What characters do I need to escape with sed to make this regex work

Question

(?<![0-9])0+(?=[0-9]+)

I need to remove unnecessary leading zeros in malformed octettes of IP addresses.

I want to do something like this but it is not working.

cat Qualys-Active-IPs.csv | awk -F';' {'print $1'} | sed 's/(?<![0-9])0+(?\=[0-9]+)//g'

Escaping won't help, lookarounds are not supported with `sed`. — Wiktor Stribiżew, Jun 13 '18 at 09:34
Try `sed -E 's/([^0-9]|^)0+([^0-9]|$)/\1\2/g'` or `perl -pe 's/(?<![0-9])0+(?=[0-9])//g'` — Wiktor Stribiżew, Jun 13 '18 at 09:40
Is there a CLI tool that I could use as an alternative or another combination of piping commands? — overflyer, Jun 13 '18 at 09:40
Unfortunately you suggesstion did not work, but this did: sed -r 's/^0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)$/\1.\2.\3.\4/' — overflyer, Jun 13 '18 at 09:43
[`sed -r 's/([^0-9]|^)0+([0-9])/\1\2/g'` would work, but the `perl` solution was already correct.](http://rextester.com/NIOE62860) — Wiktor Stribiżew, Jun 13 '18 at 09:48
Possible duplicate of [Does lookbehind work in sed?](https://stackoverflow.com/questions/26110266/does-lookbehind-work-in-sed) — Wiktor Stribiżew, Jun 13 '18 at 09:50

score 0 · Answer 1 · answered Jun 13 '18 at 09:44

0

The solution is:

 sed -r 's/^0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)$/\1.\2.\3.\4/'

answered Jun 13 '18 at 09:44

overflyer

score 0 · Answer 2 · edited Jun 14 '18 at 14:54

0

You may try this code:

sed -r 's/^0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)-0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+)\.0*([0-9]+),...,(.*)$/\9:\1.\2.\3.\4-\5.\6.\7.\8/'

edited Jun 14 '18 at 14:54

J...S

answered Jun 13 '18 at 10:09

James Andreson

2 Answers2