replace.
Date sent: Wed, 16 Mar 2022 20:41:43 +0000
>
>
>>>> I am looking for a find & replace solution in Writer, where there is a
>>>> blank space as the last character of a paragraph (eg. last word
>>>> <space><LF><CR>). In the original OOo and most of the text editors I
>>>> use the simple solution that worked/works perfectly is: Find =
>>>> <space>$ Replace = $ (Note: I use <space> here to represent a single
>>>> space character). This does not work in LO Writer, so I am forced to
>>>> fiddle around copying from LO and pasting into AOO, running F&R in
>>>> AOO, then copy/paste back again. Any pointers would be welcome.
>>> I'm surprised that exactly what you say works in OpenOffice. Don't you
>>> need to find space-dollar, just as you describe, but to replace with
>>> nothing? Replacing with dollar will surely insert an unwanted dollar
>>> character?
>>>
>>> Or have you forgotten to click "Other options" in the Find and Replace
>>> dialogue and to tick "Regular expressions"?
>>>
>>> I trust this helps.
>>>
>>> Brian Barker
>> Hi Brian,
>>
>> After 22 Years of you and I being involved in this project, I genuinely
>> bow to your superior knowledge in this area. However, I do assure you
>> that the convoluted LO->AOO->LO nonsense works for me, exactly as it
>> did in the days of OOo.
>>
>> No, I have not forgotten to click "Other options in the Find and Replace
>> dialogue and to tick "Regular expressions", please see my screen capture:
>> https://www.mediafire.com/view/80552jkp3qw6o8x/F%26R_LO.png/file
>>
>> Here is a very simplistic file illustrating the <space><CR><LF>
>> paragraph endings:
>> https://www.mediafire.com/file/vmixa9r4dhjpc9e/F%26R_LO.odt/file
>>
>> Version: 7.3.1.3 (x64) / LibreOffice Community
>> Build ID: a69ca51ded25f3eefd52d7bf9a5fad8c90b87951
>> CPU threads: 8; OS: Windows 10.0 Build 19043; UI render: default; VCL:
>> win
>> Locale: es-ES (en_IE); UI: en-US
>> Calc: threaded
>>
>> If you can please show me where I am going wrong, I would be extremely
>> grateful.
>>
>> Dave Barton
>>
>>
> May be this is sorted and I have received the emails out of order, but
> assuming you just want to remove the trailing space from each line, both
> Johnny's and Michael's methods work for me on your sample file.
>
> Find= [:space:]{1,}$
> Replace= empty
>
> Find= <space>$
> Replace= empty
>
> Use regular expression= checked
> Steve
Thanks Steve, but no it's NOT sorted.
You have "Kinda" given me 1/2 a clue.
What does {.1.} mean?
file:///opt/libreoffice7.3/help/en-US/text/shared/01/02100001.html?&DbPAR=WRITER
{N}
The post-fix repetition operator that specifies an exact
number of occurrences ("N") of the regular expression
term immediately preceding it must be present for a
match to occur. For example, "tre{2}" matches "tree".
{N,M}
The post-fix repetition operator that specifies a range
(minimum of "N" to a maximum of "M") of occurrences of
the regular expression term immediately preceding it
that can be present for a match to occur. For example,
"tre{1,2}" matches "tre" and "tree".
{N,}
The post-fix repetition operator that specifies a range
(minimum "N" to an unspecified maximum) of
occurrences of the regular expression term immediately
preceding it that can be present for a match to occur.
(The maximum number of occurrences is limited only by
the size of the document). For example, "tre{2,}"
matches "tree", "treee", and "treeeee".
So, [:space:]{1,}$ matches one or more spaces at end of
line.
<space>{1,}$ does the same with <space> actually
being a single space.
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