Find multiple substrings in a string

Question

I want to check if there are two or more values in a string, regardless of their positions within said string. For example, if I want a condition of "OR" in regex, I would do so:

/(a|b)/.test("a") // true

But what I need is an "AND"; something like this:

/(a&b)/.test("a") // false
/(a&b)/.test("b") // false
/(a&b)/.test("a b") // true
/(b&a)/.test("a b") // true
/(a&b&c)/.test("a b") // false
/(a&b&c)/.test("a c b") // true

Obviously this syntax is not correct...

These values a, b, c, etc. are pulled from an array. I've tried using a combination of eval() and indexOf(a) !== -1 && indexOf(b) !== -1 but that was too slow, which is why I'm turning to regexes.

discussion moved to chat

NullUserException
– NullUserException

2012-10-12 19:15:48 +00:00
Commented Oct 12, 2012 at 19:15 — NullUserException
– NullUserException, Commented Oct 12, 2012 at 19:15

DiverseAndRemote.com · Accepted Answer · 2012-10-12 16:23:04Z

2

You can do it like this:

/(?=.*a)(?=.*b)/.test("a")

edited Oct 12, 2012 at 16:23

answered Oct 12, 2012 at 16:21

DiverseAndRemote.com

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2 Comments

NullUserException Over a year ago

To those wondering what this is: Lookahead (zero-width assertions)

Korvo Over a year ago

You do not like the sign "?" . Thanks

NullUserException · Accepted Answer · 2012-10-12 20:13:25Z

1

Since you're matching fixed strings, you could just use:

function matchAll(str, arr)
{
  for (var i=0; i<arr.length; ++i) {
    if (str.indexOf(arr[i]) === -1) {
      return false;
    }
  }
  return true;
}

matchAll('a', ['a']); // true
matchAll('a', ['a', 'b']); // false
matchAll('a b', ['a', 'b']); // true
matchAll('a b c', ['a', 'b']); // true
matchAll('a b', ['a', 'b', 'c']); // false
matchAll('a c b', ['a', 'b', 'c']); // true
matchAll('c a b', ['a', 'b', 'c']); // true

If you're looking for fixed strings, .indexOf() will be faster than regexes.

edited Oct 12, 2012 at 20:13

answered Oct 12, 2012 at 19:55

NullUserException

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Comments

ean5533 · Accepted Answer · 2012-10-12 16:42:38Z

1

The answer @OmarJackman posted will do what you're asking for. However, it's worth noting that his solution uses lookarounds which require more processing than simpler regexes. If you're looking for good performance I'd recommend that you just run two separate regexes (test for case a and then for case b). Two simple regexes will run orders of magnitude faster than one complex one, especially as the search text becomes larger.

Edit: As noted in the comments, "orders of magnitude" is an unfair exaggeration of the performance impact, but performance should be considered regardless.

edited Oct 12, 2012 at 16:42

answered Oct 12, 2012 at 16:25

ean5533

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6 Comments

NullUserException Over a year ago

I'm not convinced lookaround would make a regex "orders of magnitude" slower.

Korvo Over a year ago

Yes worry too much about the performance, if you have any examples without "?" 'd be grateful. Thanks

ean5533 Over a year ago

@NullUserException You're probably right, "orders of magnitude" is almost certainly an exaggeration, but regardless the performance point is worth noting (especially when a simpler solution like CaffGeek's is just as readable while being more performant)

NullUserException Over a year ago

Benchmarks: jsperf.com/regex-lookahead ; here it is with shorter input: jsperf.com/regex-lookahead/2 CC: @GuilhermeNascimento

ean5533 Over a year ago

@NullUserException Wow, that is really interesting. At least in my browser (Chrome, webkit) lookahead's performance for 2000 char strings is negligible, only 7% slower. But for the short 100 char strings, lookahead's performance is 72% slower! That's just weird. Opposite from what I would have assumed.

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