处理后台工作程序错误

Question

在执行耗时的python脚本时,我会用后台工作者管理IU以显示进度条.

当我不需要事件时OutputDataReceived,我已成功使用后台工作程序,但我正在使用的脚本打印了一些进度值,如("10","80",..),所以我必须听取事件OutputDataReceived.

我得到这个错误:This operation has already had OperationCompleted called on it and further calls are illegal.在这一行progress.bw.ReportProgress(v);.

我试图使用2个后台工作器实例,一个执行而另一个监听,它没有给出任何错误,但它似乎没有调用事件'OutputDataReceived'所以我没有在进度条中看到任何进展.

在我使用的代码下面:

    private void execute_script()
    {
             progress.bw.DoWork += new DoWorkEventHandler( //progress.bw is reference to the background worker instance
        delegate(object o, DoWorkEventArgs args)
        {

        System.Diagnostics.Process proc = new System.Diagnostics.Process();
        proc.StartInfo.FileName = "python.exe";
        proc.StartInfo.UseShellExecute = false;
        proc.StartInfo.Arguments = @".\scripts\script1.py " + file_path + " " + txtscale.Text;
        //proc.StartInfo.CreateNoWindow = true;
        //proc.StartInfo.WindowStyle = System.Diagnostics.ProcessWindowStyle.Hidden;
        proc.StartInfo.RedirectStandardOutput = true;
        //proc.EnableRaisingEvents = true;
        proc.StartInfo.RedirectStandardError = true;
        proc.StartInfo.RedirectStandardError = true; 
        proc.OutputDataReceived += new System.Diagnostics.DataReceivedEventHandler(proc_OutputDataReceived);
        proc.Start();
        proc.BeginOutputReadLine();

      //proc.WaitForExit();
        //proc.Close();
                   });

           progress.bw.RunWorkerAsync();
        }

 ///the function called in the event OutputDataReceived 
 void proc_OutputDataReceived(object sender, System.Diagnostics.DataReceivedEventArgs e)
    {
        //throw new NotImplementedException();
        if (e.Data != null)
        {
            int v = Convert.ToInt32(e.Data.ToString()); 
            MessageBox.Show(v.ToString());
         //   report(v);
            progress.bw.ReportProgress(v);

        }
        else
            MessageBox.Show("null received"); 


    }

Answer 1

问题是,BackgroundWorker的DoWork处理程序一旦启动过程结束,因为没有什么'等待’(因为你注释掉proc.WaitForExit()的过程来完成).一旦BackgroundWorker工作处理程序完成后,您可以使用该实例不再报告进度.

由于Process.Start已经是异步的,因此根本没有理由使用后台工作程序.您可以OutputDataReceived自己将调用编组到UI线程上:

///the function called in the event OutputDataReceived 
void proc_OutputDataReceived(object sender, System.Diagnostics.DataReceivedEventArgs e)
{
    //throw new NotImplementedException();
    if (e.Data != null)
    {
        int v = Convert.ToInt32(e.Data.ToString()); 
        // MessageBox.Show(v.ToString());
        // progress.bw.ReportProgress(v);
        this.BeginInvoke( new Action( () => {
             this.progressBar.Value = v;
        }));
    }
}

如果您使用此功能,请不要创建它BackgroundWorker.