调用时 UI 冻结。在新窗口上显示

Question

在我的应用程序中，我从 LoginWindow 打开 MainWindow。当我调用 MainWindow.Show() 时，LoginWindow UI 冻结。我试过了：

• await Task.Run() 但这不起作用，因为我需要在 UI 线程上打开窗口

• 等待 Application.Current.Dispatcher.BeginInvoke 但这似乎不能异步工作

我一定误解了异步编程的基本原理。

public async void LoginAsync(object pbx)
    {
        VisLoginLoading = Visibility.Visible;
        var passwordBox = pbx as PasswordBox;
        string enteredPassword = passwordBox.Password;

        string cmdQuery = $"SELECT USERNAME, PASSWORD, AccessLevel, ALERTS, STORES FROM Tbl_UserAccounts WHERE USERNAME = @Username";

        try
        {
            string errorMsg = await Task.Run(() =>
            {
                using (SqlConnection con = new SqlConnection(PosConString))
                using (SqlCommand cmd = new SqlCommand(cmdQuery, con))
                {
                    cmd.Parameters.AddWithValue("@Username", SqlDbType.NVarChar).Value = Username;
                    cmd.CommandTimeout = 30;
                    con.Open();
                    SqlDataReader reader = cmd.ExecuteReader();

                    if (reader.HasRows)
                    {
                        while (reader.Read())
                        {
                            string password = reader.GetString(1);

                            if (DoesPasswordMatch(password, enteredPassword))
                            {
                                UserAccount.Username = (reader[0] == DBNull.Value) ? null : reader.GetString(0);
                                UserAccount.Access = Enum.IsDefined(typeof(UserAccess), reader.GetInt32(2)) ? (UserAccess)reader.GetInt32(2) : UserAccess.Unknown;
                                UserAccount.Alerts = (reader[3] == DBNull.Value) ? false : reader.GetBoolean(3);
                                UserAccount.Stores = (reader[4] == DBNull.Value) ? null : HelperMethods.ReturnWordList(reader.GetString(4));
                                return String.Empty;
                            }
                            else
                            {
                                return "Password is incorrect";
                            }
                        }
                    }
                    else
                    {
                        return "User does not exist";
                    }
                }

                return "Failed to login";
            });

            if (String.IsNullOrEmpty(errorMsg))
            {
                LoadApp();
            }
            else
            {
                MessageBox.Show(errorMsg);
            }
        }
        catch (Exception ex)
        {
            MessageBox.Show($"Failed to login\n\n{ex}");
        }

        VisLoginLoading = Visibility.Hidden;
    }

    public async void LoadApp()
    {
        VisLoginLoading = Visibility.Visible;

        await Application.Current.Dispatcher.BeginInvoke(new Action(() =>
            {
                MainWindow mainWindow = new MainWindow();
                mainWindow.Show();
            }));

        foreach (Window window in Application.Current.Windows)
        {
            if (window.Title == "Login") window.Close();
        }
        VisLoginLoading = Visibility.Hidden;
    }

Answer 1

从逻辑上讲，GUI 线程会冻结。
看，您使用Application.Current.Dispatcher.BeginInvoke 在 GUI 线程上启动了一个长时间运行的操作。所以这个线程的所有其他“用户”必须等到它可用，才能处理他们的（LoginWindow）操作。在这种情况下你能做什么？

您可以启动另一个 GUI 线程，该线程将处理新窗口：

public void LoadApp()
{
    VisLoginLoading = Visibility.Visible;

    var thr = new Thread(() =>
    {
        var mw = new MainWindow();
        mw.Show();
        mw.Closed += (sender, e) => mw.Dispatcher.InvokeShutdown();
        System.Windows.Threading.Dispatcher.Run();
    });
    thr.SetApartmentState(ApartmentState.STA);
    thr.Start();

    //Move this logic to the MainWindow, after initialization is done.
    //foreach (Window window in Application.Current.Windows)
    //{
    //    if (window.Title == "Login") window.Close();
    //}
    VisLoginLoading = Visibility.Hidden;
}

现在您可以启动MainWindow，但您必须解决另一个挑战（我认为）- 通知VisLoginLoading 初始化完成。但这是另一个问题。所以你的用户界面不会冻结。
另见Threading Model.

更新：

在主窗口中使用：

Application.Current.Dispatcher.BeginInvoke((Action)(()=>
{
    foreach (Window window in Application.Current.Windows)
    {
        if (window.Title == "Login") window.Close();
    }
}));