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[Solved] OS [Errno 22] invalid argument when use open() in Python

Hello Guys, How are you all? Hope You all Are Fine. Today I get the following error OS [Errno 22] invalid argument when use open() in Python in python. So Here I am Explain to you all the possible solutions here.

Without wasting your time, Let’s start This Article to Solve This Error.

How OS [Errno 22] invalid argument when use open() in Python Error Occurs?

Today I get the following error OS [Errno 22] invalid argument when use open() in Python in python.

How To Solve OS [Errno 22] invalid argument when use open() in Python Error ?

  1. How To Solve OS [Errno 22] invalid argument when use open() in Python Error ?

    To Solve OS [Errno 22] invalid argument when use open() in Python Error I also ran into this fault when I used open(file_path). My reason for this fault was that my file_path had a special character like "?" or "<".

  2. OS [Errno 22] invalid argument when use open() in Python

    To Solve OS [Errno 22] invalid argument when use open() in Python Error I also ran into this fault when I used open(file_path). My reason for this fault was that my file_path had a special character like "?" or "<".

Solution 1

That is not a valid file path. You must either use a full path

open(r"C:\description_files\program_description.txt","r")

Or a relative path

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

Solution 2

I also ran into this fault when I used open(file_path). My reason for this fault was that my file_path had a special character like "?" or "<".

Summery

It’s all About this issue. Hope all solution helped you a lot. Comment below Your thoughts and your queries. Also, Comment below which solution worked for you? Thank You.

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