'Convert XML Attributes To Elements XSLT
I am trying to convert attributes to sub-elements, ie turn the following:
<WP featured="yes" player="no" dancers="no" series="logos" archive="no" fanart="no" id="eclipse_logos_">
<seriesName>LOGOS</seriesName>
<selection>ECLIPSE</selection>
<imgurl>http://www.nba.com/warriors/photos/eclipse_logos_</imgurl>
<res>1024x1024r(iPad/iPhone)?1280x1024r(Regular)?1440x900r(Widescreen)?1920x1080r(HDTV)?1920x1200r(Widescreen)</res>
</WP>
Into:
<WP>
<featured>yes</featured>
<player>no</player>
<dancers>no</dancers>
<series>logos</series>
<archive>no</archive>
<fanart>no></fanart>
<id>eclipse_logos_</id>
<seriesName>LOGOS</seriesName>
<selection>ECLIPSE</selection>
<imgurl>http://www.nba.com/warriors/photos/eclipse_logos_</imgurl>
<res>1024x1024r(iPad/iPhone)?1280x1024r(Regular)?1440x900r(Widescreen)?1920x1080r(HDTV)?1920x1200r(Widescreen)</res>
</WP>
Solution 1:[1]
Try this:
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:msxsl="urn:schemas-microsoft-com:xslt" exclude-result-prefixes="msxsl"
>
<xsl:output method="xml" indent="yes"/>
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="@*">
<xsl:element name="{name()}"><xsl:value-of select="."/></xsl:element>
</xsl:template>
</xsl:stylesheet>
Solution 2:[2]
dradu's code is generic, the transformation will be applicable to all the attributes, below code is more specific about WP element: only those attributes coming under WP element will be converted to elements.
<?xml version="1.0" encoding="utf-8"?>
<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" indent="yes"/>
<xsl:template match="@* | node()">
<xsl:copy>
<xsl:apply-templates select="@* | node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="WP/@*">
<xsl:element name="{name()}">
<xsl:value-of select="."/>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | dan radu |
| Solution 2 | InfantPro'Aravind' |
