I have a regex which extracts the file name from path but if I don't specify path then it returns an error.

Code:

PATH = '.\var\pth\index.txt'
filename = re.search('.*/(.*)\\.\\w ', PATH).group(1)
print(filename)

Above regex has been tested on these examples:

• .\var\pth\index.txt
• .\index.php
• index.txt (FAILED)

If I just pass index.txt then I get the error

AttributeError: 'NoneType' object has no attribute 'group'

Is there any way I can modify my above regex which can take any path and extract the filename?

CodePudding user response：

All too often, regex is the wrong tool. Just use either pathlib.Path for an OO approach or os.path.basename for a more functional approach:

In [1]: import os.path

In [2]: os.path.basename("/foo/bar/baz")
Out[2]: 'baz'

In [3]: os.path.basename??
Signature: os.path.basename(p)
Source:   
def basename(p):
    """Returns the final component of a pathname"""
    p = os.fspath(p)
    sep = _get_sep(p)
    i = p.rfind(sep)   1
    return p[i:]

CodePudding user response：

Another approach can be using rsplit as filename = PATH.rsplit("\\", 1)[-1]:

PATH = "\\var\\pth\\index.txt"
filename = PATH.rsplit("\\", 1)[-1]
print (filename)
PATH = "index.php"
filename = PATH.rsplit("\\", 1)[-1]
print (filename)

Output:

index.txt
index.php

If you want this code to work on Linux and Windows as well then below should be the approach:

import os
PATH = "/var/pth/index.txt"
filename = PATH.rsplit(f"{os.sep}", 1)[-1]
print (filename)

Output:

index.txt