How do I omit certain parts of a string with python's re?

Question

I have this string:

url = '/justicefor/404/1nirmala5.jpg'

I want to extract it as 404.jpg. I tried something like:

pattern = re.compile(
         r"./justicefor/(\d+/.\.\w+)",
         re.IGNORECASE
    )

But this selects the text between 404 and jpg too. How do I fix this?

I'm new to regular expressions so

score 1 · Answer 1 · answered Aug 19 '20 at 17:20

1

Here is a solution,

Regex Demo

import re

re.sub("/justicefor/(.*)/.*(\.\w+)", r"\1\2", "/justicefor/404/1nirmala5.jpg")

'404.jpg'

answered Aug 19 '20 at 17:20

sushanth

8,275
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I am using `findall(url)`. So I want 404 and jpg to be in a single string with find_all. Is that possible? – Saurav Pathak Aug 19 '20 at 17:28
Yes you can try, ``[i + j for i, j in re.findall("", "")]`` – sushanth Aug 19 '20 at 17:35

score 1 · Answer 2 · answered Aug 19 '20 at 17:21

1

You can use the os module

Ex:

import os

url = '/justicefor/404/1nirmala5.jpg'

path, ext = os.path.splitext(url)
print(os.path.basename(os.path.dirname(path)) + ext)  #--> 404.jpg

answered Aug 19 '20 at 17:21

Rakesh

81,458
17
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113

How do I omit certain parts of a string with python's re?

2 Answers2