我需要从dll项目打开一个窗口(在后台运行或在调用时触发)。为此,我需要从标记为[STAThread]的方法打开一个窗口来创建窗口的实例。
我使用MVVM模式将ViewModel绑定到View,并将视图作为usercontrol添加到MainWindow。
在视图上有一个按钮绑定到一个关闭'关闭的命令。窗户。 ViewModel中的命令调用CloseCommand(),然后调用window.Close()方法。
代码段:
MethodShowingTheWindow() {
var idVM = new IDWindowViewModel();
ShowForm<MainWindow>(idVM); // Works fine, Close Command closes the window.
ShowForm<MainWindow>(idVM); // hits the ShowDialog() line, then just jumps to the while loop
ShowForm<MainWindow>(idVM); // waiting for the thread to finish. Never displays the window.
ShowForm<MainWindow>(idVM);
}
[STAThread]
public void ShowForm<T>(IViewModel vm) {
Thread th = new Thread(new ThreadStart(delegate {
var window = new MainWindow();
window.DataContext = vm;
vm.CloseAction = new Action(() => window.Close()); // Does this when the Close Action is called from the viewmodel (Close the window).
try {
window.ShowDialog();
} catch (Exception ex) {
throw ex;
}
}));
th.ApartmentState = ApartmentState.STA;
th.Start();
while (th.IsAlive) {
//Wait for thread to finish
}
}
我需要知道的是为什么窗口不会出现在第二个ShowWindow(idVM)调用中?我能想到的是窗口未正确关闭/处置。我尝试将以下内容添加到CloseAction操作:
...
vm.CloseAction = new Action(() => {
if (window is IDisposable) (window as IDisposable).Dispose();
window.Close();
});
但是(窗口是IDisposable)== false?
非常感谢任何帮助。
答案 0 :(得分:0)
这是我针对您的方案的实现。使用Dispatcher以避免STA例外:
async Task MethodShowingTheWindow()
{
var idVM = new IDWindowViewModel();
await ShowForm<MainWindow>(idVM); // Works fine, Close Command closes the window.
await ShowForm<MainWindow>(idVM); // hits the ShowDialog() line, then just jumps to the while loop
await ShowForm<MainWindow>(idVM); // waiting for the thread to finish. Never displays the window.
await ShowForm<MainWindow>(idVM);
}
public await Task ShowForm<T>(IViewModel vm)
{
Dispatcher.InvokeAsync(() =>
{
var window = new MainWindow();
window.DataContext = vm;
System.Windows.Application.Current.Dispatcher.Invoke(() =>
{
vm.CloseAction = new Action(() => window.Close());
});
window.ShowDialog();
});
}