PythonPosted on 28th April 2025|3203 views
How to resolve the error called OSError [Errno 22] invalid argument when use open() in python?
Posted on 28th April 2025| views
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")