Subject: [xsl] sort order="descending" question From: "Jessica P. Hekman" <jphekman@xxxxxxxxxxxx> Date: Thu, 7 Aug 2003 16:49:28 -0400 (EDT) |
I'm getting some weird behavior, and I've tried using three different XSLT processors, all of which do the same thing; so it must be me misunderstanding XSLT. Basically, I have a list of elements in the input document, and I'm trying to reverse their order in the output document. Input: <bar> <baz>1</baz> <baz>2</baz> <baz>3</baz> <baz>4</baz> <baz>5</baz> <baz>6</baz> <baz>7</baz> <baz>8</baz> <baz>9</baz> <baz>10</baz> <baz>11</baz> <baz>12</baz> </bar> XSL: <xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="1.0"> <xsl:template match="/"> <bar> <xsl:apply-templates select="/bar/baz"> <xsl:sort select="position()" order="descending"/> </xsl:apply-templates> </bar> </xsl:template> <xsl:template match="baz"> <baz><xsl:apply-templates/></baz> </xsl:template> </xsl:stylesheet> Output: <bar> <baz>9</baz> <baz>8</baz> <baz>7</baz> <baz>6</baz> <baz>5</baz> <baz>4</baz> <baz>3</baz> <baz>2</baz> <baz>12</baz> <baz>11</baz> <baz>10</baz> <baz>1</baz> </bar> So I'd expect the output to be 12, 11, 10, 9, 8... But it's out of order. If I have only 9 elements, they are output in order; it's when I add the tenth that this misordering starts. All processors I tried (Sablotron, xsltproc, and Xalan) produced exactly the same output. Does anyone know what's going on? Thanks very much, Jessica XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
Current Thread |
---|
|
<- Previous | Index | Next -> |
---|---|---|
Re: [xsl] Finding last node in recu, Jeff Kenton | Thread | Re: [xsl] sort order="descending" q, Abhijit Junnare |
RE: [xsl] Sorting Upper-Case first., David Carlisle | Date | RE: [xsl] XPATH question, Michael Kay |
Month |