Question

我正在尝试启动一个程序并读取它的标准输出。然而事件永远不会提出。我开始的过程正在运行，并在控制台中使用相同的参数创建输出。任何想法，我做错了什么？

    public void StartProcess(string Filename, string Arguments)
    {
        currentProcess = new Process();
        currentProcess.StartInfo.FileName = Programm;
        currentProcess.StartInfo.Arguments = Arguments;
        currentProcess.StartInfo.UseShellExecute = false;
        currentProcess.StartInfo.RedirectStandardOutput = true;
        currentProcess.StartInfo.RedirectStandardError = true;
        currentProcess.OutputDataReceived += OutputReceivedEvent;
        currentProcess.EnableRaisingEvents = true;

        string path = DateTime.Now.ToString("yyyy-MM-dd-HH-mm-ss") + "_Result.txt";
        LastResult = path;
        resultfile = File.CreateText(path);

        currentProcess.Start();
        currentProcess.BeginOutputReadLine();
        response.command = Command.ACK;
        SendMessage(response);

    }
    private void OutputReceivedEvent(object sender, DataReceivedEventArgs e)
    {
        resultfile.WriteLine(e.Data);
    }

编辑：我发现了一些奇怪的东西：我开始的过程是mcast。如果我开始像ping这样的东西，我的代码就可以了。所以mcast似乎做了一些有趣的事情！

EDIT2：所以下面的代码工作正常，但只读取一定大小的块，如果向流写入的字节数更少，则偶数不会发生，.ReadBlock也不返回任何内容。

EDIT3：再一次更新，问题是，mcast没有刷新它的输出流。我最终编写了自己的工具，工作得很好。

Answer 1

根据流程的不同，您可能还需要使用currentProcess.ErrorDataReceived，然后再使用currentProcess.BeginErrorReadLine()，因为某些流程只是重定向到一个输出，而不是两者都重定向。无论如何都值得尝试。

所以它看起来像这样：

public void StartProcess(string Filename, string Arguments)
{
    currentProcess = new Process();
    currentProcess.StartInfo.FileName = Programm;
    currentProcess.StartInfo.Arguments = Arguments;
    currentProcess.StartInfo.UseShellExecute = false;
    currentProcess.StartInfo.RedirectStandardOutput = true;
    currentProcess.StartInfo.RedirectStandardError = true;
    currentProcess.OutputDataReceived += OutputReceivedEvent;
    currentProcess.ErrorDataReceived += ErrorReceivedEvent; 

    currentProcess.EnableRaisingEvents = true;

    string path = DateTime.Now.ToString("yyyy-MM-dd-HH-mm-ss") + "_Result.txt";
    LastResult = path;
    resultfile = File.CreateText(path);

    currentProcess.Start();
    currentProcess.BeginOutputReadLine();
    currentProcess.BeginErrorReadLine();
    response.command = Command.ACK;
    SendMessage(response);

}
private void OutputReceivedEvent(object sender, DataReceivedEventArgs e)
{
    resultfile.WriteLine(e.Data);
}

/*This second event handler is to prevent a lock occuring from reading two separate streams. 
Not always an issue, but good to protect yourself either way. */
private void ErrorReceivedEvent(object sender, DataReceivedEventArgs e)
{
    resultfile.WriteLine(e.Data);
}

重定向标准输出，不会引发事件

1 个答案: