Re: [libreoffice-users] Re: Percentage with bracket as negative

Dear Andreas,
First of all.. thank you for your reply :-D

Your 1st method doesn't work (turn on special number), you're right.. the
datas are too bad.

Your 2nd method almost works.. but i think need some modification
Search: ^\s*\(([[:digit:].]+)\s*%\s*\)\s*$
Replace: $1%

I tested it, if the data is (25.24 % ) will replace it with 25.24%
The problem is LibreOffice only recoqnize 25,24% instead 25.24% (so must
use "," than ".")
For example if i change the data into (25 % ), then your regular expression
correct.

Do you know how to replace . (dot) with , (comma) ?

I already tried use English (UK) but can not

Thanks!

Best Regards,
Yudi


Best Regards,
Yudi Wijaya

On Thu, Oct 8, 2015 at 6:49 AM, Andreas Säger <villeroy@t-online.de> wrote:

> Am 08.10.2015 um 01:22 schrieb Andreas Säger:
> > Always turn on "special numbers" option. This will also import correct
> > dates.
>
> This is not enough. These data are too bad.
> menu:Find&Replace...
> Additional Options
> [X] Regular Expressions
> Search: ^\s*\(([[:digit:].]+)\s*%\s*\)\s*$
> Replace: $1%
> [Replace All]
>
> And replace all Rp with nothing in order to get the numbers.
>
>
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