Skip to main content
M
myDavey
Inspiring
June 10, 2014
Answered

Format long numbers with commas

  • June 10, 2014
  • 2 replies
  • 2435 views

Hi

I have a table with a lot of 4-digit+ numbers

I would like to add commas to them in the correct places, E.G:

     2000  should be  2,000

     23783  should be  23,783

and so on.

Is there a built in method to do this?

If not, what is the best way to accomplish this?

Thanks,

Davey

This topic has been closed for replies.
Correct answer Jump_Over

Hi,

Use find...replace with Grep:

find what: (\d+)(\d{3})

change to: $1,$2

and repeat it till each multinumber (7+) is matched

Jarek

2 replies

H
HarveyLiu
Inspiring
June 18, 2014

Hi, try this

grep {findWhat:"(\\d)(\\d\\d\\d)\\b"} {changeTo:"$1,$2"}

repeat n times

Jump_Over
Jump_Over
Legend
June 10, 2014

23783  should be  23,783

What about 2,3783? It never happends?

Jarek

M
myDaveyAuthor
Inspiring
June 10, 2014

Hi Jarek

Thanks for replying

What about 2,3783? It never happends?

I am not sure what you mean... at least not where I am from!

Its always only 3 digits per comma

3,000

30,000

300,000

3,000,000

L
Community Expert
LaubenderCommunity Expert
Community Expert
June 16, 2014

To make it cool, using Multi-Find/Change, save the regex and place 10 times in a MFC set. So, with 1 click, you treat all until:

@ Obi-wan: To make it really cool, write it in the way like Jeffrey Friedl did it: \d(?=(\d{3})+\b) Notice here the Plus-repeater after the digits group.

@ myDavey: This one works for me with .replace()

var string = "1000000";

var result = string.replace(/(\d)(?=(\d{3})+\b)/g, "$1,");

alert (result);

–Kai


@Kai – yes, really cool. Obi-wan is using the marker for "left dilimiter of a word" \> instead of \b which marks both, the right and the left delimiter of a word. Both are working fine here.This is working not only in a RegEx with a replace() function, but also with the GREP functionality in InDesign.

Searching for a single digit that is followed by a multitude of 3 digits delimited by a word boundary. The "followed-by" is expressed as a positive look ahead. Replace the found digit by itself adding a comma (or with a German text you would use a dot instead of the comma).

Instead of $1 one could also write $0.

Works the same in GREP land.

Not so with using replace() and a regular expression!

String = "1000000"

Result with GREP, using Kai's example string with $0 would be:

"1,000,000"

Result with replace() using $0 would be:

"$0,00$0,000"

It seems, that $0 is not working at all with replace().
It's just interpreted as a normal string.

Can someone verify this?

Uwe

Terms & ConditionsAccessibility statement
Using the community Terms of use Privacy Cookie preferences Do not sell or share my personal information ad choicesAdChoices