Question

我真的不知道为什么这段代码不起作用。每次我收到错误无法启动进程因为没有提供文件名。即使我提供了EXE所在的路径并对其进行了验证。

using System;
using System.Diagnostics;
using System.Diagnostics.Contracts;

namespace ProcessExitSample
{
    class testsandboxprogram
    {
        static void Main(string[] args)
        {
            Contract.Requires(args != null);
            try
            {
                var firstProc = new Process();
                Process.Start(@"PATH TO EXE I WANT TO LAUNCH");
                firstProc.EnableRaisingEvents = true;

                firstProc.Start();

                firstProc.WaitForExit();

                //so upon exit should run the second program here
                Console.WriteLine("First process exited: " + firstProc.ExitCode);

                var secondProc = new Process();
                Process.Start(@"PATH TO PROGRAM I WANT TO LAUNCH");
                secondProc.Start();

            }
            catch (Exception ex)
            {
                Console.WriteLine("Something went wrong sorry :(: " + ex.Message);
                return;
            }
        }
    }
}

Answer 1

StatisticItemDto

您不提供流程启动的路径。我注释掉了不相关的代码，因为它与var firstProc = new Process(); // Process.Start(@"PATH TO EXE I WANT TO LAUNCH"); firstProc.EnableRaisingEvents = true; firstProc.Start();变量无关。

你可能想要：

firstProc

Answer 2

String myexepath = @"C:\Program Files (x86)\Steam\steamapps\common\BattleBlock Theater\BattleBlockTheater.exe"

当此路径包含时，空格将其括在双引号之间：

Process.Start("\""+myexepath+"\"");

Answer 3

最明显的是尝试在命令行环境中运行PATH TO EXE I WANT TO LAUNCH，看看是否出现自我解释的错误。

如果您的路径包含空格，您将看到您正在尝试执行一些空格问题，然后您可以使用Graffito给出的答案。

C＃代码无法启动.exe获取错误无法找到文件名

3 个答案: