Currency Regular Expression

Question

I think I created a working regular expression for what I need. Just wondering if anyone can break it or see a shorter way to write it.

The regular expression shoul

Answer 1

Given your examples, the following regular expression will work:

/^(\$?(?(?=\()(\())\d+(?:,\d+)?(?:\.\d+)?(?(2)\)))$/gm

(note: flags and delimiters are language dependent)

This regex sets an unnecessary backreference merely to save regex-length. You can ignore the second backreference. If this is intolerable the expression will become quite a bit longer.

Have a look here: http://regex101.com/r/fH3lV1

Answer 2

Here is one shorter alternative (56 chars to your 114), which will work in almost all regex flavors:

^\$?(?=\(.*\)|[^()]*$)\(?\d{1,3}(,?\d{3})?(\.\d\d?)?\)?$

Example: http://www.rubular.com/r/qtYHEVzVK7

Explanation:

^                # start of string anchor
\$?              # optional '$'
(?=              # only match if inner regex can match (lookahead)
   \(.*\)          # both '(' and ')' are present
   |               # OR
   [^()]*$         # niether '(' or ')' are present
)                # end of lookaheand
\(?              # optional '('
\d{1,3}          # match 1 to 3 digits
(,?\d{3})?       # optionally match another 3 digits, preceeded by an optional ','
(\.\d\d?)?       # optionally match '.' followed by 1 or 2 digits
\)?              # optional ')'
$                # end of string anchor

Answer 3

You can express "between one and six digits; comma before the last three digits is optional" a bit more tersely as \d{1,3}(,?\d{3})?. This also allows you to include only two copies of (\.\d{1,2})?: one for positive and one for negative, instead of one for positive-without-comma, one for positive-with-comma, etc.

Also, \d{1,2} can be shortened slightly to \d\d?, though I'm not sure if that's an improvement.

So, barring some notation like (?(1)) to test if a backreference is set, here's the shortest version I see:

^(\$?\d{1,3}(,?\d{3})?(\.\d\d?)?|\(\$?\d{1,3}(,?\d{3})?(\.\d\d?)?\))$

One perhaps-undesirable aspect of your regex, and of this one, is that they will allow something like $00,012.7, even though no one uses leading zeroes that way. You can address that by requiring the first digit to be nonzero, and then adding a special case to handle $0 and (0.12) and so on:

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

Edited to add: using a lookahead assertion like F.J suggests in his/her answer, the latter can be shortened to:

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