Reputation: 2228
Please i'm beginner in XSLT. Can some one explain for me how transform this XMI file to XML file ?
<?xml version="1.0" encoding="UTF-8"?>
<projet:Config xmi:version="2.0" xmlns:xmi="http://www.omg.org/XMI" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xmlns:test="http://projet.org/test" xmlns:occi="http://schemas.ogf.org/projet">
<use href="extensions/plugin.xmi#/"/>
<group id="group1">
<type href="extensions/plugin#//[term='host']"/>
<variable name="v1" value="x86"/>
<variable name="v2" value="Linux"/>
<variable name="v3" value="Xen"/>
</group>
</projet:Config>
to tansform the previous XMI to xml file such as:
<host v1="x86" v2="Linux" v3="Xen">
Upvotes: 1
Views: 599
Reputation: 21641
There are many many ways to do this. Here's one: Start with a template to match the root node to give you a valid root tag (this is assuming your group
nodes are repeatable); have a template below that which matches the group
nodes, and a template below that that matches the variable
nodes in that group. Build your attributes in the variable
template.
Note that your input XML is invalid and may cause problems - that projet
prefix is undeclared, which is why I'm referencing it using the local-name()
syntax.
<?xml version="1.0" encoding="UTF-8" ?>
<xsl:transform xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0">
<xsl:output method="xml" omit-xml-declaration="yes" encoding="UTF-8" indent="yes" />
<xsl:template match="/">
<root>
<xsl:apply-templates />
</root>
</xsl:template>
<xsl:template match="/*[local-name()='Config']/group">
<xsl:element name="{type/substring-before(substring-after(@href,'term='''), ''']')}">
<xsl:apply-templates />
</xsl:element>
</xsl:template>
<xsl:template match="variable">
<xsl:attribute name="{@name}"><xsl:value-of select="@value" /></xsl:attribute>
</xsl:template>
<xsl:template match="text()" />
</xsl:transform>
Upvotes: 1