C# Web服务请求超时问题

C# Web服务请求超时问题,c#,web-services,C#,Web Services,我想与web服务交互,为此我使用了以下代码 HttpWebRequest req = (HttpWebRequest)WebRequest.Create(@"http://192.168.2.51/loodappSrv/LoodAppsrv.svc/company/insertcompany?Validation_Token=dc6f3d5e-22c7-405f-abb6-4491de140e7e"); req.Method = "POST"; req.ContentType = @"text/

我想与web服务交互,为此我使用了以下代码

HttpWebRequest req = (HttpWebRequest)WebRequest.Create(@"http://192.168.2.51/loodappSrv/LoodAppsrv.svc/company/insertcompany?Validation_Token=dc6f3d5e-22c7-405f-abb6-4491de140e7e");
req.Method = "POST";
req.ContentType = @"text/json";
JsonComapnyFormat jcf = new JsonComapnyFormat();
string data = jcf.data();//data in json Format {"companyName":"Alpha","departmentId":3}


using (Stream requestStream = req.GetRequestStream())
{
    StreamWriter streamWriter = null;

    try
    {
        //streamWriter = new StreamWriter(requestStream, System.Text.Encoding.Default);

        streamWriter = new StreamWriter(requestStream, System.Text.Encoding.Default);
        streamWriter.Write(data);

    }
    catch (Exception ex)
    {
        throw ex;
    }
    finally
    {
        try
        {
            streamWriter.Close();
            requestStream.Close();
            streamWriter.Dispose();
            streamWriter = null;
            requestStream.Dispose();
        }
        catch
        {
        }
    }
}

HttpWebResponse response = (HttpWebResponse)req.GetResponse();
Stream stream = response.GetResponseStream();
StreamReader reader = new StreamReader(stream);

var result = reader.ReadToEnd();
ViewBag.ABC = result;
return View();

如果我使用Fiddler在给定的URL上发送POST数据,这是完美的(突然响应)。但是,当我在同一URL上发送相同的日期时,消息在
HttpWebResponse=(HttpWebResponse)req.GetResponse()处返回异常“操作已超时”。请提出解决方案

你试过SLaks提到的吗

JSON文本的MIME媒体类型为application/JSON


如图所示

去掉所有的
尝试
捕获
,最后
块,然后使用
使用
。您的
内容类型
是错误的。