# new regex question

since my old question seemed to confuse a few of you(I'm terrible at wording things lol), here's a new one I need a regular expression that will match any amount of zeros at the end of a number yes, this is for the challenge An example: Number: 6768097 Trailing Zeros: 0 Number: 267700 Trailing Zeros: 2 number: 577880 Trailing Zeros: 1 You can send your full code, as I already have a code that produces the number that will be tested Actually, I'll just include my code. https://code.sololearn.com/Wm0CPovm9N4L/?ref=app I saved it after experimenting a bit, so if you remove the zeros from the output it should work.

9/23/2018 8:04:53 PM

Daniel Cooper7 Answers

New AnswerTo match everything before the zeroes, you can express it like ([0-9]+[1-9]) Which means, a number, but at the end a digit that's not 0. Then you match all the zeroes after. We also need to consider the case when the number is all zeroes. The full regex then is: ([0-9]+[1-9])?(0*) And in group 2 you have the trailing zeroes. EDIT: Though, looking at Anna's answer on the last post, /0+$/ or /0*$/ is perfect if you know that the number is at the end of the line. Nothing wrong with that!

IMHO, a RegExp is not cut out to do that. It's used for matching patterns, not counting things.

Also, the number of zeros at the end is NOT known The program takes all of the digits in the array and multiplies them. I want it to count the trailing zeros and display them. But i'm terrible at regular expressions xD