Regular Expression : Finding the last 9 digits of a number and if the number is less than 9 digits than take the entire number

Question

I have been working on an regular expression which say consists of more than 9 digits ( 12345678910111213 )

With the help of regex \d{9}(?!\d) i am able to find the last 9 digits of the number.

But when the entire number is less than 9 digits how do i take the entire number as pass it. is there any such regular expression.

The question is not related to Swift/Objective-C, please do not add the `nsregularexpression` tag. — Wiktor Stribiżew, Mar 02 '18 at 11:11

score 2 · Answer 1 · answered Mar 02 '18 at 11:09

2

\d{1,9}(?!\d) will take as many as possible from the end, up to 9.

answered Mar 02 '18 at 11:09

tripleee

Awesome... \d{1,9}$ also this works fine for me... thanks a ton – Pranjal Gohain Mar 02 '18 at 11:14
1

@PranjalGohain: in this case you can accept the answer. – Casimir et Hippolyte Mar 02 '18 at 17:38
Result1 = scriptletInput.slice(-1); Result2 = scriptletInput.slice(-9, -1); scriptletResult = ""+Result2+""+Result1+"" – Pranjal Gohain May 18 '18 at 09:17
With this scriplet also we can find the last 9 digits and in case the input is less than 9 then it will take the number of items in the inputs. – Pranjal Gohain May 18 '18 at 09:18

1 Answers1