How to open windows application from ignition?

Good afternoon,

What would be the script if I wanted to open a windows program (for example skipe).

I have to link this action to a button on the scada.

Thank you

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Take a look at system.util.execute.


You need to be careful as system.util.execute will launch new instances of the application each time.
We found this after having over 100 calculators open in the background :open_mouth::sweat_smile:

So we now issue a taskkill before the launch.

import time


What if it is not a default windows app like for example -

C:\Program Files\uvnc bvba\UltraVNC.

Just pass it the full path to the exe

This should open each file or application exactly how the OS would.

import subprocess

subprocess.Popen(["C:\\Shared Gateway\\IgnitionTagHistory.pdf"],shell=True)
subprocess.Popen(["C:\\Program Files (x86)\\Google\\\Chrome\\Application\\chrome.exe"],shell=True)
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thanks for your answer

I didn’t notice the double backslash at first. That should have been in the manual.

Thank you samngeru. This is exactly what I needed.

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No problem.