Question

我将xml / cxml文档发布到网上的网址。当我发布它时，我收到错误“连接被重置”。我想发布代码，以确保他们没有错误。 stXML是xml文档。

HttpWebRequest req = (HttpWebRequest)WebRequest.Create(uri);

        byte[] postDataBytes = Encoding.ASCII.GetBytes(stXML);

        req.Method = "POST";

        req.ContentLength = postDataBytes.Length;

        // req.ContentType = "text/XML-urlencoded";

        Stream requestStream = req.GetRequestStream();

        requestStream.Write(postDataBytes, 0, postDataBytes.Length);

        requestStream.Close();

        HttpWebResponse resp = (HttpWebResponse)request.GetResponse();

        StreamReader responseReader = new StreamReader(resp.GetResponseStream(), Encoding.Default);

        string strRet = responseReader.ReadToEnd();

        Response.Write(strRet);

        Response.Close();

Answer 1

也许您的标题存在问题。我会避免使用HttpWebRequest增加复杂性，只使用System.Net.WebClient。

string response = new WebClient().UploadString(uri, stXML);

XML Post Connection重置错误

1 个答案: