I am trying to start a process using os.startfile
. But when I run it, I get the following error!
---------------------------------------------------------------------------
FileNotFoundError Traceback (most recent call last)
<ipython-input-4-3c26e36152b3> in <module>
1 import os
----> 2 os.startfile(r"%ProgramData%\Microsoft\DeviceSync\app.exe")
FileNotFoundError: [WinError 2] The system cannot find the file specified: '%ProgramData%\\Microsoft\\DeviceSync\\app.exe'
But if I use the full path
import os
os.startfile(r"C:\ProgramData\Microsoft\DeviceSync\app.exe")
It runs fine.
What am I doing wrong? :(
For Windows
So looking at your problem what you can do is actuall create a short of the file you want to start
For example, create a shortcut like. Right-click your EXE file and from the drop-down menu select the Create Shortcut option.
App.exe
App.exe.lnk
Then should Try
import os
x=os.path.expandvars(r"%ProgramData%\Microsoft\DeviceSync\app.exe.lnk")
os.startfile(x)
It should Work Normally and give you the output