C# 在运行时获取进程输出

Question

是否有重定向生成进程的标准输出并在其发生时捕获它的方法。我所看到的一切都在该过程完成后执行 ReadToEnd。我希望能够在打印时获得输出。

编辑:

    private void ConvertToMPEG()
    {
        // Start the child process.
        Process p = new Process();
        // Redirect the output stream of the child process.
        p.StartInfo.UseShellExecute = false;
        p.StartInfo.RedirectStandardOutput = true;
        //Setup filename and arguments
        p.StartInfo.Arguments = String.Format("-y -i \"{0}\" -target ntsc-dvd -sameq -s 720x480 \"{1}\"", tempDir + "out.avi", tempDir + "out.mpg");
        p.StartInfo.FileName = "ffmpeg.exe";
        //Handle data received
        p.OutputDataReceived += new DataReceivedEventHandler(p_OutputDataReceived);
        p.Start();
    }

    void p_OutputDataReceived(object sender, DataReceivedEventArgs e)
    {
        Debug.WriteLine(e.Data);
    }

Answer 1

使用Process.OutputDataReceived来自流程的事件，以接收您需要的数据。

示例:

var myProc= new Process();

...            
myProc.StartInfo.RedirectStandardOutput = true;
myProc.OutputDataReceived += new DataReceivedEventHandler(MyProcOutputHandler);

...

private static void MyProcOutputHandler(object sendingProcess, 
            DataReceivedEventArgs outLine)
{
            // Collect the sort command output. 
    if (!String.IsNullOrEmpty(outLine.Data))
    {
      ....    
    }
}