CODEWITHSUNDEEP
javascriptregex
Korvo
Korvo

Reputation: 9753

Find multiple substrings in a string

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.

Upvotes: 2

Views: 1799

Answers (3)

NullUserException
NullUserException

Reputation: 85478

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.

Upvotes: 1

ean5533
ean5533

Reputation: 8994

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.

Upvotes: 1

DiverseAndRemote.com
DiverseAndRemote.com

Reputation: 19888

You can do it like this:

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

Upvotes: 2

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