Unix grep using Regex

Question

In Unix, I need an input validation which use grep:

echo $INPUT| grep -E -q '^foo1foo2foo3' || echo "no"

What I really need is if input doesn't match at least one of these values: foo1 foo2 or foo3, exit the program.

Be careful to use `^` and `$` to anchor your matches if you want the *entire* value to match. Otherwise you could pass values like "foo14" and "aaafoo3zzz". — shawnt00, Jun 02 '16 at 19:52
I was just copying from the source I found, I just need to match those words. — Casper, Jun 02 '16 at 21:29

score 4 · Answer 1 · answered Jun 02 '16 at 19:45

4

You need to use alternation:

echo "$INPUT" | grep -Eq 'foo1|foo2|foo3' || echo "no"

answered Jun 02 '16 at 19:45

anubhava

You missed `^`, I guess the OP needs a match on the beginning of the line. – Pedro Lobito Jun 02 '16 at 19:59
Yes because OP wrote `if input doesn't match`, question didn't say if these values must be at the start of each line. – anubhava Jun 02 '16 at 20:00
Thanks, this works perfect, but the other answer is shorter. I was just copying the code from the other source, hence the "echo" – Casper Jun 02 '16 at 21:30
If you're doing this against a file then `grep -Eq 'foo[1-3]' file` would be shortest – anubhava Jun 02 '16 at 21:34

score 2 · Answer 2 · answered Jun 02 '16 at 19:45

2

Does this solve your problem:

echo $INPUT | grep -E 'foo1|foo2|foo3' || echo "no"

?

answered Jun 02 '16 at 19:45

Sasha Pachev

score 2 · Accepted Answer · answered Jun 02 '16 at 20:41

2

Do you really need grep? If you're scripting in bash:

[[ $INPUT == @(foo1|foo2|foo3) ]] || echo "no"

or

[[ $INPUT == foo[123] ]] || echo "no"

If you want "$INPUT contains one of those patterns

[[ $INPUT == *@(foo1|foo2|foo3)* ]] || echo "no"

answered Jun 02 '16 at 20:41

glenn jackman

Thank you, works great. I want to match exact so I'll use this `[[ $INPUT == foo[123] ]] || echo "no"` – Casper Jun 02 '16 at 21:31

3 Answers3