[xsl] <xsl:sort> question.

Subject: [xsl] <xsl:sort> question.
From: "Dung, Ming-tzung" <Ming-tzung_Dung@xxxxxxxxxxxxx>
Date: Fri, 19 Apr 2002 10:54:24 -0600
   The version 1 will work and the version 2 will not sort the data, even
though I think that these two are logically equivalent.
  Please let me know what you think?  Thanks in advance!!
------
**Version1 - output the sorted last name**
<xsl:template match="/">
     <xsl:for-each select="addressbook/address">
      <xsl:sort select="name/last-name"/>
      <p>
      <xsl:value-of select="name/last-name"/>      
      </p>      
    </xsl:for-each>
</xsl:template>

**Version 2 - output the sorted last name**
  <xsl:template match="/addressbook/address">
    <xsl:for-each select=".">
      <xsl:sort select="name/last-name"/>
      <p>
      <xsl:value-of select="name/last-name"/>      
      </p>      
    </xsl:for-each>
  </xsl:template>

**Input xml data
-----
<?xml version="1.0"?>
<addressbook>
  <address>
    <name>
      <title>Mr.</title>
      <first-name>Chester Hasbrouck</first-name>
      <last-name>Frisby</last-name>
    </name>
  </address>
  <address>
    <name>
      <first-name>Harry</first-name>
      <last-name>Backstayge</last-name>
    </name>
  </address>
  <address>
    <name>
      <first-name>Mary</first-name>
      <last-name>McGoon</last-name>
    </name>
  </address>
  <address>
    <name>
      <title>Ms.</title>
      <first-name>Amanda</first-name>
      <last-name>Reckonwith</last-name>
    </name>
  </address>
</addressbook>
-----

Ming

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