在 WPF 和 C# 中构建 MainWindow() 和 InitializeComponent()

Question

This code is from my C# WPF application:

public MainWindow()
{
    InitializeComponent();
    status("Getting dl links");
    getLinks();
}

The procedure getLinks currently displays a couple of links in a messagebox. Those links are displayed in a messagebox before the WPF application becomes visible. In this is case not a problem, but how would I show progress (like a progressbar) of any procedure I want to load at startup?

Answer 1

Here is an example on how you can do it. To simplify it a bit, I added the controls directly in the MainWindow constructor, but I would prefer to do this with XAML.

public MainWindow()
{
    InitializeComponent();

    var progressBar = new ProgressBar();
    progressBar.Height = 40;
    progressBar.Width = 200;
    progressBar.Margin = new Thickness(100, 100, 100, 100);

    Task.Factory.StartNew(() =>
    {
        // getLinks();
        for (var i = 0; i < 5; i++)
        {
            Dispatcher.Invoke(new Action(() => { progressBar.Value += 20; }));

            Thread.Sleep(500);
        }
    });

    var stackPanel = new StackPanel();
    stackPanel.Children.Add(progressBar);

    Content = stackPanel;

}

I first add a ProgressBar somewhere on the interface to make it visible for this demo and then I add it to a new StackPanel , it could be any panel at all, in this case it doesn't matter.

To load the links on another thread, I create a new Task , this is a part of the TPL ( Task Parallel Library ) in .NET 4.0. In this case I am simulating that getLinks() takes 5 * 500 milliseconds to run and that it in fact is five links that will be loaded, hence 20% each iteration.

What I do then is that I add 20 to the progressBar value, which indicates that it increased with 20%.

This line might confuse you a bit

Dispatcher.Invoke(new Action(() => { progressBar.Value += 20; }));

But it is in fact quite common when you do cross-thread programming with GUI. So the problem is that you are on another thread here, we started of a Task that will run on a separate thread, and you cannot update your UI thread from another thread. So what you need is something called a Dispatcher , and this is accessable from within your Window -class.

Then you Invoke an action on it, which means that you simply say "Run this piece of code on this thread for me".

And if you want to display a MessageBox when everything is loaded, you can simply add a MessageBox.Show("Loaded;"); after the for -loop.

Answer 2

Any 'loading' tasks need to happen on a background thread (see the BackgroundWorker class - google for lots of examples). That way, the UI thread is free to show your window and update your window with status messages. Otherwise, the UI thread is blocked from doing anything until your loading is complete.