RE: [xsl] How to find all parents having the same child?

Subject: RE: [xsl] How to find all parents having the same child?
From: Jarno.Elovirta@xxxxxxxxx
Date: Wed, 18 Jun 2003 11:12:15 +0300

> Here an example:
> <Quries>
>  <Query name="query1">
>    <Query name="query2">
>      <Table name="table1"/>
>      <Table name="table2"/>
>      <Query name="query3">
>        <Table name="table3"/>
>      </Query>
>    </Query> 
>  </Query>
>  <Query name="query4">
>    <Table name="table1"/>
>    <Table name="table2"/>
>    <Table name="table3"/>
>  </Query> 
> </Queries>
> Given that, I would like to extract all parents for
> each leaf like the following:
> table3: query4, query3
> table1: query4, query1

did you mean

  table1: query4, query2

as query1 is not a direct parent of table1.

> query3: query2
> query2: query1
> An example from you on how to get that cross-reference
> list would help me a lot.

Hopefully this will get you started.

  <xsl:key name="x" match="Table | Query" use="@name"/>
  <xsl:output method="text"/>
  <xsl:template match="Queries">
    <xsl:for-each select="descendant::*[generate-id(.) = generate-id(key('x', @name))]">
      <xsl:if test="key('x', @name)[parent::Query]">
        <xsl:value-of select="@name"/>: <xsl:text/>
        <xsl:for-each select="key('x', @name)">
          <xsl:if test="not(position() = 1)">, </xsl:if>
          <xsl:value-of select="../@name"/>


Jarno - Chris C: Vengeance Is Mine

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