如何为进程提供参数

Question

嗨，我正在尝试创建一个用 c# 编写的 Windows 表单应用程序，它将启动、停止并监视它是否正在运行另一个控制台应用程序。

现在我可以在我的 Windows 应用程序上启动控制台应用程序，但是我如何为其 Console.ReadLine 提供值

这是我的代码：

windows 应用程序：

 int ProcessIDDaca = 111111111;
        private void btnDacaStart_Click(object sender, EventArgs e)
        {

            try
            {
                using (Process myprocess = new Process())
                {
                    myprocess.StartInfo.UseShellExecute = false;
                    myprocess.StartInfo.FileName = @"C:\Users\nx011116\Documents\MachineMonitor\CopyChimp_Server\bin\Debug\CopyChimpServer.exe";
                    //myprocess.StartInfo.FileName = @"C:\Program Files\Eltima Software\Virtual Serial Port Driver 9.0\vspdconfig.exe";
                    myprocess.StartInfo.CreateNoWindow = true;

                    if (btnDacaStart.Text == "Stop")
                    {
                        Process proc = Process.GetProcessById(ProcessIDDaca);
                        proc.Kill();
                        btnDacaStart.Text = "Start";
                        lblDacaID.Text = "";
                    }
                    else
                    {
                        myprocess.Start();
                        var s = myprocess.Id;
                        ProcessIDDaca = myprocess.Id;
                        lblDacaID.Text = s.ToString();
                        btnDacaStart.Text = "Stop";
                    }

                }
            }

            catch (Exception ex)
            {

            }

        }

控制台应用程序：

  public static string copychimp_server = "";
  public static int port_number = 0;

  static void Main(string[] args)
        {
            Console.Write("Enter copychimp server: ");
            //Here i need to provide
            copychimp_server = Console.ReadLine();
            Console.Write("Enter port number: ");
            //Here i need to provide
            port_number = Convert.ToInt16(Console.ReadLine());


        }

还有为什么没有显示，但我可以看到它正在我的任务管理器上运行？ 像编程初学者一样，我非常感谢您的帮助。 先感谢您

Answer 1

我已经解决了我的最新代码：

int ProcessIDDaca = 111111111;
        private void btnDacaStart_Click(object sender, EventArgs e)
        {

            try
            {
                using (Process myprocess = new Process())
                {
                    myprocess.StartInfo.UseShellExecute = false;
                    myprocess.StartInfo.FileName = @"C:\Users\nx011116\Documents\MachineMonitor\CopyChimp_Server\bin\Debug\CopyChimpServer.exe";
                    //myprocess.StartInfo.FileName = @"C:\Program Files\Eltima Software\Virtual Serial Port Driver 9.0\vspdconfig.exe";
                    myprocess.StartInfo.CreateNoWindow = true;

                    if (btnDacaStart.Text == "Stop")
                    {
                        Process proc = Process.GetProcessById(ProcessIDDaca);
                        proc.Kill();
                        btnDacaStart.Text = "Start";
                        lblDacaID.Text = "";
                    }
                    else
                    {
                        myprocess.StartInfo.Arguments = "test";
                        myprocess.Start();
                        var s = myprocess.Id;
                        ProcessIDDaca = myprocess.Id;
                        lblDacaID.Text = s.ToString();
                        btnDacaStart.Text = "Stop";
                    }

                }
            }

            catch (Exception ex)
            {

            }

        }

public static string copychimp_server = "";
  public static int port_number = 0;

  static void Main(string[] args)
        {
            Console.Write("Enter copychimp server: ");
            //Here i need to provide
            copychimp_server = args[0];
            Console.Write("Enter port number: ");
            //Here i need to provide
            port_number = Convert.ToInt16(Console.ReadLine());


        }

参考链接c#中如何将参数传递给另一个进程

Answer 2

为什么将 myprocess.StartInfo.CreateNoWindow 设置为 true？ 将其设置为 false 并重试。