How do I launch application one from another in C#?

Question

I have two desktop applications. After closing the first application, the first application will start the second application.

How do I start the second application after finishing first application?

My first application creates a separate desktop.

Answer 1

Use the Process class when you are exiting your first application.

var p = new Process();
p.StartInfo.FileName   = "notepad.exe";  // just for example, you can use yours.
p.Start();

Answer 2

You can use .NET's Process Class to start a process as other people described. Then the question is when to call.

In most cases, using either Form.Closing or Form.Closed event seems to be an easy choice.

However, if someone else can handle the event and can set CancelEventArgs.Cancel to true, this may not be the right place to do this. Also, Form.Closing and Form.Closed events will not be raised when Application.Exit() is called. I am not sure whether either of events will be raised if any unhandled exceptions occur. (Also, you have to decide whether you want to launch the second application in case of Application.Exit() or any unhandled exception).

If you really want to make sure the second application (App2) launches after the first application (App1) exited, you can play a trick:

1. Create a separate application (App0)
2. App0 launches App1
3. App0 waits for App1 to exit with Process.WaitExit()
4. App0 launches App2 and exits itself

The sample console app attached below shows a very simple case: my sample app launches the notepad first. Then, when the notepad exits, it launches mspaint and exits itself.

If you want to hide the console, you can simply set the 'Output Type' property from 'Console Application' to 'Windows Application' under 'Application' tab of Project Property.

Sample code:

using System;
using System.Diagnostics;

namespace ProcessExitSample
{
    class Program
    {
        static void Main(string[] args)
        {
            try
            {

                Process firstProc = new Process();
                firstProc.StartInfo.FileName = "notepad.exe";
                firstProc.EnableRaisingEvents = true;

                firstProc.Start();

                firstProc.WaitForExit();

                //You may want to perform different actions depending on the exit code.
                Console.WriteLine("First process exited: " + firstProc.ExitCode);

                Process secondProc = new Process();
                secondProc.StartInfo.FileName = "mspaint.exe";
                secondProc.Start();                

            }
            catch (Exception ex)
            {
                Console.WriteLine("An error occurred!!!: " + ex.Message);
                return;
            }
        }
    }
}

Answer 3

您可以关闭它，所以当您即将退出第一个应用程序时，只需启动第二个应用程序：

System.Diagnostics.Process.Start(@"PATH\NAME.EXE");

Answer 4

使用.NET的Process类。

Answer 5

Some sample code:

try
{
  stateMainLayout b = new stateMainLayout();
 b.Location = Screen.AllScreens[1].WorkingArea.Location;
 b.ShowDialog();
 }
catch
{
 stateMainLayout b = new stateMainLayout();
b.ShowDialog();
}

Answer 6

CSharp / PowerShell调用另一个程序并发送/接收数据： https ：//huseyincakir.wordpress.com/2014/12/23/sending-input-from-csharppowershell-to-another-program/

Answer 7

in Some cases its necessary to add Working directory to your code in order to make the app works perfectly. especially when the app has dependency to DLL and other resources.

 TestProcess.StartInfo.FileName = "notepad.exe"; 
 TestProcess.StartInfo.WorkingDirectory = @"C:\\blah\blah\Directory of notepad.exe\";
 TestProcess.Start();

Answer 8

Here ProcName means the name of the application you want to start but it can only start system application and some other application

        public void Startapp(String ProcName)
        {
            try
            {
                Process firstProc = new Process();
                firstProc.StartInfo.FileName = ProcName;
                firstProc.EnableRaisingEvents = true;
                firstProc.Start();
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }