How to create a single JavaScript regex to match both length and multiple constraints?

Question

I need to build a JavaScript regular expression with the following constraints:

The input string needs to be at least 6 characters long
The input string needs to contain at least 1 alphabetical character
The input string needs to contain at least 1 non-alphabetical character

I'm seriously lacking a lookback feature in JavaScript. The thing I came up with:

((([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z]))....)|
(.(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z]))...)|
(..(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z]))..)|
(...(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z])).)|
(....(([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z])))

This looks pretty long. Is there a better way?

How I came to this:

Regex for alphabetical character is [a-zA-Z]
Regex for non-alphabetical character is [^a-zA-Z]
So I need to look for a [a-zA-Z][^a-zA-Z] or [^a-zA-Z][a-zA-Z]
so (([a-zA-Z][^a-zA-Z])|([^a-zA-Z][a-zA-Z])).
I need to check for n preceding characters and 6-n succeeding characters.

Why didn't you accept any of the answer. Is there still a problem ? — Denys Séguret
– Denys Séguret, Commented Nov 29, 2012 at 11:18

melpomene · Accepted Answer · 2012-11-28 14:27:31Z

5

/^(?=.{6})(?=.*[a-zA-Z])(?=.*[^a-zA-Z])/

This means:

^ - start of the string
(?= ... ) - followed by (i.e. an independent submatch; it won't move the current match position)
.{6} - six characters ("start of string followed by six characters" implements the "must be at least six characters long" rule)
.* - 0 or more of any character (except newline - may need to fix this?)
[a-zA-Z] - a letter (.*[a-zA-Z] therefore finds any string with a letter anywhere in it (technically it finds the last letter in it))
[^a-zA-Z] - a non-letter character

In summary: Starting from the beginning of the string, we try to match each of the following in turn:

6 characters (if we find those, the string must be 6 characters long (or more))
an arbitrary string followed by a letter
an arbitrary string followed by a non-letter

edited Nov 28, 2012 at 14:27

answered Nov 28, 2012 at 14:17

melpomene

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9 Comments

Denys Séguret Over a year ago

Pardon me, I'm a regex noob : Is the ^ at the start needed ?

Kees C. Bakker Over a year ago

Can you explain your regex a bit?

melpomene Over a year ago

@dystroy Technically, no. But it's an optimization: If the regex engine isn't very clever, it will have to repeat the check at every position of the input string otherwise.

garyh Over a year ago

This has nothing to match, just look-aheads

Denys Séguret Over a year ago

@Some1.Kill.The.DJ As there is no $ at the end, I see no problem with {6}. The group doesn't have to match the whole string.

|

Anirudha · Accepted Answer · 2012-11-28 14:33:45Z

3

Use this regex...

/^(?=.{6,})(?=.*[a-zA-Z])(?=.*[^a-zA-Z]).*$/
  -------- ------------- --------------
    ^          ^              ^
    |          |              |->checks for a single non-alphabet
    |          |->checks for a single alphabet
    |->checks for 6 to many characters

(?=) is a zero width look ahead which checks for a match.It doesn't consume characters.This is the reason why we can use multiple lookaheads back to back

edited Nov 28, 2012 at 14:33

answered Nov 28, 2012 at 14:17

Anirudha

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2 Comments

Kees C. Bakker Over a year ago

Can you explain your regex a bit?

garyh Over a year ago

@KeesC.Bakker (?=) does look-ahead

Gabber · Accepted Answer · 2012-11-28 14:41:37Z

0

Similar answer to others, thus this doesn't need much explanation, I think the best way is to do

/^(?=.*[a-zA-Z])(?=.*[^a-zA-Z]).{6,}$/

This starts at the beginning of the string, looks ahead for an alphabetical character, looks ahead for a non-alphabetical character and, in the end, it finds a string of 6+ chars, I think there's no need for lookaheads about length

answered Nov 28, 2012 at 14:41

Gabber

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4 Comments

melpomene Over a year ago

Sure, you can always reorder the checks and remove the (?= ) around the last one.

Gabber Over a year ago

That's not just about reordering, it's about removing a redundant check and obtaining a string as a match. Using only lookaheads doesn't output a match string even if the input string matches.

melpomene Over a year ago

OK, you can also always add .*$ to the regex, which is effectively what you've done: 1) /^(?=.{6})(?=.*[a-zA-Z])(?=.*[^a-zA-Z]).*$/ 2) Reorder: /^(?=.*[a-zA-Z])(?=.*[^a-zA-Z])(?=.{6}).*$/ 3) Remove: /^(?=.*[a-zA-Z])(?=.*[^a-zA-Z]).{6}.*$/ 4) Fuse: /^(?=.*[a-zA-Z])(?=.*[^a-zA-Z]).{6,}$/ If you'd skipped step 2, you'd've ended up with /^(?=.{6})(?=.*[a-zA-Z]).*[^a-zA-Z].*$/, which works the same.

Gabber Over a year ago

Sorry if I was not clear. My intended purpose with this answer was to match the whole string. I did it. Also @Some1.Kill.The.DJ did it, but I think the lookahead to check the string's length is useless and can be effortlessly joined with the string's match itself resulting in two lookaheads instead of three which I think (and might be disproved) results in slightly better performance.

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