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Question

GREP reference; need end-of-paragraph expression

Forum|Forum|18 years ago
March 28, 2008
18 replies
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GREP reference; need end-of-paragraph expression

I'm doing a canned GREP search to delete all trailing zeros. So if I have a number like 8.2500, it changes it to 8.25.
I have the code to find zeros leading up to a space, line break, and paragraph return but I can't find the expression for an end of paragraph. With hidden characters on, it looks like a hash ( # ).

Does anyone know what this is?

If anyone knows a more elegant way to do this, please advise. Here's what I have:

0+$|00+$|000+$|0000+$|.0000+$|0+\s|00+\s|000+\s|0000+\s|.0000+\s|0+\n|00+\n|000+\n|0000+\n|.0000+\n|0+\r|00+\r|000+\r|0000+\r|.0000+\r

Scripting

This topic has been closed for replies.

Peter Kahrel

Community Expert

X,

/g is a switch in JavaScript GREP, but the OP wanted scripted Indesign. In fact InDesign's UI (and scripted GREPs) default to /g.

Peter

_

_jxswm_

Known Participant

(\d+)(((\.\d*[1-9])(0+))|(\.(0+)))(?=\s)

this will be better than \D?

OR:
(\d+)(((\.\d*[1-9])(0+))|(\.(0+)))(?=\D)

OR: THIS WILL MATCH END OF THE STORY OR CELL END
(\d+)(((\.\d*[1-9])(0+))|(\.(0+)))(?=(\D|\Z))

jxswm

R

Really RandyAuthor

Known Participant

This GREP search:
FIND = (\d+)(((\.\d*[1-9])(0+))|(\.(0+)))(?=\D?)
REPLACE = $1$4

turns this:
1.0000 x 1.2000
3.1200 x 4.1230
1.2090 x 1.2001

into this:
1x 1.2000
3.12x 4.1230
1.209x 1.2001

deleting the spaces before the x and missing the trailing zeros before the line breaks or paragraph returns

almost there, I can't tell you guys how much I appreciate (and learn from) your efforts on this forum :-)

_

_xbytor_

Known Participant

really_randy@adobeforums.com wrote:

> deleting the spaces before the x and missing the trailing zeros before the line breaks or paragraph returns
>
> almost there, I can't tell you guys how much I appreciate (and learn from) your efforts on this forum :-)

I don't recall what the notation is for a GREP search and replace, but you need
to have it operate globally.

In raw JS regex it would look like:
txt = txt.replace(/(\d+)(((\.\d*[1-9])(0+))|(\.(0+)))(?=\D?)/g, "$1$4");

where the "g" means to replace all occurrences.

-X

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_jxswm_

Known Participant

FIND = (\d+)(((\.\d*[1-9])(0+))|(\.(0+)))(?=\D?)
REPLACE = $1$4
THIS WILL trim LEFT the end ZEROS(0);

jxswm

Peter Kahrel

Community Expert

jxswm,

Your GREPs don't find 1.2090

Peter

_

_jxswm_

Known Participant

OR:

(\d+)(((\.[1-9]+)(0+))|(\.(0+)))(?=\D+)

This will be match all not number after the numbers.

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_jxswm_

Known Participant

(\d+)(((\.[1-9]+)(0+))|(\.(0+)))(?=\s+)

this assume at least one space(includeing hard return) after the numbers;

if maybe no space after the number, but a x:
(\d+)(((\.[1-9]+)(0+))|(\.(0+)))(?=(\s|x)+)

Peter Kahrel

Community Expert

So much for reflection. Step one didn't do anything because it works only for numbers at the end of a paragraph ($). What is required is "at word boundary" (\b). Here's the correct expression:

Find: \.0+\b
change to: (nothing)

>this is harder than I thought it would be

It often works out like that. In your case, the complication is that sometimes the decimal separator should be deleted, sometimes not.

Peter

R

Really RandyAuthor

Known Participant

Step 2 did the trick for the 1.2001 scenario but step one didn't do anything, this is harder than I thought it would be.

Peter Kahrel

Community Expert

On reflection, the first step should be this:

find: \.0+$
change to: (nothing)

To ensure that numbers such as 4.0300 and 5.0002 are dealt with correctly.

Peter

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