Why doesn't a regex containing $ (built from a string) work?

Question

This works:

var r="xS";
var regex = new RegExp(r); // Anchor at the end
var s="axS";
s = s.replace( regex, "Z" );
// Now, s is "aZ"

But this doesn't

var r="x$";
var regex = new RegExp(r); // Anchor at the end
var s="ax$";
s = s.replace( regex, "Z" );
// Now, s is STILL "ax$". NOT "aZ".

This doesn't work no matter where "$" is in the string r - e.g even if it's not at the end.

In the second bit of code, r is "x$$". Why would the the two $ symbols behave differently from each other? — Michael Myers
– Michael Myers ♦, Commented Feb 15, 2012 at 16:26
@MichaelMyers - I removed the $ from the end of the regex. It was a red herring from exiting code (right-side anchor) but didn't affect the problem. — DVK
– DVK, Commented Feb 15, 2012 at 16:29

gen_Eric · Accepted Answer · 2012-02-15 16:24:35Z

4

If you want to look for a $ in a string, you need to escape it. The $ is a special character in regexes meaning "end of string".

var r="x\$";
var regex = new RegExp( r + "$" ); // Anchor at the end

answered Feb 15, 2012 at 16:24

gen_Eric

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3 Comments

DVK Over a year ago

Gotcha. For some reason I thought (errorneously) that RegExp would escape it automagically.

gen_Eric Over a year ago

@DVK: It can't, how would it know what you're trying to do? And if it did, how would you use special characters? :-P

DVK Over a year ago

I said "for some reason", not for "for sane reason" :)

Jimmy · Accepted Answer · 2012-02-15 16:27:10Z

1

In the second case, "ax$" is a literal string which contains the character '$'. The regex (r) does not contain the literal character, but instead contains two anchors. You need to escape the '$' in the regex to match a literal value.

var r = "x\$"; should do the trick.

answered Feb 15, 2012 at 16:27

Jimmy

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