Using grep style to add a 3 digit number in front of any 6 digit number

[Backbone22]

Im looking to be able to add a 3 digit number in front of any 6 digit number that i find for example 380005 becomes 12538005 and 690567 becomes 125690567 and so on.

I have this as the find

(\d\d\d\d\d\d)

But am struggling to find a style that will say place 125 or whatever the 3 digit code is in front of the 6 that i have.

Any help appreciated

[Obi-wan Kenobi]

You can't with a Grep style!

Try a F/R:

Find: \b\d{6}\b

Replace by: 125$0

(^/)

[Backbone22]

Just got the answer to this and apparently you can. I think this is basically doing the same thing that you have outlined the $1 refers to the bracketed six digits

(\d\d\d\d\d\d)

125$1

[Obi-wan Kenobi]

Not exactly the same! 

try with: 0123456789

… and the parenthesis aren't not useful if not other group(s)!

Think simple and global! (even if you don't need it!)

(^/)

[Jongware]

Backbone22  wrote

Just got the answer to this and apparently you can.

No you cannot. A GREP style can only apply formatting to a text, found with a GREP. See the online help: Drop caps and nested styles in InDesign

[Erica Gamet]

You say GREP style, which is why the answers above say you can't do it. You can't do this as a GREP style within a paragraph style. Or did you just generally mean a GREP expression?

To clarify, you can only (as Jongware stated) apply formatting via a character style to text found with GREP style in a paragraph style. If you need to add, subtract, re-arrange, or otherwise modify the found text, then you need to use the Find/Change dialog box...which is where you would need to work for your needs here.

[Backbone22]

I guess i meant Grep expression as i used the Grep function in the Find/Change dialog box which worked for me.