CODEWITHSUNDEEP
javascriptregex
Juan
Juan

Reputation: 15715

Can't make sense of the way javascript's regex works

Why does this code (here's the JSBin):

var text = "T: 01202 870738";

var regex1 = /T: (.*)/;
var matches1 = text.match(regex1);

for(var i = 0; i < matches1.length; i++) {
  log("[" + i + "]: " + matches1[i]);
}

logs this:

[0]: T: 01202 870738
[1]: 01202 870738

and this code (note I've added the g option):

var regex2 = /T: (.*)/g;
var matches2 = text.match(regex2);

for(var i = 0; i < matches2.length; i++) {
  log("[" + i + "]: " + matches2[i]);
}

logs this:

[0]: T: 01202 870738

I actually don't even understand why is the first code logging 01202 870738 as the second match. How is that a match for /T: (.*)/ if it doesn't include a T:?

Upvotes: 1

Views: 70

Answers (2)

NitayArt
NitayArt

Reputation: 456

What throws you off is the different behavior you get from a regex with and without the g flag. Calling String.match with a g flagged regex will return an array of all instances of the pattern within the String object. For example, the expression:

"Hello World!".match(/l/g);

Will return this array:

["l", "l", "l"]

However, calling the same function without the g flag will return an array whose first element is the matched pattern. Any element thereafter will match each expression within parentheses. For example, the expression:

"Hello World!".match(/(Hell)o World(!)/);

Will conveniently return this array:

["Hello World!", "Hell", "!"]

Upvotes: 1

Ry-
Ry-

Reputation: 224867

The second one is a global regular expression, so the array returned is a list of all the matches for the expression in the string. The first one isn’t, so it’s a list of groups, like you would get from exec. (Group zero being the entire match, and group one being the only parenthesized... group.)

Upvotes: 3

Related Questions