等待多个命令行进程完成

本文关键字：进程 命令行 等待 | 更新日期: 2023-09-27 18:03:59

我需要执行许多命令行脚本。它们当前存储在List中。我想同时运行它们，并在它们全部完成后仅在继续下一步。

我尝试了下面展示的方法，但发现它不够，因为最后一个命令不一定是end last。事实上，我发现最后一个命令甚至可以是第一个要完成的。因此，我认为我需要像WaitForExit()这样的东西，但它在所有执行进程完成之前不会返回。

for (int i = 0; i < commands.Count; i++)
{
    string strCmdText = commands[i];
    var process = System.Diagnostics.Process.Start("CMD.exe", strCmdText);
    if (i == (commands.Count - 1))
    {
        process.WaitForExit();
    }
}
//next course of action once all the above is done

等待多个命令行进程完成

由于每次调用Process.Start都会启动一个新进程，因此您可以分别跟踪它们，如下所示:

var processes = new List<Process>();
for (int i = 0; i < commands.Count; i++)
{
    string strCmdText = commands[i];
    processes.Add(System.Diagnostics.Process.Start("CMD.exe", strCmdText));
}
foreach(var process in processes)
{
    process.WaitForExit();
    process.Close();
}

编辑

Process.Close()添加到注释

使用一个任务数组并等待它们全部完成。

var tasks = new Task[commands.Count];
for (int i = 0; i < commands.Count; i++)
{
    tasks[i] = Task.Factory.StartNew(() => {
       string strCmdText = commands[i];
       var process = System.Diagnostics.Process.Start("CMD.exe", strCmdText);
       process.WaitForExit();
    });
}
Task.WaitAll(tasks);

或者，更像这样:

var tasks = commands.Select(strCmdText => Task.Factory.StartNew(() => {
    var process = System.Diagnostics.Process.Start("CMD.exe", strCmdText);
    process.WaitForExit();
})).ToArray();
Task.WaitAll(tasks);

至少在Windows上，您可以使用WaitHandle.WaitAll()。

using System;
using System.Diagnostics;
using System.Threading;
using Microsoft.Win32.SafeHandles;
using static System.FormattableString;
public class ProcessWaitHandle : WaitHandle
{
    public ProcessWaitHandle(Process process) =>
        this.SafeWaitHandle = new SafeWaitHandle(process.Handle, false);
}
class Program
{
    static void Main(string[] args)
    {
        int processesCount = 42;
        var processes = new Process[processesCount];
        var waitHandles = new WaitHandle[processesCount];
        try
        {
            for (int i = 0; processesCount > i; ++i)
            {
                // exit immediately with return code i
                Process process = Process.Start(
                    "cmd.exe",
                    Invariant($"/C '"exit {i}'""));
                processes[i] = process;
                waitHandles[i] = new ProcessWaitHandle(process);
            }
            WaitHandle.WaitAll(waitHandles);
            foreach (Process p in processes)
            {
                Console.Error.WriteLine(
                    Invariant($"process with Id {p.Id} exited with code {p.ExitCode}"));
            }
        }
        finally
        {
            foreach (Process p in processes)
            {
                p?.Dispose();
            }
            foreach (WaitHandle h in waitHandles)
            {
                h?.Dispose();
            }
        }
        Console.WriteLine("Press any key to continue...");
        Console.ReadKey(false);
    }
}

这种方法还提供了使用其他WaitAll重载和等待超时的可能性，例如

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