XSLT - Matching nodes that have that a certain node inside them (like jQuery ":has")

Question

Let's say I have the following XML file

<a id="123">
   <b type="foo" value="1" />
   <b type="baz" value="1" />
</a>
<a id="789">
  <b type="bar" value="12" />
</a>
<a id="999">
   <b type="foo", value="2" />
</a>

I want to get a list of all the 'a' nodes that have a 'b' subnode with a type="foo" and value="1". You can do something similar in jQuery with the ":has" selector.

For the record I'm planning on using xmlstarlet on the command line (but I'm not married to doing it that way), so a xslt that works that way would be best.

Answer 1

something like this:

a[b[@type='foo'][@value='1']]

should do the trick

Answer 2

<xsl:variable name="nodeList" select="a[b[@type='foo' and @value=1]]"/>

<xsl:for-each select="$nodeList">
    <xsl:value-of select="."/>
</xsl:for-each>

Answer 3

I think this would be: /b[@type='foo' and @value=1]/parent::a

Answer 4

This can be done using a single XPath expression as pointed out in gizmo's answer.

Because the question is specifically for XSLT, here is an efficient XSLT solution using keys:

<xsl:stylesheet version="1.0"
 xmlns:xsl="http://www.w3.org/1999/XSL/Transform"&gt;
 <xsl:output omit-xml-declaration="yes"/>
<!--                                      --> 
 <xsl:key name="kAByBTypeVal" 
  match="a"
  use="concat(b/@type,'+',b/@value)"/>
<!--                                      -->       
    <xsl:template match="/">
      <xsl:copy-of select=
       "key('kAByBTypeVal', 'bar+12')"/>
    </xsl:template>
</xsl:stylesheet>

When the above transformation is applied on this XML document:

<t>
    <a id="123">
     <b type="foo" value="1" />
     <b type="baz" value="1" />
    </a>
    <a id="789">
     <b type="bar" value="12" />
    </a>
    <a id="999">
     <b type="foo" value="2" />
    </a>
</t>

the correct result is produced:

<a id="789">
  <b type="bar" value="12"/>
</a>