Why does this javascript regex give me a syntax error?

Question

I'm trying to use this regexp from jeff Atwood's blog post on detecting links:

\(?\bhttp://[-A-Za-z0-9+&@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#/%=~_()|]

However, this JS code breaks and gives a SyntaxError: Unexpected token ILLEGAL

var myRe = \(?\bhttp://[-A-Za-z0-9+&@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#/%=~_()|];
var myArray = myRe.exec("http://en.wikipedia.org/wiki/PC_Tools_(Central_Point_Software)");
console.log(myArray);

As an aside comment, you can reduce the size of your character classes using ranges and \w: var myRe = /\(?\bhttp:\/\/[!#%&()+-;=?-Z_a-z~|]*[-\w#%&()+\/=@|~]/; — Casimir et Hippolyte
– Casimir et Hippolyte, Commented Oct 23, 2013 at 3:48

Marcus Stade · Accepted Answer · 2013-10-23 01:21:58Z

3

This is because your JS is in fact invalid. This is not a valid RegExp literal:

var myRe = \(?\bhttp://[-A-Za-z0-9+&@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#/%=~_()|];

A RegExp literal starts with / and similarly ends with / in JS, so you can change the code to this:

var myRe = /\(?\bhttp:\/\/[-A-Za-z0-9+&@#\/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#\/%=~_()|]/;

And it should work (you may need to turn on the console tab.) Note that because / is a literal terminal, you have to escape any / characters in your RegExp with a \ character.

answered Oct 23, 2013 at 1:21

Marcus Stade

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Josh · Accepted Answer · 2013-10-23 01:26:21Z

1

Try declaring it as a regex variable (inside //):

var myRe = /\(?\bhttp:\/\/[-A-Za-z0-9+&@#/%?=~_()|!:,.;]*[-A-Za-z0-9+&@#/%=~_()|]/;

And check out some documentation.

edited Oct 23, 2013 at 1:26

answered Oct 23, 2013 at 1:18

Josh

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2 Comments

Ry- Over a year ago

Please don’t use new RegExp. (e.g. you forgot to escape \b)

Josh Over a year ago

Yeah, I'll get rid of the dysfunctional first one. Second one's fine.

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