CODEWITHSUNDEEP
pythonpython-2.6
PeXXeR
PeXXeR

Reputation: 101

How to manipulate directory paths to work across multiple OS?

I'm writing a python script which has to internally create output path from the input path. However, I am facing issues to create the path which I can use irrespective of OS.

I have tried to use os.path.join and it has its own limitations. Apart from that, I think simple string concatenation is not the way to go. Pathlib can be an option but I am not allowed to use it.

import os
inputpath = "C:\projects\django\hereisinput"

lastSlash = left.rfind("\\")

# This won't work as os path join stops at a slash
outputDir = os.path.join(left[:lastSlash], "\internal\morelevel\outputpath")

OR

OutDir = left[:lastSlash] + "\internal\morelevel\outputpath"

Expected output path : C:\projects\django\internal\morelevel\outputpath

Also, the above code doesn't do it OS Specific where in Linux, the slash will be different.

Is os.sep() some option ?

Upvotes: 0

Views: 47

Answers (2)

BoboDarph
BoboDarph

Reputation: 2891

Assuming your original path is "C:\projects\django\hereisinput", your other part of the path as "internal\morelevel\outputpath" (notice this is a relative path, not absolute), you could always move your primary back one folder (or more) and then append the second part. Do note that your first path needs to contain only folders and can be absolute or relative, while your second path must always be relative for this hack to work

path_1 = r"C:\projects\django\hereisinput"
path_2 = r"internal\morelevel\outputpath"
path_1_one_folder_down = os.path.join(path_1, os.path.pardir)
final_path = os.path.join(path_1_one_folder_down, path_2)
'C:\\projects\\django\\hereisinput\\..\\internal\\morelevel\\outputpath'

Upvotes: 0

D Malan
D Malan

Reputation: 11414

From the documentation os.path.join can join "one or more path components...". So you could split "\internal\morelevel\outputpath" up into each of its components and pass all of them to your os.path.join function instead. That way you don't need to "hard-code" the separator between the path components. For example:

paths = ("internal", "morelevel", "outputpath")
outputDir = os.path.join(left[:lastSlash], *paths)

Remember that the backslash (\) is a special character in Python so your strings containing singular backslashes wouldn't work as you expect them to! You need to escape them with another \ in front.

This part of your code lastSlash = left.rfind("\\") might also not work on any operating system. You could rather use something like os.path.split to get the last part of the path that you need. For example, _, lastSlash = os.path.split(left).

Upvotes: 1

Related Questions