CODEWITHSUNDEEP
xslt
srinannapa
srinannapa

Reputation: 3207

Sort Using XSLT

I need to sort the below xml by /parentList/employeesList/personalInfo/middleName

using XSLT. I'm unable to generate the format I'm expecting.The one with Middle name "John" should come up in the output xml. Can you someone help me on this. Below are my XML and XSL's

XML :

<parentList>

    <employeesList>
        <personalInfo>
            <middleName>Mike</middleName>
            <lastName>S</lastName>
        </personalInfo>
        <mailingAddress>
            <postalCode>12345</postalCode>
            <cityName>CoEmployee CityName</cityName>
            <state>PA</state>
            <addressLineText>CoEmployee Full Address</addressLineText>
        </mailingAddress>
    </employeesList>
    <employeesList>
        <personalInfo>
            <middleName>John</middleName>
            <lastName>G</lastName>
        </personalInfo>
        <mailingAddress>
            <postalCode>12345</postalCode>
            <cityName>CoEmployee CityName</cityName>
            <state>PA</state>
            <addressLineText>CoEmployee Full Address</addressLineText>
        </mailingAddress>
    </employeesList>
</parentList>

XSL:

 <xsl:template match="parentList">
    <xsl:copy>
        <xsl:apply-templates>

            <xsl:sort select="/employeesList/personalInfo/middleName" order="ascending"/> 
        </xsl:apply-templates>
    </xsl:copy>
</xsl:template>

Upvotes: 1

Views: 50

Answers (2)

helderdarocha
helderdarocha

Reputation: 23627

The problem is in the XPath expression that you are using in your select attribute. It's an absolute expression (starts with /) but there your root element is not employeesList.

When you match parentList, you apply templates to its children, which provide the context for the sort element. That means that a relative XPath expression in the select attribute of sort should be in the context of employeesList.

It should work if you simply change your template to this:

<xsl:template match="parentList">
    <xsl:copy>
        <xsl:apply-templates>
            <xsl:sort select="personalInfo/middleName" order="ascending"/> 
        </xsl:apply-templates>
    </xsl:copy>
</xsl:template>

Upvotes: 1

PhillyNJ
PhillyNJ

Reputation: 3901

You need to put the sort inside a for-each:

<xsl:stylesheet version="1.0" xmlns:xsl="http://www.w3.org/1999/XSL/Transform" xmlns:fo="http://www.w3.org/1999/XSL/Format">
<xsl:output method="xml" indent="yes"/>
    <xsl:template match="/">
            <xsl:apply-templates/>
    </xsl:template>
    <xsl:template match="parentList">
        <xsl:for-each select="employeesList">
            <xsl:sort select="personalInfo/middleName" order="ascending"/>
            <xsl:copy>
                <xsl:apply-templates/>
            </xsl:copy>
        </xsl:for-each>
    </xsl:template>
</xsl:stylesheet>

Upvotes: 1

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