Question

我在Windows Phone 8.1应用程序中使用RestSharp。当服务器返回代码不同于200的响应时，RestClient抛出异常。Wiki表示我应该使用正确的状态代码获得响应。我希望获得响应内容，因为服务器返回错误消息。

private async Task<T> ExecuteAsync<T>(IRestRequest request)
    {
        if (!_networkAvailableService.IsNetworkAvailable)
        {
            throw new NoInternetException();
        }

        request.AddHeader("Accept", "application/json");

        IRestResponse<T> response;
        try
        {
            response = await _client.Execute<T>(request); //here I get exception
        }
        catch (Exception ex)
        {
            throw new ApiException();
        }

        HandleApiException(response);

        return response.Data;
    }

    private void HandleApiException(IRestResponse response)
    {
        if (response.StatusCode == HttpStatusCode.OK)
        {
            return;
        }
//never reach here :( 
        ApiException apiException;
        try
        {
            var apiError = _deserializer.Deserialize<ApiErrorResponse>(response);
            apiException = new ApiException(apiError);
        }
        catch (Exception)
        {
            throw new ApiException();
        }

        throw apiException;
    }

服务器响应示例：

HTTP/1.1 400 Bad Request
Cache-Control: no-cache
Pragma: no-cache
Content-Length: 86
Content-Type: application/json;charset=UTF-8
Expires: -1
Server: Microsoft-IIS/8.0
Access-Control-Allow-Origin: *
X-Powered-By: ASP.NET
Date: Mon, 28 Sep 2015 12:30:10 GMT
Connection: close

{"error":"invalid_token","error_description":"The user name or password is incorrect"}

Answer 1

如果您在Windows Phone 8.1下工作，则可能正在使用RestSharp Portable（https://github.com/FubarDevelopment/restsharp.portable）（可能）。使用此：

var client = new RestClient();
client.IgnoreResponseStatusCode = true;

有了这个，例如你不会因404而异常。我希望它会有所帮助:)）

RestSharp - 如何处理非200响应？ RestClient在Execute上抛出异常

1 个答案: