Subject: Re: [xsl] Newbie Q: Style sheet for simple transform From: Joerg Heinicke <joerg.heinicke@xxxxxx> Date: Wed, 05 Jun 2002 23:53:49 +0200 |
<xsl:template match="node()"> <xsl:copy> <xsl:copy-of select="@*"/> <xsl:apply-templates select="node()"/> </xsl:copy> </xsl:template>
I recieve a XML file structured like so:
<records>
<agencies>
<agency>
<name>Agency name</name> <description>test data - description of agency here</description>
<landmarks>landmark information here</landmarks>
<zips_served> <zip>32215</zip>
<zip>32202</zip>
</zips_served>
...(other data fields)
<custom_label1>customlabel1 name</custom_label1>
<custom_label2>customlabel2 name</custom_label2>
<custom_label3>label3</custom_label3>
<custom_label4>label4</custom_label4>
<custom_label5>label5</custom_label5>
...(other date fields)
</agency>
...(more agency records)
</agencies>
...(other types of records go here)
</records>
I would like to transform this file so that the custom_label fields are contained in a <custom_labels> record like so: ... <custom_labels> <custom_label1>customlabel1 name</custom_label1> <custom_label2>customlabel2 name</custom_label2> <custom_label3>label3</custom_label3> <custom_label4>label4</custom_label4> <custom_label5>label5</custom_label5> </custom_labels> ...
The rest of the file should just "pass through" unchanged. I'm positive this is possible using an XSL processor (in my case Xalan), and I'm fairly sure it will be very simple... But I'm having a hard time grasping exactly what the style sheet should look like.
TIA Richard Rowell richard@xxxxxxxxxxxxxxxxx
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