Regex, get what's after the second occurence of a string

Question

I have a string of the following format:

TEXT####TEXT####SPECIALTEXT

I need to get the SPECIALTEXT, basically what is after the second occurrence of the ####. I can't get it done. Thanks

You also said you "can't get it done", what have you tried? – Matthew Warman Dec 11 '12 at 14:44 — Matthew Warman, Dec 11 '12 at 14:44

sp00m · Accepted Answer · 2015-12-18T15:44:26.087

15

The regex (?:.*?####){2}(.*) contains what you're looking for in its first group.

edited Dec 18 '15 at 15:44

answered Dec 11 '12 at 14:43

sp00m

score 0 · Answer 2 · answered Dec 11 '12 at 14:44

0

If you are using shell and can use awk for it:

From a file:

awk 'BEGIN{FS="####"} {print $3}' input_file

From a variable:

awk 'BEGIN{FS="####"} {print $3}' <<< "$input_variable"

answered Dec 11 '12 at 14:44

sampson-chen

2 Answers2