异步显示对话框
本文关键字:对话框 显示 异步 | 更新日期: 2023-09-27 18:19:40
我使用async/await从数据库异步加载数据,在加载过程中,我想弹出一个加载表单,这只是一个简单的表单,带有运行进度条,指示有一个正在运行的进程。加载数据后,对话框将自动关闭。我怎样才能做到这一点?以下是我当前的代码:
protected async void LoadData()
{
ProgressForm _progress = new ProgressForm();
_progress.ShowDialog() // not working
var data = await GetData();
_progress.Close();
}
更新:
我设法通过更改代码使其工作:
protected async void LoadData()
{
ProgressForm _progress = new ProgressForm();
_progress.BeginInvoke(new System.Action(()=>_progress.ShowDialog()));
var data = await GetData();
_progress.Close();
}
这是正确的方法还是有更好的方法?
谢谢你的帮助。
使用Task.Yield
很容易实现,如下所示(WinForms,为了简单起见,没有异常处理)。重要的是要理解执行流如何跳到此处的新嵌套消息循环(模式对话框的消息循环),然后返回到原始消息循环(这就是await progressFormTask
的作用):
namespace WinFormsApp
{
internal static class DialogExt
{
public static async Task<DialogResult> ShowDialogAsync(this Form @this)
{
await Task.Yield();
if (@this.IsDisposed)
return DialogResult.Cancel;
return @this.ShowDialog();
}
}
public partial class MainForm : Form
{
public MainForm()
{
InitializeComponent();
}
async Task<int> LoadDataAsync()
{
await Task.Delay(2000);
return 42;
}
private async void button1_Click(object sender, EventArgs e)
{
var progressForm = new Form() {
Width = 300, Height = 100, Text = "Please wait... " };
object data;
var progressFormTask = progressForm.ShowDialogAsync();
try
{
data = await LoadDataAsync();
}
finally
{
progressForm.Close();
await progressFormTask;
}
// we got the data and the progress dialog is closed here
MessageBox.Show(data.ToString());
}
}
}
下面是一个使用Task.ContinueWith的模式,使用模态ProgressForm:应该可以避免任何竞争条件
protected async void LoadDataAsync()
{
var progressForm = new ProgressForm();
// 'await' long-running method by wrapping inside Task.Run
await Task.Run(new Action(() =>
{
// Display dialog modally
// Use BeginInvoke here to avoid blocking
// and illegal cross threading exception
this.BeginInvoke(new Action(() =>
{
progressForm.ShowDialog();
}));
// Begin long-running method here
LoadData();
})).ContinueWith(new Action<Task>(task =>
{
// Close modal dialog
// No need to use BeginInvoke here
// because ContinueWith was called with TaskScheduler.FromCurrentSynchronizationContext()
progressForm.Close();
}), TaskScheduler.FromCurrentSynchronizationContext());
}
ShowDialog()
是一个阻塞调用;在用户关闭对话框之前,执行不会前进到await
语句。请改用Show()
。不幸的是,您的对话框将不是模态的,但它将正确地跟踪异步操作的进度。
总是调用ShowDialog()
、LoadDataAsync
和Close()
,以避免像@noseration的答案中那样调用IsDisposed
。所以使用Task.Yield()
来延迟LoadDataAsync()
而不是ShowDialog()
public partial class Form1 : Form
{
public Form1()
{
InitializeComponent();
}
async Task<int> LoadDataAsync()
{
Console.WriteLine("Load");
await Task.Delay(2000);
return 42;
}
private async void button1_Click(object sender, EventArgs e)
{
var progressForm = new Form()
{
Width = 300,
Height = 100,
Text = "Please wait... "
};
async Task<int> work()
{
try
{
await Task.Yield();
return await LoadDataAsync();
}
finally
{
Console.WriteLine("Close");
progressForm.Close();
}
}
var task = work();
Console.WriteLine("ShowDialog");
progressForm.ShowDialog();
object data = await task;
// we got the data and the progress dialog is closed here
Console.WriteLine("MessageBox");
MessageBox.Show(data.ToString());
}
}
您可以尝试以下操作:
protected async void LoadData()
{
ProgressForm _progress = new ProgressForm();
var loadDataTask = GetData();
loadDataTask.ContinueWith(a =>
this.Invoke((MethodInvoker)delegate
{
_progress.Close();
}));
_progress.ShowDialog();
}