[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: super basic xsl question
Don't have time for a long reply, but try xsl:copy-of Jay Bryant Bryant Communication Services (on contract at Synergistic Solution Technologies) Jeb Boniakowski <jeb@xxxxxxxxxxx> 01/13/2005 10:58 AM Please respond to xsl-list@xxxxxxxxxxxxxxxxxxxxxx To xsl-list@xxxxxxxxxxxxxxxxxxxxxx cc Subject super basic xsl question Thanks to everyone on the list that helped me with my last problem. I still have some fundamental misunderstanding of how xsl is supposed to work, because I can't figure out how to do this: <xmlroot> <child>some text</child> <child><a href="">a link</child> </xmlroot> and convert it to say... <ul> <li>some text</li> <li><a href="">a link</child> </ul> I've tried various combos like: <xsl:template match="child"> <xsl:value-of select="."/> </xsl:template> but that means I lose the <a> tags. I thought something with <xsl:copy> would work, but that gives me the context node, and furthermore, it doesn't give me the children, so it does the opposite. Then I thought fiddling with the 'select' expr in the value-of tag would do it, but I can't figure out if there's a magical combination of XPath slang that means, "whatever the hell is below here, be it tags or text, i want them". jeb.
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