CODEWITHSUNDEEP
javascriptregex
Dan
Dan

Reputation: 7724

Regular expression to check word and word length

In JavaScript I am using the regular expression /^([a-z]){3}^(foo)/i to try and match a specific word and the word's length. The regex should also be case insensitive. So I added an i modifier on the end.

What I think this says it should do is

^          : Start at beginning of string
([a-z]){3} : Match [a-z] exactly three times
^          : Go back to start of string
(foo)      : Match the word foo exactly

However when I tested this with the following strings fOo, foo, Foo, FoO it didn't return any matches.

I would appreciate it if someone could explain what I am doing wrong and help me fix it.

Regex101

Edit for Sukima

Example strings which should work:

fOo
FoO
foo
FOO

Example string which shouldn't work

f o o
adfsFoO
fooFoe
FfOoF
:fdFoo:

The aim of the regex is to check that the string matches the word foo exactly and is length of 3 exactly.

Upvotes: 2

Views: 659

Answers (2)

ivoputzer
ivoputzer

Reputation: 6469

Hence foo actually is a subset of your prior regex, the following should do the trick:

/^\w{3})/i

or one of the following if numbers are not allowed:

/^[a-z]{3}/i
/^[A-a]{3}/

This regex matches every word of exactly three characters at the beginning of the string (or line if you specify the /m modifier) regardless if its upper- or lowercase.

Edit (responds to comment):

if you need to match the word foo precisely regardless of its case just go for:

/^foo$/i

Example matches: fOo, fOo, foo, Foo, FoO... more available on regex101

Upvotes: 5

Sukima
Sukima

Reputation: 10064

You have two issues in that regexp. First the ^ means start of line which is meaningless unless it is the first character in the regexp pattern.

Second the pattern [a-z]{3} will match foo anyway so the second pattern is just noise.

Think what you want is simply ^[a-z]{3} which will match foo in foo and foobar.

As an aside if you want to match more than one pattern at start (or end) of line then you have to separate the patterns with a pipe |.

^foo|^bar will match foo and bar, but not bazbar or bazfoo.

Upvotes: 0

Related Questions