Regex to remove part of a string

Question

Here is my string http://www.example.com/dl/r66h5icmsyrh

I need to remove http://www.example.com/dl/

I am using autoblogged a wordpress plugin; which can handle regex in post template

This is what i tried.

%regex("%link%", "http://www.example.com/dl/", "0")%

[What have you tried?](http://whathaveyoutried.com) – Mark Reed May 29 '12 at 15:50 — Mark Reed, May 29 '12 at 15:50

score 0 · Answer 1 · edited Jun 07 '12 at 04:57

0

Use this:

%regex("%link%", "\\([^\\]+)$\", "0")%

edited Jun 07 '12 at 04:57

Somnath Muluk

answered May 29 '12 at 15:51

dockeryZ

Yes, i do. Autoblogged template only allows for it. http://autoblogged.com/support/online-manual/advanced-usage/post-template-reference/ – user1068544 May 29 '12 at 15:59
Try this then, for your regex – dockeryZ May 29 '12 at 16:01
([^\\]+)$ Should grab everything after the last \ – dockeryZ May 29 '12 at 16:01
%regex("%link%",([^\]+)$) - Would this work; %link% is the dynamic link from autoblogged. – user1068544 May 29 '12 at 16:04
the regex pattern needs to be in quotes. – dockeryZ May 29 '12 at 16:08
I'll try %regex("%link%","([^]+)$") – user1068544 May 29 '12 at 16:14
%regex("%link%", "\([^\]+)$\", "0")% didn't work i get the following error in autoblogged: preg_match_all() [function.preg-match-all]: No ending delimiter '/' found in /home/content/46/8529846/html/wp-content/plugins/autoblogged_2-8-51/autoblogged.php at line 2654 I only got that error when i tried using the regex – user1068544 May 29 '12 at 22:05
Apparently, that delimiter doesn't work. Try using a hashtag/number sign (#) for the delimiter ---> #([^\]+)$# – dockeryZ May 30 '12 at 15:36

1 Answers1