c# - 异步任务运行时如何更改进度条值

Question

我正在使用异步等待任务来运行这段代码，我想在提取时更改进度条

public async Task<string> DownloadAndExtractFile(string source, string destination, string ItemDownload) //source = File Location //destination = Restore Location
    {
        string zPath = @"C:\Program Files\7-Zip\7zG.exe";
        ProcessStartInfo pro = new ProcessStartInfo();
        pro.WindowStyle = ProcessWindowStyle.Hidden;
        pro.FileName = zPath;
        pro.Arguments = "x \"" + source + "\" -o" + destination;

        await Task.Run(() =>
        {
            Restore.frmRestore.progressBar1.Value = 50; //already set to public
            try
            {
                Process x = Process.Start(pro);
                Task.WaitAll();
                Restore.frmRestore.progressBar1.Value = 100;//already set to public
                x.Close();
                Console.WriteLine("Extract Successful.");
            }
            catch (Exception ex)
            {
                Console.WriteLine(ex.ToString());
            }

        }
           );

        return "Success";
    }

任务运行时如何改变进度条值。这是错误“跨线程操作无效:从创建它的线程以外的线程访问控件‘progressBar1’。”

Answer 1

Use the Progress<T> type to report progress ，正如我在我的博客中所描述的那样:

public async Task<string> DownloadAndExtractFile(string source, string destination, string ItemDownload)
{
  string zPath = @"C:\Program Files\7-Zip\7zG.exe";
  ProcessStartInfo pro = new ProcessStartInfo();
  pro.WindowStyle = ProcessWindowStyle.Hidden;
  pro.FileName = zPath;
  pro.Arguments = "x \"" + source + "\" -o" + destination;

  IProgress<int> progress = new Progress<int>(
      value => { Restore.frmRestore.progressBar1.Value = value; });

  await Task.Run(() =>
  {
    progress.Report(50);
    try
    {
      Process x = Process.Start(pro);
      Task.WaitAll();
      progress.Report(100);
      x.Close();
      Console.WriteLine("Extract Successful.");
    }
    catch (Exception ex)
    {
      Console.WriteLine(ex.ToString());
    }
  });
  return "Success";
}