okay that was actually the first thing I'd tried. later I used this:
<replaceregexp file="temptest.csv" match="^(([^;]*;){30})([^;]*)(;)$"
replace="\1\3" byline="true" />
This worked fine (it removed the 31rst semi-colon from each line). Then I
wanted to add 00 if the 31rst column of my CSV file was missing, so I think
I tried (sorry I deleted it) :
<replaceregexp file="temptest.csv" match="^(([^;]*;){30})$" replace="\100"
byline="true" />
now if I put replace="\1'00'", it added '00' to the end of each line where
the 32rst column was missing, but I didn't want the single quotes, but with
just \100 (as in \1 and 00), it didn't work. I supposed it interpreted it as
"give me the 100th result".
Now I read somewhere that you could match a caracter by using it's octal
value as \nnn, or its hexadecimal value with \x, but this is supposedly for
matching, and not for the replacement.
As i don't know much about regular expressions, it's got me beat. Any ideas?
Roland.
-----Message d'origine-----
De : Stefan Bodewig [mailto:[EMAIL PROTECTED]]
Envoye : jeudi 26 septembre 2002 15:42
A : [EMAIL PROTECTED]
Objet : Re: Using ReplaceRegExp
On Tue, 24 Sep 2002, <[EMAIL PROTECTED]> wrote:
> Thanks for that. now I wanted to add a value at the end
>
> <replaceregexp file="temptest.csv" match="(.);$"
> replace="\1<a value>" byline="true" />
>
> now if i put replace="\100" to add a '00' (reusing the result '\1'),
> it doesn't work.
What is the result?
\1 will only match the very last character of the line (in front of
the ";"), you probably want to use (.*) instead.
Stefan
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