RegEx false negative with .test()

Question

I'm making a Chrome extension that searches a page for a dollar amount (a number with no more then two decimal places immediately preceded by a "$") then tacks on a bit with how much that value would be in another currency. I found a commonly used regex that matches exactly those parameters.

/^\$?\-?([1-9]{1}[0-9]{0,2}(\,\d{3})*(\.\d{0,2})?|[1-9]{1}\d{0,}(\.\d{0,2})?|0(\.\d{0,2})?|(\.\d{1,2}))$|^\-?\$?([1-9]{1}\d{0,2}(\,\d{3})*(\.\d{0,2})?|[1-9]{1}\d{0,}(\.\d{0,2})?|0(\.\d{0,2})?|(\.\d{1,2}))$|^\(\$?([1-9]{1}\d{0,2}(\,\d{3})*(\.\d{0,2})?|[1-9]{1}\d{0,}(\.\d{0,2})?|0(\.\d{0,2})?|(\.\d{1,2}))\)$/g

so I'm thinking I have a nice headstart. I've only been coding a couple of months and of all the concepts I've encountered, regex's give me the most headache. I test out my shiny new expression with:

var regex = /^\$?\-?([1-9]{1}[0-9]{0,2}(\,\d{3})*(\.\d{0,2})?|[1-9]{1}\d{0,}(\.\d{0,2})?|0(\.\d{0,2})?|(\.\d{1,2}))$|^\-?\$?([1-9]{1}\d{0,2}(\,\d{3})*(\.\d{0,2})?|[1-9]{1}\d{0,}(\.\d{0,2})?|0(\.\d{0,2})?|(\.\d{1,2}))$|^\(\$?([1-9]{1}\d{0,2}(\,\d{3})*(\.\d{0,2})?|[1-9]{1}\d{0,}(\.\d{0,2})?|0(\.\d{0,2})?|(\.\d{1,2}))\)$/g;

var str = "The total it $2.25 Would you like paper or plastic?";

r = regex.test(str);
console.log(r);

and of course that sucker returns false! I tried a few more strings with "2.25" or "$2" or "$2.256" just to be sure and they all returned false.

I am thoroughly stumped. The expression came recommended, I'm using .test() correctly. All I can think of is it's probably some small newbish detail that has nothing to do with regex's.

Thanks for your time.

That regex looks like it matches at the start of the string, so it would only match if you tested it on ONLY a dollar value (without the rest of the string). My regex-fu isn't strong enough to fix that, but that's somewhere to start (the `^` at the start of the regex matches the start of the string). — Chris O'Kelly, Dec 02 '15 at 22:34
Well, the last `$2.256` [is not matched](https://regex101.com/r/qW3sT8/1). Not sure it is [that dupe I chose](http://stackoverflow.com/questions/34034770/javascript-regexp-test-method-weird-behaviour). At any rate, you must remove the `/g` flag if you want to use the regex with `RegExp#test()`. — Wiktor Stribiżew, Dec 02 '15 at 22:36

score 1 · Accepted Answer · answered Dec 02 '15 at 22:50

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Your overly complex regular expression is checking the entire string. Remove the ^ and $ which denote the beginning and end of the string, respectively. Then remove the /g flag, which is used to search for multiple matches.

What's wrong with checking for /\$\d+\.\d\d/?

I find http://regex101.com/ to be a helpful resource.

answered Dec 02 '15 at 22:50

miken32

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Thank you. I thought the ^ and \$ were combined and it meant "start looking for everything else that begins with a $". I didn't even see that last $. I just discovered regex101 today while working on this problem. I'll have to spend some serious time with it. – Wes Schumaker-Reid Dec 02 '15 at 23:09
One thing though. Your expression workswonderfully for sites like amazon but if someone just posts "$35" with no cents it still returns false which is enough to get started but I'll have to work something out later. – Wes Schumaker-Reid Dec 02 '15 at 23:17
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`/\$\d+(?:\.\d\d)?/` makes the decimal portion optional – Chris O'Kelly Dec 02 '15 at 23:20

RegEx false negative with .test()

1 Answers1