Regular expression get last characters

Question

I want to write a regular expression which will take the characters from the last "slash" "/" to the right.

Exemple1 : http:// thesite.com/tester/blabla/this_is_the.jpg - I want to extract only the "this_is_the.jpg" Exemple2 : http:// thesite.com/tester/blabla/this_is_2.jpg - Only the "this_is_2.jpg"

@Amine: In some languages, the regex requires to be wrapped in delimiters. `/` is the most commonly used delimiter — if the delimiter is a `/`, all occurrences of `/` must be escaped inside the pattern. — Amal Murali, Jul 03 '14 at 19:31

score 6 · Accepted Answer · answered Jul 03 '14 at 19:27

6

It's as simple as:

[^/]+$

Which basically says: "give me all characters at the end that are not slashes".

Example:

$ echo 'http:// thesite.com/tester/blabla/this_is_the.jpg' | egrep -o '[^/]+$'
this_is_the.jpg

answered Jul 03 '14 at 19:27

xaizek

4,575
1
24
48

a user · Answer 2 · 2014-07-03T20:57:50.710

2

short answer:

/[^\/]*$/

explaination

$ match end of the string
[^\/] any character except /
* any number of things matched by previous token

edited Jul 03 '14 at 20:57

answered Jul 03 '14 at 19:27

a user

20,998
4
53
83

score 1 · Answer 3 · answered Jul 03 '14 at 19:28

1

Here it is using sed:

sed 's#.*/\([^/]*\)$#\1#'

answered Jul 03 '14 at 19:28

unxnut

7,506
2
23
36

ArtisticPhoenix · Answer 4 · 2014-07-03T21:11:32.377

I would anchor it on the extension, personally - then your validating that all in one go.

"/bla bala/this_is_the.jpg" 

\/([^\/]+\.jpg)$ 
Array
(
    [0] => /this_is_the.jpg
    [1] => this_is_the.jpg
)

for other file types

\/([^\/]+\.(jpg|png|bmp))$

Array
(
    [0] => /this_is_the.jpg
    [1] => this_is_the.jpg
    [2] => jpg
)

for the other guys ( cant comment yet ) I would avoid the * and go with +, do you really want it to match on this "/bla bala/this_is_the/" ~ no file

Regular expression get last characters

4 Answers4