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Home/ Questions/Q 7182127
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T17:42:14+00:00 2026-05-28T17:42:14+00:00

I have next JS-code: var str = ‘some string’; var rexp = /^([^#]+)/; var

  • 0

I have next JS-code:

var str = 'some string';
var rexp = /^([^#]+)/;
var matchArr = str.match(rexp);

matchArr then contains two items matchArr[0] = ‘some string’ and
matchArr[1] = ‘some string’; Whereas I’m expecting the array with only one item.

I can’t understand this behavior. When I remove parentheses, then matchArr contains only one match. Why this happens, does anybody can explain?

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  1. Editorial Team
    Editorial Team
    2026-05-28T17:42:15+00:00Added an answer on May 28, 2026 at 5:42 pm

    When you use capturing parentheses in your regular expression, it adds an item to the returned results array for the overall match and each set of capturing parentheses.

    matchArr[0] is everything that was matched.

    matchArr[1] is what matches in the first set of capturing parentheses

    matchArr[2] is what matches in the second set of capturing parentheses
    and so on

    So, in your regex /^([^#]+)/, there is no difference between matchArr[0] and matchArr[1] because everything that matches is in the capturing parentheses.

    If you did this:

    var str = 'some string';
    var rexp = /^some([^#]+)/;
    var matchArr = str.match(rexp);
    

    You would find that:

    matchArr[0] == "some string";
    matchArr[1] == " string";
    

    because there are parts of the match that are not in the capturing parentheses.

    Or if you did this:

    var str = 'some string';
    var rexp = /^(some)([^#]+)/;
    var matchArr = str.match(rexp);
    

    You would find that:

    matchArr[0] == "some string";
    matchArr[1] == "some"
    matchArr[2] == " string";
    
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