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[xsl] copying attributes
Subject: [xsl] copying attributes From: "Russ Holmes" <rholmes@xxxxxxxxx> Date: Tue, 9 Jan 2001 11:13:45 +1300 |
When converting the result of three SQL queries to xml I generate something like; <breakfasts> <life_to_date> <z breakfast="bacon" servings="2000"/> <z breakfast="eggs" servings="2000"/> <z breakfast="cereal" servings="7000"/> </life_to_date> <y2000> <z breakfast="bacon" servings="130"> <z breakfast="eggs" servings="100"> </y2000> <january> <z breakfast="eggs" servings="1"> </january> </breakfasts> ..you can assume that any 'breakfast' types in the y2000 and january nodes will show in the life_to_date node.. I would like to load this xml into a Data Source Object using the following format; <z breakfast="bacon" january_servings="" y2000_servings="130" life_to_date_servings="2000"/> <z breakfast="eggs" january_servings="1" y2000_servings="100" life_to_date_servings="2000"/> <z breakfast="cereal" january_servings="" y2000_servings="" life_to_date_servings="7000"/> Can I transform into this format? If so how? In trying to create the 'january_servings' node I've tried; <xsl:template match="/breakfasts/life_to_date/z"> <z> <xsl:for-each select="@breakfast" > <xsl:attribute name="breakfast"><xsl:value-of select="."/></xsl:attribute> <xsl:attribute name="january_servings"><xsl:value-of select="/breakfasts/january[@breakfast=.]/@servings"/> </xsl:attribute> </xsl:for-each> </z> </xsl:template> But with no success...could anyone help me out here? I'm trying to say for each life_to_date/z node, iterate through the 'breakfast' attributes and create new attributes with values that match the value of the corresponding 'january' node..where the 'breakfast' attibutes match.. Thanks heaps, Russ XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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