatusCode.OK)
14 {
15 throw new Exception("上传文件返回结果错误!");
16 }
17 }
请注意看第10行及以下部分代码,就是获取请求响应的代码,而次过程为非常关键的步骤,不可缺少。