Matching last occurrence before a previous match

Question

How do I match the last occurrence of foo before the match of some number?

foo: A
  1
  2
foo: B
  1
foo: C
  2

A search for pattern 2 should return:

foo: A
foo: C

anubhava · Accepted Answer · 2014-02-06T06:17:08.247

4

Using awk:

awk -v s='2' '/^foo:/{line=$0;next} $1==s{print line}' file
foo: A
foo: C

edited Feb 06 '14 at 06:17

answered Feb 05 '14 at 19:36

anubhava

1

+1 but you should never use the letter `l` as a variable as it looks far too much like the number `1`, completely indistinguishable in some fonts. – Ed Morton Feb 05 '14 at 23:54
1

I had to copy and paste that to Google just to be sure. – Christopher Markieta Feb 06 '14 at 02:07
Advise well taken, changed `l` to `line` for readability purpose. – anubhava Feb 06 '14 at 06:17

score 0 · Answer 2 · answered Feb 05 '14 at 19:36

0

This regex could definitely use some optimization, but it should work:

foo: [A-Z](?=(?:(?!foo)[^2])*2)

answered Feb 05 '14 at 19:36

2 Answers2