相关疑难解决方法(0)

我的项目有问题.我想启动一个进程,7z.exe(控制台版).我尝试过三种不同的东西:

• Process.StandardOutput.ReadToEnd();
• OutputDataReceived和BeginOutputReadLine
• 的StreamWriter

什么都行不通.它始终"等待"过程结束以显示我想要的内容.我没有任何代码可以放,只要你想要我的代码与其中一个列出的东西.谢谢.

编辑:我的代码:

        process.StartInfo.UseShellExecute = false;
        process.StartInfo.RedirectStandardOutput = true;
        process.StartInfo.CreateNoWindow = true;
        process.Start();

        this.sr = process.StandardOutput;
        while (!sr.EndOfStream)
        {
            String s = sr.ReadLine();
            if (s != "")
            {
                System.Console.WriteLine(DateTime.Now + " - " + s);
            }
        }

要么

process.StartInfo.UseShellExecute = false;
process.StartInfo.RedirectStandardOutput = true;
process.OutputDataReceived += new DataReceivedEventHandler(recieve);
process.StartInfo.CreateNoWindow = true;
process.Start();
process.BeginOutputReadLine();
process.WaitForExit();
public void recieve(object e, DataReceivedEventArgs outLine)
{
    System.Console.WriteLine(DateTime.Now + " - " + outLine.Data);
}

要么

process.StartInfo.UseShellExecute = false;
process.StartInfo.RedirectStandardOutput = true;
process.Start(); …