Reputation: 1723
I have a source XML file that I want to convert into another XML file with a different format. My source XML file uses attributes (I think, correct me if I'm wrong here) and the target XML format should be plain parent-child XML tags. I have no issues with a straight forward approach.
However, I need to find a way to concatenate 2 or more fields into a single element using XSLT. I tried using variables then concatenating them however, I am having some issues with variables and they can only be assigned once.
My XML is like this:
<?xml version="1.0" encoding="UTF-8"?>
<Document xmlns="http://www.taleo.com/ws/integration/toolkit/2005/07" xmlns:ns1="http://www.taleo.com/ws/integration/toolkit/2005/07" xmlns:soapenv="http://schemas.xmlsoap.org/soap/envelope/" xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"><Attributes><Attribute name="count">1</Attribute><Attribute name="duration">0:00:00.109</Attribute><Attribute name="entity">SourcingRequest</Attribute><Attribute name="mode">XML</Attribute><Attribute name="version">http://www.taleo.com/ws/tee800/2009/01</Attribute></Attributes><Content>
<ExportXML xmlns="http://www.taleo.com/ws/integration/toolkit/2005/07">
<record>
<field name="Title">Project Analyst</field>
<field name="Position_ID">64057</field>
<field name="RequisitionNumber">180767</field>
<field name="LocationCode">HQ</field>
<field name="LocationName">Headquarters</field>
<field name="LocationCountry">Country</field>
<field name="LocationCity">City</field>
</record>
</ExportXML></Content></Document>
Then my XSL is like this:
<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform" version="2.0" xmlns:itk="http://www.taleo.com/ws/integration/toolkit/2005/07">
<xsl:output method="xml" encoding="UTF-8" omit-xml-declaration="yes" cdata-section-elements="DescriptionInternal DescriptionExternal QualificationExternal QualificationInternal"/>
<xsl:template match="text()"/>
<xsl:template match="itk:record">
<xsl:element name="Job">
<xsl:for-each select="itk:field">
<!-- Set variables -->
<xsl:variable name="location_code">
<xsl:if test="@name='LocationCode'">
<xsl:value-of select="." />
</xsl:if>
</xsl:variable>
<xsl:variable name="location_name">
<xsl:if test="@name='LocationName'">
<xsl:value-of select="." />
</xsl:if>
</xsl:variable>
<xsl:variable name="location_country">
<xsl:if test="@name='LocationCountry'">
<xsl:value-of select="." />
</xsl:if>
</xsl:variable>
<xsl:variable name="location_city">
<xsl:if test="@name='LocationCity'">
<xsl:value-of select="." />
</xsl:if>
</xsl:variable>
<xsl:choose>
<xsl:when test="@name='LocationCode' or @name='LocationName' or @name='LocationCountry' or @name='LocationCity'">
<xsl:if test="@name='LocationCity'">
<xsl:element name="Location">
<xsl:value-of select="concat($location_code,$location_name,$location_country,$location_city)" />
</xsl:element>
<xsl:element name="LocationCode">
<xsl:value-of select="$location_code" />
</xsl:element>
<xsl:element name="LocationName">
<xsl:value-of select="$location_code" />
</xsl:element>
</xsl:if>
</xsl:when>
<xsl:otherwise>
<!-- Set variables. -->
<xsl:variable name="NodeName" select="@name"/>
<xsl:element name="{@name}">
<xsl:value-of select="."/>
</xsl:element>
</xsl:otherwise>
</xsl:choose>
</xsl:for-each>
</xsl:element>
</xsl:template>
</xsl:stylesheet>
However, my output is like this:
<Job>
<Title>Project Analyst</Title>
<Position_ID>64057</Position_ID>
<RequisitionNumber>180767</RequisitionNumber>
<Location>City</Location>
</Job>
But my desired output should be like this:
<Job>
<Title>Project Analyst</Title>
<Position_ID>64057</Position_ID>
<RequisitionNumber>180767</RequisitionNumber>
<Location>HQ Headquarters Country City</Location>
</Job>
If my guess is right, I think the loop resets the values of variables. So whatever was last will be the value.
Is there any other way I can achieve the desired output. My options are limited to XSL although my source file can be changed into CSV format.
Appreciate any help.
Upvotes: 2
Views: 722
Reputation: 117165
Is there a reason why you can't do simply:
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xmlns:itk="http://www.taleo.com/ws/integration/toolkit/2005/07"
exclude-result-prefixes="itk">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" omit-xml-declaration="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/itk:Document">
<xsl:apply-templates select="itk:Content/itk:ExportXML/itk:record"/>
</xsl:template>
<xsl:template match="itk:record">
<Job>
<Title>
<xsl:value-of select="itk:field[@name='Title']" />
</Title>
<Position_ID>
<xsl:value-of select="itk:field[@name='Position_ID']" />
</Position_ID>
<RequisitionNumber>
<xsl:value-of select="itk:field[@name='RequisitionNumber']" />
</RequisitionNumber>
<Location>
<xsl:value-of select="itk:field[@name='LocationCode']" />
<xsl:text> </xsl:text>
<xsl:value-of select="itk:field[@name='LocationName']" />
<xsl:text> </xsl:text>
<xsl:value-of select="itk:field[@name='LocationCountry']" />
<xsl:text> </xsl:text>
<xsl:value-of select="itk:field[@name='LocationCity']" />
</Location>
</Job>
</xsl:template>
</xsl:stylesheet>
Or, if using XSLT 2.0, even simpler:
<xsl:stylesheet version="2.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
xpath-default-namespace="http://www.taleo.com/ws/integration/toolkit/2005/07" >
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes" omit-xml-declaration="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/Document">
<xsl:apply-templates select="Content/ExportXML/record"/>
</xsl:template>
<xsl:template match="record">
<Job>
<xsl:for-each select="field[not(starts-with(@name, 'Location'))]">
<xsl:element name="{@name}">
<xsl:value-of select="." />
</xsl:element>
</xsl:for-each>
<Location>
<xsl:value-of select="field[starts-with(@name, 'Location')]" />
</Location>
</Job>
</xsl:template>
</xsl:stylesheet>
Upvotes: 1