问题
This functional method takes a number and returns the same value separated with commas, as is the common convention in the US.
The only way I could get it to work with regex was to reverse the string before and after the expression. Is there a regex that can help me eliminate the need to call String#reverse twice for method functionality?
def separate_comma(number)
raise "You must enter a number." if number.is_a?(Numeric) == false
number.to_s.reverse.gsub(/(\d{3})(?=\d{1,3})/, "\\1,").reverse
end
回答1:
"1234556".gsub(/\d(?=\d{3}+\b)/,'\\0,')
# => "1,234,556"
This doesn't handle long fractional values, but this wasn't a concern for the OP's regex either.
回答2:
Other libraries have already solved this problem - ActiveSupport
for one.
require "active_support/number_helper"
ActiveSupport::NumberHelper.number_to_delimited(1234567890)
#=> "1,234,567,890"
You can even change the delimiter if you wish:
ActiveSupport::NumberHelper.number_to_delimited(1234567890, delimiter: "|")
#=> "1|234|567|890"
回答3:
The established way to do it is:
string.gsub(/(?<=\d)(?=(?:\d{3})+\z)/, ",")
If you want to do it with floats:
string.gsub(/(?<=\d)(?=(?:\d{3})+[.\z])/, ",")
来源:https://stackoverflow.com/questions/21892046/how-do-i-improve-regex-to-eliminate-unnecessary-method-chaining