后台工作者在WPF中无法正常工作 [英] Background worker does not work properly in WPF

问题描述

大家好...

我想在datagrid中显示临时文件，因此这是一个长期的过程，我在C#.net WPF应用程序中使用后台工作程序.

我的代码是

Hi Everyone...

i want to show Temporary files in a datagrid , hence it is a long term process i use background worker in my C# .net WPF application .

my Code is

private System.ComponentModel.BackgroundWorker _background = new System.ComponentModel.BackgroundWorker();

   private void button1_Click(object sender, RoutedEventArgs e)
        {
          _background.RunWorkerAsync();
        }

    public MainWindow()
       {
           InitializeComponent();
           this._background.DoWork += new DoWorkEventHandler(_background_DoWork);
           this._background.RunWorkerCompleted += new       
           RunWorkerCompletedEventHandler(_background_RunWorkerCompleted);
           this._background.WorkerReportsProgress = true;
           _background.WorkerSupportsCancellation = true;

       }

void _background_DoWork(object sender, DoWorkEventArgs e)
        {

this.Dispatcher.Invoke((Action)(() =>
    {
        try
        {
            FileInfo[] files = new     
            DirectoryInfo(System.IO.Path.GetTempPath()).GetFiles();
           
            foreach (FileInfo fi in files)
            {
                if (fi != null)              
                 {                 
                    dataGrid1.Items.Add(fi);           
                                         
                }
            }           
        }
        catch { }
    }));
}

void _background_RunWorkerCompleted(object sen, RunWorkerCompletedEventArgs e)
      {

          if (e.Cancelled)
          {
             MessageBox.Show("Cancelled");
          }
          else if (e.Error != null)
          {
               MessageBox.Show("Exception Thrown");
          }

     }

所有代码都在运行，但是在加载datagrid时挂起，这意味着程序运行时我的UI没有响应.

要在上述条件下平稳地运行后台工作者，需要进行哪些修改?

谢谢

All the code is running but it hangs when datagrid is loading means my UI does not response when program is running .

What modification is needed to run background worker smoothly in the above condition ?

Thank You

推荐答案

跟随代码，您完全破坏了在后台线程上进行工作的目的.

在后台线程的DoWork方法中，您告诉UI线程(this.Dispatcher.Invoke)执行长时间运行的代码，完全删除了后台线程的要点.

后台线程应该是调用将文件添加到datagrid的方法.

我相信您正在寻找这样的东西:

Following your code, you completely wrecked the purpose of doing work on a background thread.

In the DoWork method, on a background thread, you told the UI thread (this.Dispatcher.Invoke) to execute the long running code, completely removing the point of the background thread.

The background thread should be Invoking a method to add the file to the datagrid.

I believe you''re looking for something like this:

void _background_DoWork(object sender, DoWorkEventArgs e)
{
        FileInfo[] files = new     
        DirectoryInfo(System.IO.Path.GetTempPath()).GetFiles();
       
        foreach (FileInfo fi in files)
        {
            if (fi != null)              
            {                 
                this.Dispatcher.BeginInvoke((Action)(() => dataGrid1.Items.Add(fi)));
            }
        }
    }
}

您好，

您应该在后台工作器的主循环中添加一小段睡眠，以允许UI线程刷新显示.

尝试在您的foreach循环中添加类似的内容

Hello,

You should add to the background worker a short sleep in its main loop to allow the UI thread to refresh the display.

Try adding something like this to your foreach loop

// This sleep will allow the UI thread to refresh the display.
totalCounter++;
if (totalCounter % 100 == 0)
    System.Threading.Thread.Sleep(80);

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