Passing command line argument with whitespace in Python

Question

I am trying to pass the command line argument with white space in it, but sys.argv[1].strip() gives me only first word of the argument

import sys, os
docname = sys.argv[1].strip()

 e.g. $ python myscript.py argument with whitespace

If I try to debug - docname gives me output as argument instead of argument with whitespace

I tried to replace the white space with .replace(" ","%20") method but that didn't help

Answer 1

You need to use argv[1:] instead of argv[1]:

docname = sys.argv[1:]

To print it as a string:

' '.join(sys.argv[1:])  # Output: argument with whitespace

sys.argv[0] is the name of the script itself, and sys.argv[1:] is a list of all arguments passed to your script.

Output:

>>> python myscript.py argument with whitespace
['argument', 'with', 'whitespace']

Answer 2

This has nothing to do with Python and everything to do with the shell. The shell has a feature called wordsplitting that makes each word in your command invocation a separate word, or arg. To pass the result to Python as a single word with spaces in it, you must either escape the spaces, or use quotes.

./myscript.py 'argument with whitespace'
./myscript.py argument\ with\ whitespace

In other words, by the time your arguments get to Python, wordsplitting has already been done, the unescaped whitespace has been eliminated and sys.argv is (basically) a list of words.