GREP reference; need end-of-paragraph expression

[Really_Randy]

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

[Jongware]

End-of-paragraph is a '$' sign (only and always at the end of your expression). Perhaps your GREP got confused because of your using dollars everywhere -- did you mean to search for a dollar sign? In that case, use '\$' (escaped). That means you can ditch all the '\r' and '\n' in your search.
You should also escape the other special character in your search: the period. As it is, it'll pick up
i any
character, and that probably will be replaced with nothing. Use '\.' to search for a literal period if you want to.

It seems you've heard about the '+' operator (one or more of the last expression) but didn't quite understand it 🙂 It's not necessary to search for "0+|00+" (i.e., a '0' followed by more
b or
two zeros followed by more).
Stringing everything together with "|" (ORs) is not necessary to include 'either with a space or without' -- you can use the Zero-or-Once operator '?'. Just a little handier would be Zero-or-Umpteen -- the '*'. Where '\s?' would only include zero or one space, '\s*' includes zero or any other number.

Putting it all together, I think this should work nicely:

>\.?0+\s*$

In English: Zero or one period, followed by any number of zeros (including a single one), followed by any number of spaces (including none), followed by the end of a paragraph. And that includes end-of-story (additionally, end-of-table-cell as well).

As it is, this will also strip off the zeros for numbers such as "10" (one without a decimal point). It takes a bit more work to ignore these, although it still is possible.

By way of answer to your question, there
i is
a special End-of-Story marker: '\Z'. That works like the '$' code, but it matches the real end of story
i only.

[Harbs.]

This might work better:
(\.[1-9]+)0+\s*$
and replace with $1

(Although that wont get 8.000)
Harbs

[Jongware]

Thanks, Harbs -- nice & simple. If you use
>(\.[1-9]*)0+\s*$

i.e.,
i zero
or more of 1..9 following the period, that'll all 0s after that. Sounds like the OP meant this.

[_jxswm_]

//
find = (\d+)(((\.[1-9]+)(0+))|(\.(0+)))(?=\s*)
replace = $1$4

jxswm

[Peter Kahrel]

What's wrong with

find: 0+$
replace with: "" (i.e. nothing)

Peter

[Harbs.]

> What's wrong with
>
> find: 0+$
> replace with: "" (i.e. nothing)
>
That will get any number ending in zeros, even integers (like 100). He
only wants to strip them off decimals.

Harbs

[Peter Kahrel]

Ah, that's what's wrong with it. Back to Jongware.

Peter

[Really_Randy]

Thank you all for your replies. Unfortunately I probably didnt think of all the different scenarios. JXSWMs solution came the closest to what I was trying to achieve. What I failed to explain was all the different combos I needed to take care of. Example:
I would like to take the following
1.0000 x 1.2000
3.1200 x 4.1230
1.2090 x 1.2001
And turn it into this:
1 x 1.2
3.12 x 4.123
1.209 x 1.2001

JXSWMs query wiped out the double zeros in 1.2001, making it 1.21
Jongwares script didnt get the 000[space] scenario.(as in 1.0000 x)
My script was very inefficient and also didnt get the 0000[space] scenario.

I could add 0\s|00\s|000\s|0000\s to my original query but that would not be very elegant.

advice please?

[Peter Kahrel]

It's probably easiest to do this in two steps. First:

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

to delete dots followed by zeroes (1.0000). Then the other cases:

find: (\.[\d]+?)0+\b
change to: $1

This way, you still understand the expressions an hour later.

Peter

[Peter Kahrel]

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

[Really_Randy]

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]

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

[_jxswm_]

(\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)+)

[_jxswm_]

OR:

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

This will be match all not number after the numbers.

[Peter Kahrel]

jxswm,

Your GREPs don't find 1.2090

Peter

[_jxswm_]

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

jxswm

[Really_Randy]

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_]

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

[_jxswm_]

(\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

[Peter Kahrel]

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