Re: Node set
Hi Yue, > The reason I asked about ordered node set was that I have to group > those title elements in the sample XML... (Now I'm clearly > understand thata node set is not ordered.... ;-)) Say I may have 8 > or 10 title elements and finally, I'm going to display them within > an HTML table, 4 on the left and 4 on the right column....without > specific ordering, it's easy. And the bad news is the rule used to > grouping them is "logic", so looks like nothing I can follow.... and > as I said, the total number of title element is not guaranteed, if > 4th element was missing, I have to move the 5th element to the left > column! ;( There's an XSLT solution to this kind of 'grouping by position'. You pick out the 'odd' title elements, apply templates to them, and get them to output a row containing themselves and their 'even' partner (if there is one). Picking out the odd titles involves looking at their position() and doing a calculation on it - if you do position() mod 2, you'll get 1 for odd items and 0 for even items - so you can pick out your odd title elements with: <xsl:apply-templates select="title[position() mod 2 = 1]" /> Now have a template that matches title elements (it'll only be applied to the odd ones) and outputs a row containing themselves and their immediately following title element: <xsl:template match="title"> <tr> <td><xsl:value-of select="." /></td> <td><xsl:value-of select="following-sibling::title" /></td> </tr> </xsl:template> This might not do precisely what you want it to do, but the principle's there. If you give a better sample of what your source XML looks like (and what you want the table to look like in the end) then we can probably give a more specific solution if you need it. 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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