C# Process.StandardOutput InvalidOperationException“无法在流程流上混合同步和异步操作。”

Question

我尝试了这个

        myProcess = new Process();

        myProcess.StartInfo.CreateNoWindow = true;
        myProcess.StartInfo.WindowStyle = ProcessWindowStyle.Hidden;

        myProcess.StartInfo.FileName = "Hello.exe";

        myProcess.StartInfo.Arguments ="-say Hello";
        myProcess.StartInfo.UseShellExecute = false;  

        myProcess.OutputDataReceived += new DataReceivedEventHandler(myProcess_OutputDataReceived);
        myProcess.ErrorDataReceived += new DataReceivedEventHandler(myProcess_OutputDataReceived);
        myProcess.Exited += new EventHandler(myProcess_Exited);
        myProcess.EnableRaisingEvents = true;

        myProcess.StartInfo.RedirectStandardOutput = true;
        myProcess.StartInfo.RedirectStandardError = true;
        myProcess.StartInfo.ErrorDialog = true;
        myProcess.StartInfo.WorkingDirectory = "D:\\Program Files\\Hello";

        myProcess.Start();

        myProcess.BeginOutputReadLine();
        myProcess.BeginErrorReadLine();

然后我收到这个错误.. 替代文本 http://img188.imageshack.us/img188/3759/errorstack.jpg< /a>

我的过程需要很长时间才能完成，因此我需要在运行时显示进度。

Answer 1

您不需要调用 ReadLine()，读取的文本行是在 DataReceivedEventArgs 对象中传递给您的属性之一。