Programming the same button to open and close an EXE file

Question

Im trying to program one single button to open and close an exe file, so when I press it once, it opens the file, when I press it again then it closes the file.

Im able to program the button to open the file but i cannot get it to close.

private void button1_click(object sender, EventArgs e)
{

Process proc = new Process();
proc.StartInfo.FileName = @"C:\Program Files (x86)\TeamViewer\TeamViwer.exe";
proc.Start();
}

Possible duplicate of [Kill some processes by .exe file name](https://stackoverflow.com/questions/3345363/kill-some-processes-by-exe-file-name) — Bùi Đức Khánh, Jun 06 '19 at 03:34

score 1 · Accepted Answer · answered Jun 06 '19 at 03:16

You can set up a boolean flag, whenever you open your program you set that flag to true, whenever you close it, you set it to false. If its true then it tries to kill the process, the other way around it just executes the code you got there.

bool running = false;
private void button1_click(object sender, EventArgs e)
{
    if (running)
    {
        var processes = Process.GetProcesses();

        foreach (Process pr in processes)
        {

            if (pr.ProcessName == "TeamViwer")
            {
                pr.Kill();
            }

        }
       running = false;
    }
    else
    {
        Process proc = new Process();
        proc.StartInfo.FileName = @"C:\Program Files (x86)\TeamViewer\TeamViwer.exe";
        proc.Start();
        running = true;
    }
}

or just check if its running – TheGeneral Jun 06 '19 at 03:21 — TheGeneral, Jun 06 '19 at 03:21
Totally there are numerous ways to go about this – nalnpir Jun 06 '19 at 03:25 — nalnpir, Jun 06 '19 at 03:25

Programming the same button to open and close an EXE file

1 Answers1