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.

Source: syntax is taken from Validating parameters to a Bash script

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
– shawnt00, Commented Jun 2, 2016 at 19:52
I was just copying from the source I found, I just need to match those words. — Casper
– Casper, Commented Jun 2, 2016 at 21:29

anubhava · Accepted Answer · 2016-06-02 19:45:31Z

4

You need to use alternation:

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

answered Jun 2, 2016 at 19:45

anubhava

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4 Comments

Pedro Lobito Over a year ago

You missed ^, I guess the OP needs a match on the beginning of the line.

anubhava Over a year ago

Yes because OP wrote if input doesn't match, question didn't say if these values must be at the start of each line.

Casper Over a year ago

Thanks, this works perfect, but the other answer is shorter. I was just copying the code from the other source, hence the "echo"

anubhava Over a year ago

If you're doing this against a file then grep -Eq 'foo[1-3]' file would be shortest

glenn jackman · Accepted Answer · 2016-06-02 20:41:35Z

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 2, 2016 at 20:41

glenn jackman

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1 Comment

Casper Over a year ago

Thank you, works great. I want to match exact so I'll use this [[ $INPUT == foo[123] ]] || echo "no"

Sasha Pachev · Accepted Answer · 2016-06-02 19:45:34Z

2

Does this solve your problem:

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

?

answered Jun 2, 2016 at 19:45

Sasha Pachev

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