On 15/10/2013 17:44, Nathan Tallman wrote:
Thank you, David. Unfortunately, I still get ". after processing...
Presumably the template never matched.
(I'm transforming XML to HML, btw.)
Also, a question. Why didn't you need to escape the period in the
third argument of replace()?
You need \. to mean a literal . in a regular expression, not in the
replacement text.
> (When I inert one, my processor throws an
error; so, I understand it's wrong, but not why. Also, I seem to be
able to delete the backslash in front of the period in the second
argument without causing problems... Why is that?)
". is a legal regex but matches a " followed by any character
replace(.,'"."','."')
would change "x to ."
David
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