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PythonPython

H
Harish Harsh

Posted on 28th November 2022|1492 views

0
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Oserror Errno 22 Invalid Argument

How to resolve the error called OSError [Errno 22] invalid argument when use open() in python?

Answers
P
James ROBERT

Posted on 28th November 2022

Check the pat address you must give the full path address like this

open(r"C:\doc_files\program_doc.txt", "r")

Or give the relative path

open("program_doc.txt","r")

 

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