Subject: RE: [xsl] sorting based on a variable From: John Pallister <jpallister@xxxxxxxxxxxx> Date: Tue, 12 Nov 2002 10:27:26 -0500 |
See FAQ at http://www.dpawson.co.uk/xsl/sect2/nono.html#d1243e407 Hope this helps, John Pallister jpallister@xxxxxxxxxxxx -----Original Message----- From: Kasper Nielsen [mailto:news@xxxxxx] Sent: Tuesday, November 12, 2002 9:34 AM To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx Subject: [xsl] sorting based on a variable hi (another question) lets say I want to some data for the $column data-column <projects> <project name="a"> <data> 1237</data> <data> 1234</data> <data> 1235</data> <data> 1236</data> </project> <project name="b"> <data> 12</data> <data> 41234</data> <data> 51235</data> <data> 71236</data> </project> <project name="c"> <data> 1</data> <data> 41234</data> <data> 51235</data> <data> 71236</data> </project> </projects> so if $column=1 then it would sort them into {project name="c", project name="b", project name="a"} however i've tried <xsl:for-each select="projects/project"> <xsl:sort select="./data[$column]" data-type="number"/> but it doesn't work? however if just use something like <xsl:sort select="./data[1]" data-type="number"/> it works fine anyone can tell me what im doing wrong? regards Kasper Nielsen XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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