deleting the parent node if child node is not present in xml using xslt

Question

I am trying to transform a given XML using xslt. The caveat is that I would have to delete a parent node if a given child node is not present. I did do some template matchin

Answer 1

A different solution is to use the 'xsl:copy-of' element.

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">

    <xsl:output method="xml" />

    <xsl:template match="Cars">
        <xsl:copy>
            <xsl:copy-of select="Car[Price]" />
        </xsl:copy>
    </xsl:template>

</xsl:stylesheet>

Answer 2

Super close. Just change your last template to:

<xsl:template match="Car[not(Price)]"/>

Also, it's not incorrect but you can combine your 2 xsl:output elements:

<xsl:output method="xml" indent="yes" omit-xml-declaration="yes"/>

Answer 3

<?xml version="1.0" encoding="UTF-8"?>
<xsl:stylesheet xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    version="1.0">

    <!--Identity template to copy all content by default-->
    <xsl:template match="node()|@*">
        <xsl:copy>
            <xsl:apply-templates select="node()|@*"/>
        </xsl:copy>
    </xsl:template>


    <xsl:template match="Car[not(Price)]"/>

</xsl:stylesheet>