Question

我想像这样调用WebClient：

using (TimeoutWebClient client = new TimeoutWebClient())
            {
                System.Collections.Specialized.NameValueCollection reqparm = new System.Collections.Specialized.NameValueCollection();
                reqparm.Add("email", email);
                reqparm.Add("pass", password);
                Debug.Log("5");

                if (!string.IsNullOrEmpty(arg))
                        reqparm.Add(arg, "please");

                Uri url = new Uri(URL);
                byte[] responsebytes = client.UploadValues(url, "POST", reqparm);
                Debug.Log("6");
                string responsebody = Encoding.UTF8.GetString(responsebytes);
                // Debug here
                return responsebody;
            }
        }
        catch (TimeoutException)
        {
            Debug.LogWarning("Timeout while request, retry !");
            Debug.Log("7");
        }
        catch (Exception e)
        {
            Debug.LogError("Exception while request: " + e.Message + e.StackTrace);
            return "Error";
        }

但是当我运行它时，有时会产生一个奇怪的异常，如下所示：

Exception while request: An error occurred performing a WebClient request.  at System.Net.WebClient.UploadValues (System.Uri address, System.String method, System.Collections.Specialized.NameValueCollection data) [0x00000] in <filename unknown>:0 
  at (wrapper remoting-invoke-with-check) System.Net.WebClient:UploadValues (System.Uri,string,System.Collections.Specialized.NameValueCollection)

我真的不知道这是什么意思所以如果有人已经遇到这样一个奇怪的例外，请告诉我：X。（我正在制作视频游戏，这是登录帖子请求）

PS：在Unity和.NET 2.0下，但它几乎相同^^

编辑：这是完整的日志：

Message =  The request timed out   Help link =    Source = System   StackTrace =   at System.Net.HttpWebRequest.EndGetResponse (IAsyncResult asyncResult) [0x00000] in <filename unknown>:0 at System.Net.HttpWebRequest.GetResponse () [0x00000] in <filename unknown>:0

编辑：这是TimeoutWebClient类：

    public class TimeoutWebClient : WebClient
    {
    private int _timeOut = 7000; // 7s
    public int TimeOut
    {
        get
        {
            return _timeOut;
        }
        set
        {
            _timeOut = value;
        }
    }
    protected override WebRequest GetWebRequest(Uri address)
    {
        WebRequest webRequest = base.GetWebRequest(address);
        webRequest.Timeout = _timeOut;
        if (webRequest is HttpWebRequest)
        {
            (webRequest as HttpWebRequest).KeepAlive = false;
            (webRequest as HttpWebRequest).Timeout = _timeOut; //(tried different values)
        }
        return webRequest;
    }
}

Answer 1

这表示您的请求连接时间超过7秒。您要连接的URL不正确或服务器存在问题。

Answer 2

好的，最后，我决定使用Unity API中的WWW类而不是.NET API，因为它似乎已经过优化。如果你想知道我正在使用的代码，这里是：

    WWWForm form = new WWWForm();
    form.AddField("email", email);
    form.AddField("pass", password);

    if (!string.IsNullOrEmpty(arg))
        form.AddField(arg, "please");

    WWW www = new WWW(URL, form);
    // Wait until the request has been sent
    yield return www;

    if (www.isDone)
    {
        Debug.Log("WWW text = " + www.text);
        if (callback(www.text))
        {
            // we are logged
        }
        else
        {
            // we arn't
        }
    }

Answer 3

为什么你不使用HttpClient，这是一个使用httpClient和Json.net的简单的Get / Post请求代码

     public async Task<T> MakeHttpClientRequestASync<T>(string requestUrl, string authenticationToken,
        Dictionary<string, string> requestContent, HttpMethod verb, Action<Exception> error)
    {
        var httpClient = new HttpClient();
        httpClient.DefaultRequestHeaders.CacheControl = new CacheControlHeaderValue() { NoCache = true };
        HttpResponseMessage response;

        var returnVal = default(T);

        try
        {
            if (verb == HttpMethod.Post)
            {
                response = await httpClient.PostAsync(requestUrl, new FormUrlEncodedContent(requestContent));
            }
            else
            {
                response = await httpClient.GetAsync(requestUrl);
            }

            var resultString = await response.Content.ReadAsStringAsync();
            returnVal = JsonConvert.DeserializeObject<T>(resultString);
        }
        catch (Exception ex)
        {
            error(ex);
        }

        return returnVal;
    }

UploadValues时的WebClient异常

3 个答案: