capturing group in regex [duplicate]

Answer 1

Here's the most obvious example:

"madhur".replace(/(madhur)?/, "$1 ahuja");   // returns "madhur ahuja"
"madhur".replace(/(?:madhur)?/, "$1 ahuja"); // returns "$1 ahuja"

Backreferences are stored in order such that the first match can be recalled with $1, the second with $2, etc. If you capture a match (i.e. (...) instead of (?:...)), you can use these, and if you don't then there's nothing special. As another example, consider the following:

/(mad)hur/.exec("madhur");   // returns an array ["madhur", "mad"]
/(?:mad)hur/.exec("madhur"); // returns an array ["madhur"]

Answer 2

The first one won't store the capturing group, e.g. $1 will be empty. The ?: prefix makes it a non capturing group. This is usually done for better performance and un-cluttering of back references.

In the second example, the characters in the capturing group will be stored in the backreference $1.

Further Reading.

Answer 3

It doesn't affect the matching at all.

It tells the regex engine

• not to store the group contents for use (as $1, $2, ...) by the replace() method
• not to return it in the return array of the exec() method and
• not to count it as a backreference (\1, \2, etc.)