How to execute an .EXE file in Python with a given filepath?

Question

I'm trying to write a bit of code to allow me to open up an exe file within Python, but I don't know the general way to make it properly. I would think that the code would look something like this:

exec(open("C:\\Users\\user\\AppData\\Local\\Programs\\file-folder\\file.exe").read())

but when I compile the code using Geany, it gives me an error saying UnicodeDecodeError: 'charmap' codec can't decode byte 0x90 in position 2: character maps to undefined

I get the same error message when I try to write the file path with just one \ between each folder, but it gives me the same error in the compiler. Any help on this is greatly appreciated.

Math chiller · Accepted Answer · 2020-05-14 22:30:42Z

2

you can use subprocess.call:

import subprocess
subprocess.call(["fullPath\\yourExe.exe"])

edited May 14, 2020 at 22:30

Math chiller

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answered May 14, 2020 at 22:10

Mit

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Comments

raven404 · Accepted Answer · 2020-05-14 22:10:43Z

1

well you need to import os in your code if you wanna execute external files

import os
os.startfile(""C:\\Users\\user\\AppData\\Local\\Programs\\file-folder\\file.exe"")

answered May 14, 2020 at 22:10

raven404

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1 Comment

Mit Over a year ago

This does the trick too, however, if the program produces any output then usually os.startfile is not recommended, unlike subprocess.call where it only continues when the output has been produced

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