Re: getting xsl to produce ill-formed xml?
I got several suggestions for this problem: 1. (John Simpson) Generate a CDATA block. I don't see how this can be done (Oren seems to confirm this in his <Opinion> element). 2. (Duane Nickul) Post-process with perl. But that is what I was already doing. It is the principle of the thing. If I was going to be using perl, this little exercise would have been done long ago. I wanted an excuse to play with xsl. 3. (Oren Ben-Kiki) Use <SCRIPT>. This works, but now I don't get the exact output I wanted -- I've got this "# <SCRIPT>" in the perl program itself. It is like letting xsl win; not acceptable :). (Again, I've already got a working system... a reasonable definition of a hacker is someone who takes a working system and "improves" it....) 4. (Duane Nickul) Use an entity. This almost works. If I put this in my style sheet: <!DOCTYPE xsl:stylesheet SYSTEM "xsl.dtd" [ <!ENTITY plarrow "=>"> ]> <xsl:stylesheet xmlns:xsl="http://www.w3.org/TR/WD-xsl"> ... <xsl:text> < &plarrow; </xsl:text> Then I get this in the output: < => I might note that this also holds if I use this stylesheet declaration: <xsl:stylesheet xmlns:xsl="http://www.w3.org/TR/WD-xsl" xmlns:html="http://www.w3.org/TR/REC-html40" result-ns="html"> In other words, I can't seem to get xt to expand all entities I insert (like "<"), and worse yet, it seems to want to convert any ">" in sight to a ">". I don't know if this is an xsl thing or an xt thing. (In general, I've yet to find that xt does much different based on the stylesheet declaration except maybe put a line like this across the top of the output: <!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN"> .) -mda XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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