How to write \r and \Z in the same expression?

[dublove]

For example, the following expression that looks for spaces at the beginning and end of a paragraph:
(^[[:blank:]]+)|([[:blank:]]+$)

I'm trying to add to this the case of end-of-article, end-of-table-cell spaces.
(^[[:blank:]]+)|([[:blank:]]+$)|([[:blank:]]+\Z)
The result failed.

But it's OK for you to separate two separate executions of ([[:blank:]]+\Z).

Translated with DeepL.com (free version)

[Mike Witherell]

Allow them to be 2 separate GREP searches chained together in Find/ChangeByList script built into InDesign. Edit the FindChangeByList.txt textfile in the FindChangeSupport subfolder to write both searches. It already has a search to clean up trailing spaces. You could add a duplication that is adjusted to look at the beginning of the paragraph.

I made a version of this that does some 32 text cleanups and it is available on my website. If you want to use the built-in FindChangeByList, add this expression into the txt file:

grep {findWhat:"(?<=\\r)\\s"} {changeTo:""} {includeFootnotes:true, includeMasterPages:true, includeHiddenLayers:true, wholeWord:false}
//Find all spaces preceded by a return and remove the space.

grep {findWhat:"\\s(?=\\r)"} {changeTo:""} {includeFootnotes:true, includeMasterPages:true, includeHiddenLayers:true, wholeWord:false}
//Find all spaces followed by a return and remove the space.

[dublove]

Thank you very much.
Can't the same regularity be realized?

[Marc Autret]

Hi @dublove 

Try this:

(\h+)(?=\r|$)|^(?1)

In the above pattern, you could replace \h by a similar space metacharacter or class (e.g. \s, [[:space:]], [ \t], etc), but \h is usually considered the safest for that purpose.

Best,

Marc

[dublove]

Hi Marc ~

Thank you very much.
Can you explain what |^(?1) means?

[Marc Autret]

Can you explain what |^(?1) means?

By @dublove

Well,

| is the alternation operator,

^ is the start-paragraph assertion,

and (?1) is the actual fun part of the GREP expression: it reproduces the literal pattern from the 1st capturing parenthesis, that is, (\h+) in my example:

(\h+)(?=\r|$)|^(?1)

But if you use a different expression in the 1st parenthesis, say (\s+), then (?1) will reflect that change. So it's a convenient way to "factor out" syntactic elements within a GREP pattern.

Best,

Marc

[FRIdNGE]

Hi Marc,

Just Grep-curious! What about "\n" to be removed in the same time with "\h"?

(^/)  The Jedi