At 17:50 07/05/2007 +1200, Rex Teague wrote:
Greetings from Aotearoa/new Zealand.
Running OOo 2.0.2 on Mepis 6.5 (Ubuntu repositories), been doing the
Linux thing for 1½ years and still learning!
I need to find several hundred instances of a line in an odt document
that have an @ as a common denominator (not an email address) and
reformat them as bold. After spending hours researching regular
expressions in the archives and on the Internet, I haven't found an
answer. Can somebody here give me guidance, I know how to do the bold
bit?
A sample of the line is: "by Howard on Mar 15th, 2007 @ 9:11pm" without
the quotation marks. They all vary, except for the @ and ideally I want
to then delete the @ and what follows it. Could be a second operation?
Cheers... Rex
The detailed answer to this will depend on what
you mean by a line. Do these lines start and
finish with hard returns? In other words, are
the lines actually, in word processor terms,
complete paragraphs? Or are they delimited by
line breaks - the sort you get by pressing
Shift+Enter? Or are they just lines in the sense
of what currently happens to surround the "@"
sign on a line in a flowing paragraph? (I guess
they are not that third possibility.) Or some
mixture of these - do they perhaps start or end a paragraph?
If they are actually paragraphs, you should be
able to achieve what you ask by putting
[EMAIL PROTECTED]
in the "Search for" box and
&
in the "Replace with" box. The ".*" means any
number of any character and the "&" replaces the
found text with an exact copy. Tick the "Regular
expressions" box (select "More Options" if
necessary). Put the cursor in the "Replace with"
box and use the "Format" button to attach the
Bold font attribute. (You say you know this bit.)
If the lines are delimited by line breaks, your search pattern should be
[EMAIL PROTECTED]
instead. But that won't find a line that happens
to start or end a paragraph. Those possibilities
are left, as they say in the best textbooks, as an exercise for the reader!
The second problem is easier: replace
@.*
with nothing. Precede the pattern with a space
if the "@" sign is always preceded by a space and you'd like that deleted too.
I don't see an easy way to combine these
replacements, but that is hardly a problem.
Incidentally, I take it that when you say "not an
email address", you mean that these are not, as
it happens, email addresses. If you mean that
you want the search pattern to catch everything
that has an "@" sign but is *not* an email
address, that's another question. If the "@"
signs you want to find are always preceded and
followed by spaces, you can probably just
interpolate spaces in the pattern suggested above.
Brian Barker
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