[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] Re: Array of all X elements in XML
The precise scenarios you mention are explained and distinguished in Roger Costello's XPath 1.0 tutorial which you can get at www.xfront.com. On Sun, May 27, 2012 at 11:21 PM, Jorge <chocolate.camera@xxxxxxxxx> wrote: > 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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