生成具有不同输出的控制台子进程

发布于 2024-09-19 16:06:06 字数 287 浏览 5 评论 0原文

我有一个用 C# 编写的控制台应用程序，它启动一个新进程。这个新进程也是一个控制台应用程序。问题是整个子进程输出都转到父控制台窗口，但我不希望它这样做。它是否创建一个新的控制台窗口并不重要，我不需要它。

编辑：

Process p = new Process(); 
p.StartInfo.FileName = @"C:\example.exe";
p.Start();

我尝试了一些额外的设置，如 CreateNoWindow、RedirectOutput 等，但没有运气

原文

I have a console application written in c# which starts a new process. This new process is also a console application. The problem is that whole child process output goes to parent console window and i don't want it to. It doesn't matter if it creates a new console window or not, I don't need it.

Edit:

Process p = new Process(); 
p.StartInfo.FileName = @"C:\example.exe";
p.Start();

I tried some additional setting like CreateNoWindow, RedirectOutput and so on but with no luck

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鲜肉鲜肉永远不皱 2024-09-26 16:06:07

通常，当您创建子进程时，会弹出一个新的控制台窗口，并且子进程的所有输出最终都在该窗口中。以下是展示此行为的示例程序：

using System;
using System.Diagnostics;
using System.IO;
using System.Reflection;

static class Program
{
    static void Main(string[] args)
    {
        Console.WriteLine("TEST");

        if (args.Length == 0)
        {
            string path = Assembly.GetExecutingAssembly().Location;
            Process.Start(path, "DontStartProcess");
        }
        Console.ReadLine();
    }
}

Normally, when you create a child process, a new console window pops up and all output of the child process ends up in that window. Here is a sample program that exhibits this behavior:

using System;
using System.Diagnostics;
using System.IO;
using System.Reflection;

static class Program
{
    static void Main(string[] args)
    {
        Console.WriteLine("TEST");

        if (args.Length == 0)
        {
            string path = Assembly.GetExecutingAssembly().Location;
            Process.Start(path, "DontStartProcess");
        }
        Console.ReadLine();
    }
}

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