在WCF REST API中,我想出于某种原因设置自己的StatusCode -
WebOperationContext.Current.OutgoingResponse.StatusCode = HttpStatusCode.Forbidden (or any other statuscode)
它在客户端完全不起作用: -
using (HttpWebResponse restResponse = (HttpWebResponse)restRequest.GetResponse())
{
if (restResponse.StatusCode != HttpStatusCode.OK)
{
retVal = String.Format("Request failed. Received HTTP {0}, Description : {1}", restResponse.StatusCode, restResponse.StatusDescription);
return retVal;
}
using (var st = restResponse.GetResponseStream())
{
StreamReader sr = new StreamReader(st);
retVal = sr.ReadToEnd();
return retVal;
}
}
但是这里的restResponse.StatusCode总是HttpStatusCode.OK。
我可以知道原因吗?这有什么不对?
除此之外: -
我尝试在休息服务方法中使用以下内容 -
throw new WebFaultException<string>(<...Your message....>, HttpStatusCode.Forbidden );
但这种方法的问题是在客户端.....它使代码出来
Try
{
using (HttpWebResponse restResponse = (HttpWebResponse)restRequest.GetResponse())
{
if (restResponse.StatusCode != HttpStatusCode.OK)
{
retVal = String.Format("Request failed. Received HTTP {0}, Description : {1}", restResponse.StatusCode, restResponse.StatusDescription);
return retVal;
}
using (var st = restResponse.GetResponseStream())
{
StreamReader sr = new StreamReader(st);
retVal = sr.ReadToEnd();
return retVal;
}
}
}
catch
{
//it comes here when using "throw new WebFaultException...."
}
没有必要使用 -
retVal = String.Format("Request failed. Received HTTP {0}, Description : {1}", restResponse.StatusCode, restResponse.StatusDescription);
return retVal;
有什么建议吗?
答案 0 :(得分:2)
在WCF Rest API中,您可以使用此
throw new WebFaultException<string>(<...Your message....>, HttpStatusCode.Forbidden );