[XSL-LIST Mailing List Archive Home] [By Thread] [By Date]

group using element in two different places in xml

---

---

Hello
Currently I have a table which uses the xml and pseudo code shown below:

<a>
 <b>
  <c>
   <d>
   <data>
	<info>group1</info>
      <info2>two</info2>
      <info3>three</info4>
   </data>
   <data>
	<info>group3</info>
      <info2>four</info2>
      <info3>five</info4>
   </data>
   <data>
	<info>group1</info>
      <info2>two</info2>
      <info3>five</info4>
   </data>
   <data2>
	<info>group1</info>
      <info2>two</info2>
      <info3>four</info4>
   </data2>
   <data2>
      <info>group2</info>
      <info2>two</info2>
      <info3>five</info4>
   </data2>
   <data2>
	<info>group3</info>
      <info2>three</info2>
      <info3>four</info4>
   </data2>
    </d>
   </c>
 </b>
</a>

<xsl:key name="entries" match="data" use="info"/>
<xsl:key name="entries2" match="data2" use="info"/>
 <xsl:template match="a"/>
  <xsl:for-each
select="d/data[generate-id(.)=generate-id(key('entries',info))] |
d/data2[generate-id(.)=generate-id(key('entries2',info))] ">
   <xsl:for-each select="key('entries',info)">
   <!--display value of each unique <info> as header and other elements in
<data>--!>
   </xsl:for-each>
   <xsl:for-each select="key('entries2',info)">
   <!--display other each unique <info> as header and other elements in
<data2>--!>
   </xsl:for-each>
</xsl:for-each>
</xsl:template>

The pseudo code above produces a table displaying each <data> and <data2> in
whatever <info> group it has the same value as.  The problem with this
method is that it produces each <info> group twice as <xsl:for-each> adds
both d/data and d/data2.  How do I list each unique <info> and the <data> or
<data2> that corresponds with it without duplicating?

Thanks for any help.
James

---