Open files, specified by a path provided via sys.argv in Python on Windows?

Question

Consider the program below, which purpose is to open some files, referenced by sys.argv[1] + "string".

import sys

def searchAndReplacePath(path):

    filesToSearch = [path + "\main\file1.txt",
                     path + "\main\folder\file2.txt"]

    for files in filesToSearch:
        with open(files, 'r') as inFile:
            filedata = inFile.readlines()

if __name__ == "__main__":
    # sys.argv[1] -- Specify which path should be used
    searchAndReplacePath(sys.argv[1])

At execution (Python 3.6, Windows 7) I receive an error:

Traceback (most recent call last):
  File "searchAndReplacePath.py", line 35, in <module>
    searchAndReplacePath(sys.argv[1])
  File "searchAndReplacePath.py", line 19, in searchAndReplacePath
    with open(file, 'r') as inFile:
FileNotFoundError: [Errno 2] No such file or directory: 'C:\\Users\\user\\main\\file1.txt'

Python adds backslashes to my backslashes! If I iterate the list and print() each item in said list, I get "single backslashes", so I don't really know what to do. I have read a lot about raw strings and how to use: r"path\to\file", but since I take my input from sys.argv[] I don't really now how to achieve the same result.

How do I open files, specified by a path provided via sys.argv[] in Python on Windows?

Answer 1

Try using python's os module: import os files_to_search = [os.path.join(path, 'main\file1.txt') ... ]

Answer 2

The issue here is seen in the following interactive session:

>>> "\f"
'\x0c'
>>> '\\f'
'\\f'
>>> r'\\f'
'\\\\f'
>>> '\f'
'\x0c'
>>> r'\f'
'\\f'
>>>

As you can see, when the interpreter prints the repr of a string, it quotes the backslashes to reflect the Python source. '\f' is a single character, not two. When printing the filename, the interpreter is using its repr().

Also, note this difference:

>>> print('\"')
"
>>> '\"'
'"'

You will see that the repr() adds quotes around the value, showing it as it would be represented in a Python program.