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http://www.openoffice.org/issues/show_bug.cgi?id=83938
User discoleo changed the following:
What |Old value |New value
================================================================================
Summary|replace all incorrect with|replace all incorrect with
| regular exp. | regular exp.
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------- Additional comments from [EMAIL PROTECTED] Sun Nov 25 11:19:06 +0000
2007 -------
The actual behaviour is correct! Are you sure you have done everything as stated
here?
Lets see what '^[^:]+:' does find:
NOTE: '^[^:]+:' is probably better than your '^.[^:]*:'
Your version will also find:
:this is some text beginning with a ':' ...further text...
^ ^
| |
Selects from the start to this second colon!
My version will skip that line. [IF you wish only the first colon, use
'^[^:]*:]' instead.]
Now, what will the regular expression '^[^:]+:]' find?
Lets say we have:
...some text: more text: even more text
It will select form the beginning to the first ':'. Now IF you replace this
with, e.g. 'different text', then you end basically with:
different text more text: even more text
so, the replace all will also find: 'different text more text:', and will
replace this, too:
different text even more text
Now it has finished.
IF, however, you replace with '*:' (note the additional ':'), then this will not
happen, so your example would have been processed correctly, as you desired it.
Also, replacing with ' &' will produce the correct result, because '&' will also
paste the ':' from '^[^:]+:]'.
In conclusion, you code works correctly and, in the examples you give, should
provide even your desired results. IF however, you intend to replace with some
text that DOES NOT have a colon (aka without the ':'), the replace all is still
correct, but it is different from what you think it ought to do. In that
situation I can't imagine a simple one-click solution to your problem. But you
can solve it in 2 steps:
- first, replace it with e.g. ':#the replacement text##'
[NOTE: we use ':#' as markers for the replacement]
[IF your text contains this particular string,
consider a different strin, like '%%'
- second: replace: ':#.*##' with the actual replacement
This should do it.
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