使用正则表达式提取文件名

Question

I want to create a regular expressions to extract the filename of an url

https://example.net/img/src/img.jpg

I want to extract img1.jpg

I use urlparse from python but it extract the path in this way

img/src/img.jpg

How I can extract the file name with a regular expression

Answer 1

Using str.split and negative indexing

url = "https://example.net/img/src/img.jpg"
print(url.split("/")[-1])

Output:

img.jpg

or using os.path.basename

import urlparse, os
url = "https://example.net/img/src/img.jpg"
a = urlparse.urlparse(url)
print(os.path.basename(a.path))   #--->img.jpg

Answer 2

If your url pattern is static you can use positive lookahead ,

import re
pattern =r'\w+(?=\.jpg)'

text="""https://example.net/img/src/img.jpg
"""


print(re.findall(pattern,text)[0])

output:

img

Answer 3

You can either use a split on / and select the last element of the returned array (the best solution in my opinion)

or if you really want to use a regex you can use the following one

(?<=\/)(?:(?:\w+\.)*\w+)$

Note that only the following filenames are accepted: DEMO

You can adapt and change the \\w to accept other characters if necessary.

Explanations:

• (?<=\\/) positive lookbehind on / and $ add the constraint that the filename string is the last element of the path
• (?:(?:\\w+\\.)*\\w+) is used to extract words that are composed of several letters/digits and eventually underscores followed by a dot, this group can be repeated as many time as necessary ( xxx.tar.gz file for example) and then followed by the final extension.