asp.net将xml发布到重定向页面

本文关键字：重定向 net xml asp | 更新日期: 2023-09-27 18:24:52

我从Desktop应用程序向asp.net wep页面发布了一个xml数据。我可以在网页上读取xml，但如果我将页面重定向到另一个页面，我会在Desktop应用程序中出错。我的代码如下。谢谢你的帮助。

桌面应用程序：

    static void Main(string[] args)
    {
        post();
        Console.WriteLine("Post process is successed...");
        Console.Read();
    }
    private static void post() {
        System.Net.WebRequest req = null;
        System.Net.WebResponse rsp = null;
        try {
            req = System.Net.WebRequest.Create("http://localhost:5318/Account/XMLReader.aspx");
            req.Method = "POST";
            req.ContentType = "text/xml";
            StreamWriter writer = new System.IO.StreamWriter(req.GetRequestStream());
            writer.WriteLine(getXmlData());
            writer.Flush();
            writer.Close();
            rsp = req.GetResponse();
        } catch {
            throw;
        } finally {
            if (req != null) req.GetRequestStream().Close(); // *****Error occures here****
            if (rsp != null) rsp.GetResponseStream().Close();
        }
    }

asp.net页面代码：

    protected void Page_Load(object sender, EventArgs e)
    {
        Page.Response.ContentType = "text/xml";
        System.IO.StreamReader reader =
        new System.IO.StreamReader(Page.Request.InputStream);
        String xmlData = reader.ReadToEnd();
        Response.Redirect("/Account/Login.aspx");
    }

asp.net将xml发布到重定向页面

试试这个代码：

private static void post()
        {
            System.Net.WebRequest req = null;
            System.Net.WebResponse rsp = null;
            Stream newStream = null;
            try
            {
                req = System.Net.WebRequest.Create("http://localhost:39384/Default.aspx");
                req.Method = "POST";
                req.ContentType = "text/xml";
                newStream = req.GetRequestStream();
                StreamWriter writer = new System.IO.StreamWriter(newStream);
                writer.WriteLine("ddd");
                writer.Flush();
                writer.Close();
                rsp = req.GetResponse();
            }
            catch(Exception e)
            {
                throw;
            }
            finally
            {
                if (req != null) newStream.Close(); // *****Error occures here****
                if (rsp != null) rsp.GetResponseStream().Close();
            }
        }
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