How to Fix: OSError: [Errno 22] Invalid argument (File Paths)

If you are coding on Windows, you have likely encountered the frustrating OSError [Errno 22] error when trying to read or write files. This error almost exclusively affects Windows users and can stop a script dead in its tracks.

The error usually looks like this: OSError: [Errno 22] Invalid argument: 'C:\Users\Name\new_folder\data.txt'.

In this guide, we will look at why this happens and providing three different ways to fix the OSError [Errno 22] invalid argument issue permanently.

The Cause: The Backslash \

In Windows, file paths use backslashes: C:\Users\Name\Documents\file.txt. In Python strings, a backslash is a “escape character”. It’s used for special things like \n (new line) or \t (tab).

If your username starts with a ‘U’ or ‘N’ or ‘T’, Python might misinterpret your file path as containing special characters instead of just a folder name.

Problem Code:

# Python sees "\new_folder" as a NEW LINE character + ew_folder
f = open("C:\Users\Name\new_folder\data.txt", "r")
# CRASH! OSError: [Errno 22] Invalid argument

The Fixes (Pick One)

Fix 1: Use Raw Strings (r"")

Put an r before your string. This tells Python “ignore all special backslash meanings.”

f = open(r"C:\Users\Name\new_folder\data.txt", "r")

Fix 2: Use Double Backslashes \\

Escape the escape character.

f = open("C:\\Users\\Name\\new_folder\\data.txt", "r")

Fix 3: Use Forward Slashes / (Recommended)

Python on Windows is smart enough to understand forward slashes, just like Mac/Linux.

f = open("C:/Users/Name/new_folder/data.txt", "r")

This is the most cross-platform compatible way to write paths.

Conclusion

The OSError [Errno 22] is simply a misunderstanding between how Windows handles paths and how Python handles strings. By using raw strings, forward slashes, or the Pathlib module, you can ensure your file paths are read correctly every time.

If you are new to handling files, check out our guide on Working with JSON in Python for more tips.