从System.Net.WebRequest获取完整的响应正文

Question

我正在使用System.Net.WebRequest从某些API获取信息.当我收到错误时,响应只包含基本的HttpStatusCode和消息,而不是返回完整的错误.为了进行比较,在POSTMAN等工具中运行相同的发布数据和标头将返回该API的完整错误.

我想知道是否有一些属性或方式我可以获得完整的错误响应？

这是我正在运行的代码:

public HttpStatusCode GetRestResponse(
    string verb,
    string requestUrl,
    string userName,
    string password,
    out string receiveContent,
    string postContent = null)
{
    var request = (HttpWebRequest)WebRequest.Create(requestUrl);
    request.Method = verb;

    if (!string.IsNullOrEmpty(userName))
    {
        string authInfo = string.Format("{0}:{1}", userName, password);
        authInfo = Convert.ToBase64String(Encoding.Default.GetBytes(authInfo));
        request.Headers.Add("Authorization", "Basic " + authInfo);
    }

    if (!string.IsNullOrEmpty(postContent))
    {
        byte[] byteArray = Encoding.UTF8.GetBytes(postContent);
        request.ContentType = "application/json; charset=utf-8";
        request.ContentLength = byteArray.Length;
        var dataStream = request.GetRequestStream();
        dataStream.Write(byteArray, 0, byteArray.Length);
        dataStream.Close();
    }

    try
    {
        using (WebResponse response = request.GetResponse())
        {
            var responseStream = response.GetResponseStream();
            if (responseStream != null)
            {
                var reader = new StreamReader(responseStream);
                receiveContent = reader.ReadToEnd();
                reader.Close();

                return ((HttpWebResponse) response).StatusCode;
            }
        }
    }
    catch (Exception ex)
    {
        receiveContent = string.Format("{0}\n{1}\nposted content = \n{2}", ex, ex.Message, postContent);
        return HttpStatusCode.BadRequest;
    }

    receiveContent = null;
    return 0;
}

当我生成一个向我显示错误的请求时,我会收到错误消息:The remote server returned an error: (400) Bad Request.并且没有InnerException,没有其他任何我可以从异常中受益.

[答案] @Rene指向正确的方向,可以像这样获得正确的反应体:

var reader = new StreamReader(ex.Response.GetResponseStream());
var content = reader.ReadToEnd();

Answer 1

您正在捕获通用异常，因此没有足够的空间来存储特定信息。

您应该捕获由几个webrequest类引发的特殊异常，即WebException

您的捕获代码可能是这样的：

catch (WebException e)
{
    var response = ((HttpWebResponse)e.Response);
    var someheader = response.Headers["X-API-ERROR"];
    // check header
    var content = response.GetResponseStream();
    // check the content if needed 

    if (e.Status == WebExceptionStatus.ProtocolError)
    {
        // protocol errors find the statuscode in the Response
        // the enum statuscode can be cast to an int.
        int code = (int) ((HttpWebResponse)e.Response).StatusCode;
        // do what ever you want to store and return to your callers
    }
}

在WebException实例中，您还可以访问Response来自主机的发送，因此您可以访问发送给您的任何内容。