Question

106

Javascript Regex - Find all possible matches, even in already captured matches

rated 0 times [ 108] [ 2] / answers: 1 / hits: 39375 / 12 Years ago, wed, february 13, 2013, 12:00:00

I'm trying to obtain all possible matches from a string using regex with javascript. It appears that my method of doing this is not matching parts of the string that have already been matched.

Variables:

var string = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y';



var reg = /A[0-9]+B[0-9]+Y:A[0-9]+B[0-9]+Y/g;

Code:

var match = string.match(reg);

All matched results I get:

A1B1Y:A1B2Y

A1B5Y:A1B6Y

A1B9Y:A1B10Y

Matched results I want:

A1B1Y:A1B2Y

A1B2Y:A1B3Y

A1B5Y:A1B6Y

A1B6Y:A1B7Y

A1B9Y:A1B10Y

A1B10Y:A1B11Y

In my head, I want A1B1Y:A1B2Y to be a match along with A1B2Y:A1B3Y, even though A1B2Y in the string will need to be part of two matches.

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susanajamiep

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answered 12 Years ago trinity · Accepted Answer

Without modifying your regex, you can set it to start matching at the beginning of the second half of the match after each match using .exec and manipulating the regex object's lastIndex property.

var string = 'A1B1Y:A1B2Y:A1B3Y:A1B4Z:A1B5Y:A1B6Y:A1B7Y:A1B8Z:A1B9Y:A1B10Y:A1B11Y';

var reg = /A[0-9]+B[0-9]+Y:A[0-9]+B[0-9]+Y/g;

var matches = [], found;

while (found = reg.exec(string)) {

    matches.push(found[0]);

    reg.lastIndex -= found[0].split(':')[1].length;

}



console.log(matches);

//[A1B1Y:A1B2Y, A1B2Y:A1B3Y, A1B5Y:A1B6Y, A1B6Y:A1B7Y, A1B9Y:A1B10Y, A1B10Y:A1B11Y]

Demo

As per Bergi's comment, you can also get the index of the last match and increment it by 1 so it instead of starting to match from the second half of the match onwards, it will start attempting to match from the second character of each match onwards:

reg.lastIndex = found.index+1;

Demo

The final outcome is the same. Though, Bergi's update has a little less code and performs slightly faster. =]