How to trigger a task each time an executable file is open with Windows Task Scheduler?

Question

I would like to trigger my own task each time an executable file is opened.

How do I do that with Task Scheduler ? Which event should I listen to ?

The only resources related to this I found are :

• https://serverfault.com/questions/277450/how-to-trigger-a-task-at-any-change-on-a-folder-with-windows-server-task-schedul
• http://blogs.technet.com/b/heyscriptingguy/archive/2005/04/04/how-can-i-monitor-for-different-types-of-events-with-just-one-script.aspx

Answer 1

As you can see here (under 'log files') you need to configure a group policy (either on local or domain level) that would make your system 'audit process tracking'.

After that, you'll need to attach a task to that event.

If you want to run an exe (or something else which is not supported Out-Of-the-Box), you'll need to edit the XML manually (to pass arguments to your target program) as demonstrated here.

Answer 2

For those that having an hard time to do what EliadTech said: To configure an audit processing track, you start Group Policy Editor (or Local Group Policy Editor), travel through Computer Configuration > Windows Settings > Local Policies > Audit Policies, RightClick on Audit process trackin and click on Properties and select Success.

Now, we need to attach a task to that: We start Event viewer, we go to Windows Logs > Security and we right click on one of the "Process created" event. Then you create your event. I haven't done the last part for now but you'll gain 30 mins