Subject: [xsl] Array of all X elements in XML From: Jorge <chocolate.camera@xxxxxxxxx> Date: Mon, 28 May 2012 00:21:23 +0200 |
Hi, When I learned that the Xpath path "A//B" returns all B nodes in A context, I assumed that appending an N index number would return the Nth node in that list. I.e. using the following snippet as the source XML to be transformed > <article> > <p>1st paragraph</p> > <p>2nd paragraph</p> > </article> > <article> > <p>3rd paragraph</p> > <p>4rth paragraph</p> > </article> I assumed //p[2] would return only "<p>2nd paragraph</p>", whatever the hierarchy looks like. I now see I was wrong, and instead returns all nodes on the Nth position in their respective tree level. (i.e. returns "<p>2nd paragraph</p>, <p>4rth paragraph</p>"). In order to mimic the behavior I originally expected, I can only think of creating a variable, assigning all p element nodes to it, and from then on use that variable to get the single Nth node inside the variable's context. I.e., if I wanted to get the "4rth paragraph" irrespectively of the tree, I could use this XSLT > <xsl:variable name="PARAGRAPHS" select="//p"/> > <copy-of> > <xsl:value-of select="$PARAGRAPHS/p[4]"/> > </copy-of> Is there any other way more straight?
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