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Python
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 24th April 2024
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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