Hints for a complex currency regular expression for Javascript?

Question

I need to match an infinite number of figures in a web page.

I need to be able to match all of the following formats:

100 $
99$
$99
$ 8
$.99
$ .8
$ 99.8
.99$
.99 $
9.2 $
1.2$

And the equivalent using commas:

444,333
22,333
1,222
11,111,111
333,333,333,333.01132

Or spaces:

444 333
22 333
1 222
11 111 111
333 333 333 333.01132

This is a really hard one for me. I am used to playing with regexp but I have totally failed to write something bullet proof. Usually http://www.regexlib.com has the solution, but not for this one.

I can't think of another way other than using regexp since it's a plain text search/replace.

Answer 1

Here's a regular expression that will match all the number formats you've provided:

^(?:\$\s*)?(?:(?:\d{0,3}(?:[, ]\d{0,3})*[, ])+\d{3}|\d+)(?:\.\d*)?(?:\s*\$)?$

To break it down:

• ^(?:\$\s*)? will look for an optional $ at the start of the string, followed by any amount of spaces
• (?:(?:\d{0,3}(?:[, ]\d{0,3])*[, ])+\d{3}|\d*) will match either a number broken down into groups separated by a comma or space ( \d{0,3}(?:[, ]\d{0,3})*[, ])+\d{3} ) or a string of numbers ( \d+ ) -- so 123,456,789, 123 456 789 and 123456789 would be all matched. The regular expression will not accept numbers with improper grouping (so 123,45,6789 will not be matched)
• (?:\.\d*)? will match a number with an optional decimal and any number of numbers after
• (?:\s*\$)?$ will match an optional $ at the end of the string, preceded by any amount of space.

Answer 2

Why write 1 regexp, when you can write several, and apply them in turn?

I'm assuming (?) that you can iterate through line-by-line. Why not try your comma-savvy regexp, followed by your space-savvy regexp etc.? If one matches, then don't bother trying the rest, and store your result and move on to the next line.

Answer 3

what about doing this in 2 steps:

first replace all spaces with ''

then, if number formatting is always the same, you can replace commas with ''

after that, its pretty easy, no?