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

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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 :

JBE

Posted 2014-11-22T15:23:55.570

Reputation: 224

Answers

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.

EliadTech

Posted 2014-11-22T15:23:55.570

Reputation: 2 076

0

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

taka

Posted 2014-11-22T15:23:55.570

Reputation: 1