Question

public async Task<ObservableCollection<CustomerModel>> GetCustomer(string customerNumber, string department)
        {
            try
            {
                progressBar.Visibility = ViewStates.Visible;
                progressBar.Progress = 0;
                listofItems = new ObservableCollection<CustomerModel>();
                string url = _client.BaseAddress + "/getcustomers(Number='" + customerNumber + "',department='" +
                             department + "')";
                var response = await _client.GetAsync(url);
                if (response.IsSuccessStatusCode)
                {
                    progressBar.Visibility = ViewStates.Invisible;
                    progressBar.Progress = 100;
                    string returnjson = await response.Content.ReadAsStringAsync();
                    ReplyCustomerModel replyCustomerModel =
                        JsonConvert.DeserializeObject<ReplyCustomerModel>(returnjson);
                    if (replyCustomerModel != null)
                    {
                        listofItems = replyCustomerModel.Customers;
                    }

                }
                else
                {

                    AlertDialog.Builder alertDiag = new AlertDialog.Builder();
                    alertDiag.SetTitle("Butikscanner App");
                    alertDiag.SetMessage("User Does not exist");
                    alertDiag.SetPositiveButton("OK",
                        (senderAlert, args) => { });
                    alertDiag.SetNegativeButton("Cancel", (senderAlert, args) => { alertDiag.Dispose(); });
                    Dialog diag = alertDiag.Create();
                    diag.Show();


                }

                return listofItems;
            }
            catch (Exception ex)
            {
                Console.WriteLine(ex);
                throw;
            }
        }

实际上，如果响应为假，这就是我正在做的事情，我试图显示不存在用户的警报对话框，我正在MVVM light中运行我的项目

Answer 1

通常，如果您也是这种情况，则使用async-await在后台线程中进行API调用，那么我建议您这样做是在UIThread上调用对话框的show方法。为此，您将需要一个活动上下文，即活动的引用。

您可以通过以下两种方法来执行此操作：直接将这种方法作为操作来调用，例如：

private void ShowDialog()
{                    
AlertDialog.Builder alertDiag = new AlertDialog.Builder();
alertDiag.SetTitle("Butikscanner App");
alertDiag.SetMessage("User Does not exist");
alertDiag.SetPositiveButton("OK",(senderAlert, args) => {  });
alertDiag.SetNegativeButton("Cancel", (senderAlert, args) => { alertDiag.Dispose(); });
Dialog diag = alertDiag.Create();
diag.Show();    
}

假设上面是您的方法定义方式，您可以在UI线程上运行它，如：

activity.RunOnUIThread(ShowDialog);

但是在您的情况下，我个人认为这不是一件明智的事情，因为应该在UIThread上的唯一代码行（至少我认为是）是dialog.Show();

您应该做的是对匿名方法使用lamba表达式，例如：

private void ShowDialog(Activity activity)
{                    
AlertDialog.Builder alertDiag = new AlertDialog.Builder();
alertDiag.SetTitle("Butikscanner App");
alertDiag.SetMessage("User Does not exist");
alertDiag.SetPositiveButton("OK",(senderAlert, args) => {  });
alertDiag.SetNegativeButton("Cancel", (senderAlert, args) => { alertDiag.Dispose(); });
Dialog diag = alertDiag.Create();
activity.RunOnUIThread(()=>
 {diag.Show();});
}

响应后如何应用警报对话框

1 个答案: