To comment on the following update, log in, then open the issue:
http://www.openoffice.org/issues/show_bug.cgi?id=107619
User er changed the following:
What |Old value |New value
================================================================================
CC|'pescetti' |'er,pescetti'
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Ever confirmed| |1
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Status|UNCONFIRMED |NEW
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OS/Version|Windows XP |All
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Platform|Unknown |All
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------- Additional comments from e...@openoffice.org Mon Dec 14 14:18:57 +0000
2009 -------
([0-9]) finds any single digit, in the number 987 it is the first digit 9, with
replacement $1 and ReplaceAll you told it to replace all with what was found,
that's what it did.
([^ ]*)[ ]*([^ ]*)
Let's analyze the expression:
First group is any number of characters except space, including 0 characters.
Followed by any number of spaces, including 0 spaces.
Followed by the second group of any number of characters except space, including
0 characters.
Actually that expression matches almost everything, it depends on greediness
what exactly is matched for the groups. The first group may already match
everything except a space, replacing with $1$2 replaces the match with itself.
Also leading and trailing spaces with one word would be matched by the
expression. If you want to match space between non-space characters then say so:
([^ ]+)[ ]+([^ ]+)
That an odd number of characters is replaced with itself twice using both, $1
and $2, as back references apparently is a bug.
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