Subject: Re: [xsl] copy-of question From: Jeni Tennison <mail@xxxxxxxxxxxxxxxx> Date: Thu, 19 Apr 2001 19:19:40 +0100 |
Hi Franz, > Jeni, I don't have a deep copy here (I want the complete original > html table inside the xml), but only the table text cells as output Sorry - I don't understand what you mean by the above. Let me try again. You said: > how can I extract only one element in original form? > copy-of seems to work here, but I have all the other unwanted tags also as > plain text. > Is this related to the default rules? > > <xsl:template match="TABLE"> > <xsl:copy-of select="."/> > </xsl:template> If this is the only template you have in your stylesheet, it will give you a (deep) copy of any TABLE elements in your document. The rest of the nodes in the stylesheet (aside from those under the TABLE elements) will have the built-in templates applied to them. The effect of these built-in templates is that you get the text of the rest of the document output. If you don't want that text to be output, then adding the empty template: <xsl:template match="text()" /> will stop it being output. You'll still get the (deep) copy of the TABLE elements, but this seemed to be what you were after (you said that was working). If you have a stylesheet with these two templates in it, and you're only getting the text of the table cells in your output, then I bet you've got an xsl:output instruction that looks like: <xsl:output method="text" /> Take out that xsl:output instruction, or replace it with one that looks like: <xsl:output method="xml" /> If you use the 'text' method, the XSLT processor doesn't output any elements that you create, only text nodes. I hope that helps, Jeni --- Jeni Tennison http://www.jenitennison.com/ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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