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我通过 webrequest POST 方法发送了一些数据。我知道如何捕获和修改数据，但我无法将其放回响应并将其发送回客户端。我尝试了 4 种方法，但没有任何效果。

服务器部分

this.Response.ContentType = "text/xml";
StreamReader reader = new StreamReader(this.Request.InputStream);
string x = reader.ReadToEnd();
if (x != "")
{
    string xml = HttpUtility.UrlDecode(x);
    XDocument xmlInput = XDocument.Parse(xml);
    EditXml(xmlInput);
    String str = "This is testing";
    System.Text.UTF8Encoding encoding = new System.Text.UTF8Encoding();
    byte[] b = encoding.GetBytes(str);
    Response.OutputStream.Write(b, 0, b.Length);
    //Response.ContentType = "text/xml";
    //Response.Clear();
    //Response.BufferOutput = true;
    //xmlInput.Save(Response.OutputStream);
    //Response.Flush();

客户部分

WebRequest request = WebRequest.Create(uri);
            request.Method = "POST";
            request.ContentLength = System.Text.Encoding.Default.GetByteCount(data);
            request.ContentType = "application/x-www-form-urlencoded";
            requestStream = request.GetRequestStream();
            StreamWriter writer = new StreamWriter(requestStream);
            writer.Write(data);
            writer.Close();
            responseStream = request.GetResponse().GetResponseStream();
            StreamReader reader = new StreamReader(responseStream);
            string received = reader.ReadToEnd();

在那里，我展示了两种尝试将数据响应的方法，但都不起作用……我完全错了，我只是在最后一步中遗漏了一些简单的东西吗？:-)

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0 回答 0