Subject: [xsl] Xsl:param from PHP to html From: "Alan Hale" <alan.hale@xxxxxxxxxxx> Date: Fri, 13 Apr 2007 16:03:18 +0100 |
I know I am doing something fundamentally wrong here, but I cannot understand what. I am trying to pass a parameter to a stylesheet from PHP, and use it to determine the sort order. I am sure the parameter is being assigned correctly in the PHP code, but when I run the xsl below, I always get the default value of $SORTBY. Many thanks Alan <html xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xsl:version='1.0'> <!--Define parameter name and set default value--> <xsl:param name = "SORTBY">species_name</xsl:param> <body> <xsl:for-each select="/fieldguide/account"> <xsl:sort select = "$SORTBY" order = "ascending"/> <td align = "center"><xsl:value-of select="species_name"/></td> <td width = "40" align = "center"><xsl:value-of select="has_text"/></td> <td width = "40" align = "center"><xsl:value-of select="has_photo"/></td> <td width = "40" align = "center"><xsl:value-of select="has_drawing"/></td> <td width = "40" align = "center"><xsl:value-of select="has_map"/></td> <td align = "right"><xsl:value-of select="format_date"/></td> </tr> </xsl:for-each> </table> </body> </html> -- No virus found in this outgoing message. Checked by AVG Free Edition. Version: 7.5.446 / Virus Database: 269.4.0/759 - Release Date: 12/04/2007 19:58
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