I would like to take a number and format it as price (as a string). For example I want to take 250000 and display $250,000. How can I accomplish this using regex?
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AFAIK, regex can't loop. Are you guaranteed an upper bound on the number?beatgammit– beatgammit2013-03-11 15:09:49 +00:00Commented Mar 11, 2013 at 15:09
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Yes. Assume the number is bounded.Addie– Addie2013-03-11 15:15:26 +00:00Commented Mar 11, 2013 at 15:15
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So, would $999,999 be an acceptable upper bound?beatgammit– beatgammit2013-03-11 15:17:06 +00:00Commented Mar 11, 2013 at 15:17
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To put the commas you need to know the number of digits, this can't be achieved with regex onlyichigolas– ichigolas2013-03-11 15:18:09 +00:00Commented Mar 11, 2013 at 15:18
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Yes that would be OK.Addie– Addie2013-03-11 15:20:18 +00:00Commented Mar 11, 2013 at 15:20
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4 Answers
For adding the commas you could give something like this a try:
/(\d)(?=(?:\d{3})+$)/
Then replace every match with \1,.
As such:
"12345512312".gsub(/(\d)(?=(?:\d{3})+$)/,'\1,') => "12,345,512,312"
This would match any digit followed by an arbitrary number of 3 digit groups.
E.g. the first 2 in the above example is followed by 3 groups: 345, 512 and 312. The first 5 is followed by 2 groups: 512 and 312, etc.
Not sure if you'll be able to add the $ in the same regex though.
2 Comments
Addie
Would you mind explaining how the
?: works? It doesn't seem like it is necessary.
rvalvik
Strictly speaking the
?: is not necessary, all it does is make the grouping \d{3} a passive group (meaning no back-reference is created). We need to make this a group because we want to match 1 or more occurrences of \d{3}. We could've done (\d{3})+, but then \2 would've back-referenced the matching numbers. In this case it would've made no difference, but it keeps it cleaner and makes the intent clearer when we use passive groups.Try this one out (disclaimer- not a regex):
def prettify(n)
int, dec = n.to_s.split('.')
int = int.reverse.scan(/.{1,3}/).join(',').reverse
[int, dec].reject(&:empty?).join('.')
end
Probably a gem out there for this kind of thing though
Comments
Matching with regex it's a pain in the #@! in this case since regex engines begin matching from the beginning of the string, not the end (finding 3 digit patterns beginning at the end of the string and going backwards). I'd suggest either doing something like this:
format_int = ->(s) do
str = s.reverse.scan(/\d{1,3}/).join(',').reverse
"$#{str}"
end
format_int['2500600'] => "$2,500,600"
... using Kernel#sprintf (this can be a little tricky) or, as you wanted:
I was wrong, it can be achieved with regex as shown here and here.
1 Comment
bjhaid
This is a cleaner solution
you should be using the number_to_currency helper
number_to_currency(1234567890.50) # => $1,234,567,890.50
number_to_currency(1234567890.506) # => $1,234,567,890.51
number_to_currency(1234567890.506, precision: 3) # => $1,234,567,890.506
number_to_currency(1234567890.506, locale: :fr) # => 1 234 567 890,51 €
number_to_currency("123a456") # => $123a456
number_to_currency("123a456", raise: true) # => InvalidNumberError
number_to_currency(-1234567890.50, negative_format: "(%u%n)")
# => ($1,234,567,890.50)
number_to_currency(1234567890.50, unit: "£", separator: ",", delimiter: "")
# => £1234567890,50
number_to_currency(1234567890.50, unit: "£", separator: ",", delimiter: "", format: "%n %u")
# => 1234567890,50 £
for more info look at http://api.rubyonrails.org/classes/ActionView/Helpers/NumberHelper.html
