从 System.Net.WebRequest 获得完整回复 body

Question

我正在使用 System.Net.WebRequest 从一些 API 获取信息。当我收到错误时，响应仅包含基本的 HttpStatusCode 和消息，而不是完整的错误 returned。为了进行比较，运行相同的 post 数据和 headers 在 POSTMAN 等工具中将 return 来自 API 的完整错误。

我想知道是否有一些属性或方法可以获得完整的错误响应？？

这是我的代码运行:

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，我收到错误消息：The remote server returned an error: (400) Bad Request. 并且没有 InnerException，并且我无法从异常中受益。

[Answer] @Rene 指出了正确的方向和正确的响应 body 可以这样获得：

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

Answer 1

您捕获的是一般异常，因此没有太多空间提供具体信息。

你应该赶上 specialized exception that is thrown by the several webrequest classes, namely 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

    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;
        string content;
        using(var reader = new StreamReader(ex.Response.GetResponseStream()))
        {
            content = reader.ReadToEnd();
        }
        // do what ever you want to store and return to your callers
    }
}

在主机发送的 WebException instance you also have access to the Response 中，因此您可以访问发送给您的任何内容。

从 System.Net.WebRequest 获得完整回复 body

Getting full response body from System.Net.WebRequest

c#

webrequest

httpwebresponse