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Javascript global match with capturing groups [duplicate]

Can anyone tells me why does't the second snippet catchs the 'groups' when use g flag ?

  "123".match(/(\d{1})(\d{1})/)    // returns  ["12", "1", "2"]
  "123".match(/(\d{1})(\d{1})/g)   // returns ["12"]   (where's 1 and 2 ?)

console.log("123".match(/(\d{1})(\d{1})/))    // returns  ["12", "1", "2"]

console.log("123".match(/(\d{1})(\d{1})/g))   // returns ["12"]   (where's 1 and 2 ?)
like image 372
johnny_trs Avatar asked Aug 11 '16 06:08

johnny_trs


1 Answers

As per MDN docs :

If the regular expression does not include the g flag, returns the same result as RegExp.exec(). The returned Array has an extra input property, which contains the original string that was parsed. In addition, it has an index property, which represents the zero-based index of the match in the string.

If the regular expression includes the g flag, the method returns an Array containing all matched substrings rather than match objects. Captured groups are not returned. If there were no matches, the method returns null.


If you want to obtain capture groups and the global flag is set, you need to use RegExp.exec() instead.

var myRe = /(\d)(\d)/g;
var str = '12 34';
var myArray;
while (myArray = myRe.exec(str)) {
  console.log(myArray);
}
like image 191
3 revs, 2 users 98% Avatar answered Oct 22 '22 17:10

3 revs, 2 users 98%