c# - 如何使用 Control.Dispatcher.BeginInvoke 修改 GUI

Question

我需要从一个需要很长时间才能完成的方法内部修改 GUI。当我阅读其他帖子时， 解决方案之一是使用 Control.Dispatcher.BeginInvoke 在工作线程内设置 GUI。 但是，我不知道如何在这里执行此操作。

public partial class MainForm : Form
{
    public MainForm()
    {
        InitializeComponent();
    }

    private void button1_Click(object sender, EventArgs e)
    {
        Task.Factory.StartNew( () =>
        {
            ProcessFilesThree();
        });
    }

    private void ProcessFilesThree()
    {
        string[] files = Directory.GetFiles(@"C:\temp\In", "*.jpg", SearchOption.AllDirectories);

        Parallel.ForEach(files, (currentFile) =>
        {
            string filename = Path.GetFileName(currentFile);

                    // the following assignment is illegal
            this.Text = string.Format("Processing {0} on thread {1}", filename,
                                        Thread.CurrentThread.ManagedThreadId); 
        });

        this.Text = "All done!"; // <- this assignment is illegal
    }
}

Answer 1

尝试以下操作:

 msg = string.Format("Processing {0} on thread {1}", filename,
            Thread.CurrentThread.ManagedThreadId);
 this.BeginInvoke( (Action) delegate ()
    {
        this.Text = msg;
    });