awk regex last occurence of character/pattern

Question

I need to get the index of the last occurrence of a pattern. Tried ideas from here and here. The below does not work if I want the index of the last : (index is 6). Tried to use the anchor $ but clearly have not grasped it (since it gives me the first occurrence, i.e. 3). Explanations really appreciated.

echo 12:45:78 | 
awk '
{
print match($1, /:.+$/)
}'

Use `/:[^:]*$/` (`[^:]*` will match 0+ chars other than `:`) — Wiktor Stribiżew, Mar 23 '17 at 07:50
Actually, [this idea](http://stackoverflow.com/questions/30147271/regex-match-everything-after-including-last-occurance-of-t) fits your scenario. How did you try it? — Wiktor Stribiżew, Mar 23 '17 at 07:52
Yes, I see that now - already looked to that. Getting wiser... — user3375672, Mar 23 '17 at 07:53
So `:[^:]` says a `:` and then any character except `:` any number of times ? — user3375672, Mar 23 '17 at 07:55
Yes. Do not forget the $ anchor. Also, please check anubhava's answer that might be more optimal for you (unless you need more things to check in your input that would really require a regex). — Wiktor Stribiżew, Mar 23 '17 at 08:18

score 3 · Accepted Answer · edited May 23 '17 at 11:46

You need to use

/:[^:]*$/

Here, [^:]* (a [^...] is a negated bracket expression) will match 0+ chars other than :, so, only the last : is matched with the first :.

Note this idea is almost right for you, the only difference being the quantifier: if you use + with [^:], you won't match the : that is at the very end of the input string. Thus, * is the quantifier you want.

Pattern details:

: - a : followed with...
[^:]* - any 0+ chars other than :
$ - end of string.

score 2 · Answer 2 · answered Mar 23 '17 at 07:56

There is no need to use any regex here as awk allows you split file using a delimiter.

After using : as input field separator subtract last field's length from total line length to get right index like this:

awk -F: '{print length($0) - length($NF)}' <<< '12:45:78'

6

More examples:

awk -F: '{print length($0) - length($NF)}' <<< '12:45:78111:123'
12

awk -F: '{print length($0) - length($NF)}' <<< '12:45:78:123'
9

awk -F: '{print length($0) - length($NF)}' <<< '12:45'
3

awk -F: '{print length($0) - length($NF)}' <<< '12:'
3

When you're using awk better to take advantage of all the awk goodies :) — anubhava, Mar 23 '17 at 08:02

awk regex last occurence of character/pattern

2 Answers2