C# Web服务请求超时问题
我想与web服务交互,为此我使用了以下代码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/
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”代码>。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。