Why doesn't this ruby regex work?

Question

I have a very simple regex task that has left me confused (and just when I thought I was starting to get the hang of them too). I just want to check that a string consists of 11 digits. The regex I have used for this is /\d{11}/. My understanding is that this will give a match if there are exactly (no more and no less than) 11 numeric characters (but clearly my understanding is wrong).

Here is what happens in irb:

ruby-1.9.2-p136 :018 > "33333333333" =~ /\d{11}/
 => 0 
ruby-1.9.2-p136 :019 > "3333333333" =~ /\d{11}/
 => nil 
ruby-1.9.2-p136 :020 > "333333333333" =~ /\d{11}/
 => 0

So while I get an appropriate match for an 11 digit string and an appropriate no-match for a 10 digit string, I am getting a match on a 12 digit string! I would have thought /\d{11,}/ would be the regex to do this.

Can anyone explain my misunderstanding?

Linus Kleen · Accepted Answer · 2011-01-25 09:39:26Z

5

Without anchors, the assumption "no more, no less" is incorrect.

/\d{5}/

matches

foo12345bar
   ^
   +---here

and

s123456yargh13337
 ^^         ^
 |+---here  |
 +----here  |
      here--+

So, instead use:

/^\d{5}$/

answered Jan 25, 2011 at 9:39

Linus Kleen

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2 Comments

brad Over a year ago

That's great, thanks. I did the recommended reading and realized that an even better regex would be /\A\d{11}\Z/ but that's just being picky.

Linus Kleen Over a year ago

@brad True. \b\d{11}\b is yet another way.

adamax · Accepted Answer · 2011-01-25 09:40:00Z

1

The 12 digit string contains the substring that matches your regexp. If you want an exact match, write the regexp like this: /^\d{11}$/.

answered Jan 25, 2011 at 9:40

adamax

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