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1、C#Post数据和接收简单示例计算机交流平台：计算机故障_计算机论坛 wy C#Post数据和接收简单示例 C# Post数据和接收简单示例 public partial class Post_Server : System.Web.UI.Page protected void Page_Load(object sender, EventArgs e) string type = ; string Re = ; Re += 数据传送方式：; if (Request.RequestType.ToUpper = POST) type = POST; Re += type + 参数分别是：; Sor
2、tedList table = Param; if (table != null) foreach (DictionaryEntry De in table) Re += 参数名： + De.Key + 值： + De.Value + ; else Re = 你没有传递任何参数过来！; else type = GET; Re += type + 参数分别是：; NameValueCollection nvc = GETInput; if (nvc.Count != 0) for (int i = 0; i nvc.Count; i+) Re += 参数名： + nvc.GetKey(i) +
3、值： + nvc.GetValues(i)0 + ; else Re = 你没有传递任何参数过来！; Response.Write(Re); /获取GET返回来的数据 private NameValueCollection GETInput return Request.QueryString; 计算机交流平台：计算机故障_计算机论坛 wy / 获取POST返回来的数据 private string PostInput try System.IO.Stream s = Request.InputStream; int count = 0; byte buffer = new byte1024;
4、 StringBuilder builder = new StringBuilder; while (count = s.Read(buffer, 0, 1024) 0) builder.Append(Encoding.UTF8.GetString(buffer, 0, count); s.Flush; s.Close; s.Dispose; return builder.ToString; catch (Exception ex) throw ex; private SortedList Param string POSTStr = PostInput; SortedList SortList = new SortedList; int index = POSTStr.IndexOf(&); string Arr = ; if (index != -1) /参数传递不只一项 Arr = POSTStr.Split(&); for (int i = 0; i 0) str += new String(read, 0, count); count = readStream.Read(read, 0, 256); readStream.Close; myResp.Close; = 计算机交流平台：计算机故障_计算机论坛 wy Response.Write(str);
