Why is my javascript regexp test function not working?

Question

I've got a Javascript function that tests a regexp that I'm intending to validate positive decimals up to a precision of two:

function isPositiveDecimalPrecisionTwo(str) {
    return /^\d*(\.\d{1,2}$)?/.test(str);
}

I'm not getting the results I'm expecting when I call it from code.

For example:

var a = isPositiveDecimalPrecisionTwo("1");         //   expect t, returns t
var b = isPositiveDecimalPrecisionTwo("1.2");       //   expect t, returns t
var c = isPositiveDecimalPrecisionTwo("1.25");      //   expect t, returns t
var d = isPositiveDecimalPrecisionTwo("1.257");     // * expect f, returns t *
var e = isPositiveDecimalPrecisionTwo("1.2575");    // * expect f, returns t *
var f = isPositiveDecimalPrecisionTwo(".25");       // * expect t, returns f *
var g = isPositiveDecimalPrecisionTwo("d");         //   expect f, returns f
var h = isPositiveDecimalPrecisionTwo("d1");        //   expect f, returns f
var i = isPositiveDecimalPrecisionTwo("1d");        // * expect f, returns t *
var j = isPositiveDecimalPrecisionTwo("1d1");       // * expect f, returns t *
var k = isPositiveDecimalPrecisionTwo("-1");        //   expect f, returns f
var l = isPositiveDecimalPrecisionTwo("-1.5");      //   expect f, returns f

There are three issues:

Validating "1.257" or "1.2575" returns true, which I though would return false due to the \d{1,2}
Validating ".25" returns false, which I thought would return true due to the ^\d*
Validating "1d" or "1d1" return true. It looks like \. is reading as any character when it looks to me like it's a properly escaped "." (dot).

However, when I use a tool like regexpal.com, the same regexp appears to be validating the way I expect:

http://regexpal.com/?flags=g&regex=%5E%5Cd%2B(%5C.%5Cd%7B1%2C2%7D%24)%3F&input=

What am I missing?

gen_Eric · Accepted Answer · 2012-11-14 16:01:39Z

7

The problem is that your $ is inside the parenthesis. You want it to be after the group.

/^\d*(\.\d{1,2})?$/

This should work the way you expect.

With the $ inside, it was matching everything that started with a number, since the end of string ($) was "optional".

edited Nov 14, 2012 at 16:01

answered Nov 14, 2012 at 15:21

gen_Eric

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7 Comments

ruakh Over a year ago

I want to +1, but there's something wrong here . . . the regex that the OP posted will actually match any string, whether or not it begins with a digit (because of the *). Your fix is right, your explanation is right, but something is just missing.

gen_Eric Over a year ago

@ruakh: The OP's original regex will not match -1. The regex will match 0 numbers, and then 0 of the group, so it will match "nothing".

ruakh Over a year ago

Re: second sentence: Right, but regexes are allowed to match "nothing". All of his examples return true, at least in Firefox.

ruakh Over a year ago

Unless I'm missing something, it looks like RegexPal doesn't actually give any indication of whether a regex matches, only of what it matches; so when the regex matches "nothing", you don't see it.

dfran02 Over a year ago

@ruakh: You're right about RegexPal. I'll keep that in mind in the future.

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garyh · Accepted Answer · 2012-11-14 16:16:03Z

2

You can also do it without grouping the decimal portion by just adding a ? after the escaped decimal point

^\d*\.?\d{1,2}$

answered Nov 14, 2012 at 16:16

garyh

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