close session in winform app after 5 seconds(idle) if no movement

Question

I am writing a c# Winform application and I need to close session if the computer is idle for 5 seconds. The application is like a restaurant application, when the waiter leaves his session open, I will close it after 5 seconds.

I found some code but I dont know how to use it and how to trigger it

using System.Runtime.InteropServices;

[DllImport("User32.dll")]
private static extern bool GetLastInputInfo(ref LASTINPUTINFO plii);   

internal struct LASTINPUTINFO
{
    public uint cbSize;

    public uint dwTime;
}

Can anybody help me with this?

I saw that link but I still dont know how to check every 5 sec — Arif YILMAZ, Mar 04 '15 at 11:05
Run a [timer](https://msdn.microsoft.com/en-us/library/system.windows.forms.timer%28v=vs.110%29.aspx) with 5-seconds interval and do your check in `Tick` event handler. — Andrey Korneyev, Mar 04 '15 at 11:07

Aria · Accepted Answer · 2019-05-19T12:08:34.963

Follow these steps:

1- Add a Timer to your Form.

2- Set its interval property to 1000 (set it in form_load or in design mode from Properties window).

3-Add this method to your Form class.

public static uint GetIdleTime()
{
     LASTINPUTINFO LastUserAction = new LASTINPUTINFO();
     LastUserAction.cbSize = (uint)System.Runtime.InteropServices.Marshal.SizeOf(LastUserAction);
     GetLastInputInfo(ref LastUserAction);
     return ((uint)Environment.TickCount - LastUserAction.dwTime);
}

4- in Form_Load start the timer:

timer1.Start();

5- in timer tick event check GetIdleTime(), for example if it is greater than 5000 means application was idled since 5 seconds ago.

private void timer1_Tick(object sender, EventArgs e)
{
    if (GetIdleTime() > 5000)  
       Application.Exit();//For Example
}

close session in winform app after 5 seconds(idle) if no movement

1 Answers1