在一个字符串中查找多个子字符串

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.

Answer 1

您可以这样做：

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

Answer 2

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.

Answer 3

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.