Question

页面显示时，我有一个需要运行的功能。这是下面的代码：

        protected async override void OnAppearing()
        {
            LeaveViewModel lvm = new LeaveViewModel();
            await lvm.FetchingAsync();
        }

        public ICommand LeaveCommand
        {
            get
            {
                return new Command(async () => await FetchingAsync());
            }
        }

        public async Task FetchingAsync()
        {

            await _apiServices.FetchLeave(this);
        }

        public async Task FetchLeave(LeaveViewModel lvm)
        {
            var keyValues = new List<KeyValuePair<string, string>>
            {
                new KeyValuePair<string, string>("UserID",App._username),
            };
            var request = new HttpRequestMessage(HttpMethod.Post, "somelink");
            request.Content = new FormUrlEncodedContent(keyValues);
            var client = new HttpClient();
            var response = await client.SendAsync(request);
            var content = await response.Content.ReadAsStringAsync();
            Debug.WriteLine(content);
            if (content == "[]")
            {
                await App.Current.MainPage.DisplayAlert("Error", "No data recorded!", "OK");
            }
            else
            {
                var posts = JsonConvert.DeserializeObject<List<LeaveViewModel>>(content);
                lvm.ViewList = new List<LeaveViewModel>(posts);
            }
        }

当我调试代码并到达

var response = await client.SendAsync(request);

调试器返回到FetchingAsync（）函数，在var响应后完全跳过其余代码。当我在按钮而不是OnAppearing（）中运行代码并将按钮绑定到Icommand时，整个代码将得到完美处理。但是我需要在页面加载时显示数据，而不是在单击按钮时显示数据。我读到这是由于异步/等待引起的，但是我对此并不陌生，我不明白这是怎么回事。我该如何实现？我在做什么错？

异步/等待跳过任务

0 个答案: