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 matching, but I am stuck. Any help would be appreciated.

The input xml :

   <Cars>
      <Car>
        <Brand>Nisan</Brand>
        <Price>12</Price>
     </Car>
     <Car>
        <Brand>Lawrence</Brand>
     </Car>
     <Car>
       <Brand>Cinrace</Brand>
       <Price>14</Price>
     </Car>
   </Cars>

I would like to delete the Car which doesn't have the price element within it. So the expected output is :

 <Cars>
      <Car>
        <Brand>Nisan</Brand>
        <Price>12</Price>
     </Car>
     <Car>
       <Brand>Cinrace</Brand>
       <Price>14</Price>
     </Car>
   </Cars>

I tried using this :

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
 <xsl:output method="xml" indent="yes"/>
<xsl:strip-space elements="*" />
 <xsl:output omit-xml-declaration="yes"/>

    <xsl:template match="node()|@*">
      <xsl:copy>
         <xsl:apply-templates select="node()|@*"/>
      </xsl:copy>
    </xsl:template>
<xsl:template match="Cars/Car[contains(Price)='false']"/>
</xsl:stylesheet>

I know the XSLT is totally wrong please advice.

UPDATE

Corrected one which works :)

<?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>

Answer 1

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"/>