Regex search without truncating repetitions

Question

Regex search without truncating repetitions

I am trying to locate patterns in strings where a character is followed by another character, and then followed by the first character again.

For instance, in the string "abab" I am looking for "aba" and "bab".

/([a-z])[a-z]{1}\1/g

But when I run this code, it only returns the first result (I am working with JavaScript).

"abab".match(/([a-z])[a-z]{1}\1/g)
["aba"]

When I try with "ababcb", it gives two matches instead of three (I should get "aba", "bab", "bcb").

"ababcb".match(/([a-z])[a-z]{1}\1/g)
["aba", "bcb"]

I suspect the regex is repeating on the truncated string, finding the first match and then proceeding with the rest. How can I prevent this behavior and find all possible matches?

javascript regex

Answer 1

Answer №1

Using a Positive Lookahead allows you to capture the following patterns:

var re = /(?=(([a-z])[a-z]\2))./g, matches = [];
while (m = re.exec('abab')) {
   matches.push(m[1]);
}
console.log(matches) //=> [ 'aba', 'bab' ]

var re = /(?=(([a-z])[a-z]\2))./g, matches = [];
while (m = re.exec('ababcb')) {
  matches.push(m[1]);
}
console.log(matches) //=> [ 'aba', 'bab', 'bcb' ]

Answer 2

Using a Positive Lookahead allows you to capture the following patterns:

var re = /(?=(([a-z])[a-z]\2))./g, matches = [];
while (m = re.exec('abab')) {
   matches.push(m[1]);
}
console.log(matches) //=> [ 'aba', 'bab' ]

var re = /(?=(([a-z])[a-z]\2))./g, matches = [];
while (m = re.exec('ababcb')) {
  matches.push(m[1]);
}
console.log(matches) //=> [ 'aba', 'bab', 'bcb' ]

Answer 3

Answer №2

Here is a regex pattern using lookahead:

var re = /(\w)(?=(\w\1))/g;
var input="abab";
var matches = [];

while (match = re.exec(input)) {
   matches.push( match[1] + match[2] );
}

console.log(matches);
//=> ["aba", "bab"]

Another Example:

input="ababcb";
matches = [];    
while (match = re.exec(input)) {
    matches.push( match[1] + match[2] );
}

console.log(matches);
//=> ["aba", "bab", "bcb"]

Check out the RegEx Demo here!

Answer 4

Here is a regex pattern using lookahead:

var re = /(\w)(?=(\w\1))/g;
var input="abab";
var matches = [];

while (match = re.exec(input)) {
   matches.push( match[1] + match[2] );
}

console.log(matches);
//=> ["aba", "bab"]

Another Example:

input="ababcb";
matches = [];    
while (match = re.exec(input)) {
    matches.push( match[1] + match[2] );
}

console.log(matches);
//=> ["aba", "bab", "bcb"]

Regex search without truncating repetitions

Answer №1

Answer №2

Check out the RegEx Demo here!

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