How to exclude lines with regexp

Question

I have a list of lines with numbers. Need to exclude from it all lines starting with 373.

For example my list is:

37322433151
37323175491
19376717186
79684480273
97246000252
37323175491
37323175491
40745108277

If i do cat ... | egrep '^[^373].*', then it excludes lines that start from 3 or 7, the output is

19376717186
97246000252
40745108277

Even if expression is ^[^(373)].*

I need too exclude only if line starts with 373. Could anyone tell me what expression should be used ?

I also tried '^(?!373).*

For the look-ahead option, read [this](http://stackoverflow.com/questions/9197814/regex-lookahead-for-not-followed-by-in-grep). — Bernhard Barker, Sep 13 '13 at 12:15

score 2 · Answer 1 · answered Sep 13 '13 at 12:14

2

If you wanna do it with a regex then you can try:

^(37[^3]|3[^7]|[^3])[0-9]+$

answered Sep 13 '13 at 12:14

Ibrahim Najjar

@ЕгорГрижук You are welcome. If this solves your problem then please accept the answer. – Ibrahim Najjar Sep 13 '13 at 12:21

anubhava · Answer 2 · 2013-09-13T12:18:08.283

1

Use grep -v:

grep -v "^373" file

Using awk:

awk '!/^373/' file

Use grep -P (PCRE): Negative Lookahead

grep -P '^(?!373)' file

edited Sep 13 '13 at 12:18

answered Sep 13 '13 at 12:09

anubhava

1

Ninja'd by 13 seconds – Foon Sep 13 '13 at 12:09
@Foon: lol :) was wondering why no obvious answer came for 5 minutes and then 2 arrived in within 13 seconds. – anubhava Sep 13 '13 at 12:10
Yes, but it is not a solution, the regexp will not be used with grep, it will be used in billing, to match numbers. Need a regexp, that excludes, not a key in grep – Егор Грижук Sep 13 '13 at 12:13
1

Not sure what you mean by `regexp will not be used with grep, it will be used in billing`. Don't you need a grep solution? – anubhava Sep 13 '13 at 12:14

2 Answers2