您将需要使用ProcessStartInfo并启用RedirectStandardOutput -然后您可以读取输出流。您可能会发现使用“>”将输出重定向到文件(通过操作系统)更容易，然后只需读取该文件。

[编辑:像Ray那样:+1]

这可能不是最好/最简单的方法，但可能是一种选择:

当从代码执行时，添加“> output.txt”，然后读入output.txt文件。

只是为了好玩，这里是我完整的获得PYTHON输出的解决方案-在单击按钮下-带有错误报告。只需添加一个名为“butPython”的按钮和一个名为“llHello”的标签…

    private void butPython(object sender, EventArgs e)
    {
        llHello.Text = "Calling Python...";
        this.Refresh();
        Tuple<String,String> python = GoPython(@"C:\Users\BLAH\Desktop\Code\Python\BLAH.py");
        llHello.Text = python.Item1; // Show result.
        if (python.Item2.Length > 0) MessageBox.Show("Sorry, there was an error:" + Environment.NewLine + python.Item2);
    }

    public Tuple<String,String> GoPython(string pythonFile, string moreArgs = "")
    {
        ProcessStartInfo PSI = new ProcessStartInfo();
        PSI.FileName = "py.exe";
        PSI.Arguments = string.Format("\"{0}\" {1}", pythonFile, moreArgs);
        PSI.CreateNoWindow = true;
        PSI.UseShellExecute = false;
        PSI.RedirectStandardError = true;
        PSI.RedirectStandardOutput = true;
        using (Process process = Process.Start(PSI))
            using (StreamReader reader = process.StandardOutput)
            {
                string stderr = process.StandardError.ReadToEnd(); // Error(s)!!
                string result = reader.ReadToEnd(); // What we want.
                return new Tuple<String,String> (result,stderr); 
            }
    }

在PublicDomain开源代码中有一个ProcessHelper类，你可能会感兴趣。

一行运行命令:

new Process() { StartInfo = new ProcessStartInfo("echo", "Hello, World") }.Start();

以最短的可读代码量读取命令输出:

    var cliProcess = new Process() {
        StartInfo = new ProcessStartInfo("echo", "Hello, World") {
            UseShellExecute = false,
            RedirectStandardOutput = true
        }
    };
    cliProcess.Start();
    string cliOut = cliProcess.StandardOutput.ReadToEnd();
    cliProcess.WaitForExit();
    cliProcess.Close();

The accepted answer on this page has a weakness that is troublesome in rare situations. There are two file handles which programs write to by convention, stdout, and stderr. If you just read a single file handle such as the answer from Ray, and the program you are starting writes enough output to stderr, it will fill up the output stderr buffer and block. Then your two processes are deadlocked. The buffer size may be 4K. This is extremely rare on short-lived programs, but if you have a long running program which repeatedly outputs to stderr, it will happen eventually. This is tricky to debug and track down.

有一些很好的方法来处理这个问题。

One way is to execute cmd.exe instead of your program and use the /c argument to cmd.exe to invoke your program along with the "2>&1" argument to cmd.exe to tell it to merge stdout and stderr. var p = new Process(); p.StartInfo.FileName = "cmd.exe"; p.StartInfo.Arguments = "/c mycmd.exe 2>&1"; Another way is to use a programming model which reads both handles at the same time. var p = new Process(); p.StartInfo.FileName = "cmd.exe"; p.StartInfo.Arguments = @"/c dir \windows"; p.StartInfo.CreateNoWindow = true; p.StartInfo.RedirectStandardError = true; p.StartInfo.RedirectStandardOutput = true; p.StartInfo.RedirectStandardInput = false; p.OutputDataReceived += (a, b) => Console.WriteLine(b.Data); p.ErrorDataReceived += (a, b) => Console.WriteLine(b.Data); p.Start(); p.BeginErrorReadLine(); p.BeginOutputReadLine(); p.WaitForExit();