Hello - I've got a doozy of an issue that I've been trying to work out for a couple days now (in the back of my head). I'd say I am a novice XSLT programmer. Here is a sample of the xml that I am transforming (i will only include a few important nodes):

<root>
    <Sections>
        <Entity>
            <EntityID/>183<EntityID>
            <Description/>description</Description>
        </Entity>
        <Entity>
            <EntityID/>184<EntityID>
            <Description/>description</Description>
        </Entity>
    </Sections>
    <SectionCounts>
        <Entity>
            <EntityID/>183<EntityID>
            <Count/>0</Count>
        </Entity>
        <Entity>
            <EntityID/>184<EntityID>
            <Count/>2</Count>
        </Entity>
    </SectionCounts>
</root>

I need to query out all /root/Section/Entities that have corresponding /root/SectionCounts/Entity/Count > 0 and then I need to do a for-each loop on them. After performing this join I need to run further xslt to format output. I can only run one transform and cannot change the given xml.

I tried to make this concise. If my description of the problem is not clear, please let me know.
Any advice is appreciated. Thanks!

You've got a bit muddled with your XML syntax - "/>" used in various inappropriate places.

You can do

<xsl:variable name="Sections" select="/root/Sections"/>
<xsl:variable name="SectionCounts" select="/root/SectionCounts"/>
<xsl:for-each select="$Sections/Entity[EntityId = $SectionCounts[Count != 0]/EntityId]">
  ...
</xsl:for-each>

You asked for an xsl:for-each solution, so I gave you one, but if you had a little more experience you would probably have asked for xsl:apply-templates, to make your code more modular.

You don't need either of the variables here, they are just for readability.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

This close but there is a bit of a problem. There may be more than one /root/SectionCounts/Entity with a count greater than zero. I will expand the set a bit to demonstrate the issue.

<root>
    <Sections>
        <Entity>
            <EntityID>183</EntityID>
            <Description>description</Description>
        </Entity>
        <Entity>
            <EntityID>184</EntityID>
            <Description>description</Description>
        </Entity>
        <Entity>
            <EntityID>185</EntityID>
            <Description>description</Description>
        </Entity>
    </Sections>
    <SectionCounts>
        <Entity>
            <EntityID>185</EntityID>
            <Count>9</Count>
        </Entity>
        <Entity>
            <EntityID>183<EntityID>
            <Count>0</Count>
        </Entity>
        <Entity>
            <EntityID>184</EntityID>
            <Count>2</Count>
        </Entity>
    </SectionCounts>
</root>

The test you offered before will return 185 every time and the 184 node will not be included in the for each loop. Any suggestions?

I feel quite silly for the malformed XML.

Yes, sorry I left out a step in the path, it should be

<xsl:variable name="Sections" select="/root/Sections"/>
<xsl:variable name="SectionCounts" select="/root/SectionCounts"/>
<xsl:for-each select="$Sections/Entity[EntityId = $SectionCounts/Entity[Count != 0]/EntityId]">
  ...
</xsl:for-each>

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

I believe that the right hand side of the outer predicate ($SectionCounts/Entity[Count != 0]/EntityId]) will always return the first node in its node set. For this reason, this for-each statement will return one node at the most. What I really need is something like this:
<xsl:for-each select="$Sections/Entity[EntityId [u]IN</u> $SectionCounts/Entity[Count != 0]/EntityId]">

Am I missing something? Please clarify.

>I believe that the right hand side of the outer predicate ($SectionCounts/Entity[Count != 0]/EntityId]) will always return the first node in its node set.

You believe wrong.

Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference

Oh goodness, it worked! Thanks very much.