Re: [xsl] Replace Special Characters: return, blank and tab

Subject: Re: [xsl] Replace Special Characters: return, blank and tab
From: David Carlisle <davidc@xxxxxxxxx>
Date: Thu, 5 Aug 2004 23:25:05 +0100
> Otherwise, I don't know how to insert "<" and ">" without escaping them,

you don't want to insert those characters you want to insert a br node,
just put a br node as the content of with-param not using the select
attribute, as I showed.

> It is a good idea to match the "&#10;" too, but now I cannot even make the
> "&#13;" working

It is _most_ unlikely that there are any #13 in the XML file, the XML
spec specifies that all line endings (dos/mac/unix format) are all
reported as #10 howeverthey are in the source.

If you are losing element markup then usual reasons are applying
xsl:value-of  which gives the string value (or string ops such as
substring-before which also force a string value) or using the text
output method. Your replace template would get rid of any markup in its
input (as that passes through substring but I thought you had it nested
such that adding <br/> was the last thing you do.

Sorry its too late for debugging your code tonight, the faq for this
list has a replace template that does work, as does Jeni's site.

David


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