任务扩展以迎合App广泛的服务呼叫

Question

我正在使用xamarin并尝试使用一种方法来使用所有服务.为此我写了一个TaskExtension.因此,从应用程序的每个页面我都可以调用该扩展方法.这是为了禁用按钮,显示加载屏幕,响应处理以及从一点开始处理异常处理.我在下面附上我的代码.需要您对此解决方案的专家意见

这是我的扩展类

public static class TaskExtensions
{
    public static async Task<ResultViewModel<U>> ExecuteAsyncOperation<U>(this Task<HttpResponseMessage> operation, object sender = null)
    {
        ResultViewModel<U> resultModel = new ResultViewModel<U>();
        Button button = BeforeAsyncCall(sender);
        try
        {
            await BackgroundOperation(operation, resultModel);
        }
        catch (Exception ex)
        {
            resultModel.Status = HttpStatusCode.InternalServerError;
            resultModel.Errors = new List<string>() { "Some error occurred. Please try again." };
        }
        finally
        {
            AfterAsyncCall(button);

        }
        return resultModel;
    }
    static async Task BackgroundOperation<U>(Task<HttpResponseMessage> operation, ResultViewModel<U> resultModel)
    {
        HttpResponseMessage RawResult = await operation;
        var Response = await RawResult.Content.ReadAsStringAsync();
        resultModel.Status = RawResult.StatusCode;
        if (RawResult.IsSuccessStatusCode)
        {
            var responseObj = await Task.Run(() => JsonConvert.DeserializeObject<U>(Response));
            resultModel.Result = responseObj;
        }
        else
        {
            var responseErrorObj = await Task.Run(() => JsonConvert.DeserializeObject<ErrorModel>(Response));
            resultModel.Errors = new List<string>();
            foreach (var modelState in responseErrorObj.ModelState)
            {
                foreach (var error in modelState.Value)
                {
                    resultModel.Errors.Add(error.ToString());
                }
            }
        }
    }
    static Button BeforeAsyncCall(object sender)
    {
        Button button = null;
        if (sender != null)
            button = (Button)sender;
        if (button != null)
            button.IsEnabled = false;
        UserDialogs.Instance.ShowLoading("Loading", MaskType.Black);
        return button;
    }
    static void AfterAsyncCall(Button button)
    {
        UserDialogs.Instance.HideLoading();
        if (button != null)
            button.IsEnabled = true;
    }
}

这是对我的扩展方法的调用

ResultViewModel<TokenModel> response = await new Service(Settings.BaseUrl).Login(loginModel).ExecuteAsyncOperation<TokenModel>(sender);

ResultViewModel

public class ResultViewModel<T> 
    {
        public HttpStatusCode Status { get; set; }
        public T Result { get; set; }
        public List<string> Errors { get; set; }
    }

异步方法

public async Task<HttpResponseMessage> Login(LoginViewModel loginModel)
        {
            try
            {

                var dataList = new List<KeyValuePair<string, string>>();
                dataList.Add(new KeyValuePair<string, string>("grant_type", "password"));
                dataList.Add(new KeyValuePair<string, string>("username", loginModel.Email));
                dataList.Add(new KeyValuePair<string, string>("password", loginModel.Password));
                var request = new HttpRequestMessage()
                {
                    RequestUri = new Uri(this.BaseUrl + "token"),
                    Method = HttpMethod.Post,
                    Content = new FormUrlEncodedContent(dataList)
                };
                var authenticateResponse = await Client.SendAsync(request);
                return authenticateResponse;
            }
            catch (Exception ex)
            {
                return null;
            }
        }

我的问题是

1)这是一个好方法吗？

2)我们可以在性能方面进行改进吗？

3)我正确使用Async吗？

Answer 1

1）这是一个好方法吗？

这种做法并没有什么问题。

2）我们可以在性能方面改进它吗？

使用扩展方法不应该有任何性能问题，但您可以进行测量以确保 100% 确定。object您正在使用to铸造创建装箱和拆箱情况button。你可以直接使用Button. ViewElement如果您想支持多种元素类型，请使用。此外，使用 也会有一些惩罚async await，但这些惩罚是最小的，并且对于不阻塞 UI 是必要的。您可以通过添加到任务中消除重新捕获上下文的需要来提高性能.ConfigureAwait(false)，但在您的情况下，您需要上下文来重新启用按钮。使用dynamic似乎没有必要并且确实有一些开销。

3）我是否正确使用了异步？

如果该方法只返回 a ，则不必awaita 。您可以通过调用方法来实现。这将减少编译器的开销，并可能提高性能。不过我之前没有测试过这个。TaskTaskawait

扩大

public static async Task<ResultViewModel<T>> ExecuteAsyncOperation<T>(this Task<HttpResponseMessage> operation, Button button)
{
    ResultViewModel<T> resultModel = new ResultViewModel<T>();
    try
    {
        if (button != null)
            button.IsEnabled = false;
        HttpResponseMessage RawResult = await operation;
        string Response = await RawResult.Content.ReadAsStringAsync();
        resultModel.Status = RawResult.StatusCode;

        if (RawResult.IsSuccessStatusCode)
        {
            var responseObj = JsonConvert.DeserializeObject<T>(Response);
            resultModel.Result = responseObj;
        }
        else
        {
            //create an error model instead of using dynamic I am guessing modelstate here
            List<ModelState> responseObj = JsonConvert.DeserializeObject<List<ModelState>>(Response);
            resultModel.Errors = new List<string>();
            foreach (ModelState modelState in responseObj)
            {
                foreach (var error in modelState.Errors)
                {
                    resultModel.Errors.Add(error.ToString());
                }
            }
        }
    }
    catch (Exception ex)
    {
        resultModel.Status = HttpStatusCode.InternalServerError;
        resultModel.Errors = new List<string>() { "Some error occurred. Please try again." };
    }
    finally
    {
        if (button != null)
            button.IsEnabled = true;
    }
    return resultModel;
}

呼唤

var button = sender as Button; 
if (button != null)
{
    ResultViewModel<TokenModel> response = await new Service(Settings.BaseUrl).Login(loginModel).ExecuteAsyncOperation<TokenModel>(sender);
}

要求

public Task<HttpResponseMessage> Login(LoginViewModel loginModel)
{
    var dataList = new List<KeyValuePair<string, string>>();
    dataList.Add(new KeyValuePair<string, string>("grant_type", "password"));
    dataList.Add(new KeyValuePair<string, string>("username", loginModel.Email));
    dataList.Add(new KeyValuePair<string, string>("password", loginModel.Password));
    var request = new HttpRequestMessage()
    {
        RequestUri = new Uri(this.BaseUrl + "token"),
        Method = HttpMethod.Post,
        Content = new FormUrlEncodedContent(dataList)
    };
    return Client.SendAsync(request);
}