xsl-list
[Top] [All Lists]

show if there is more than one node

2004-05-01 11:27:04
heloow.
i have this xml:
<?xml version="1.0" encoding="ISO-8859-1" standalone="yes"?>
<sn_servicioproyecto>
  <sn_masterinformacion>
    <paginas>
      <sn_paginas>
        <pagina_id><![CDATA[1691]]></pagina_id>
      </sn_paginas>
      <sn_paginas>
        <pagina_id><![CDATA[1692]]></pagina_id>
      </sn_paginas>
      <sn_paginas>
        <pagina_id><![CDATA[1693]]></pagina_id>
      </sn_paginas>
    </paginas>
  </sn_masterinformacion>
</sn_servicioproyecto>

whith:
<?xml version="1.0" encoding="ISO-8859-1"?>
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform";>
<xsl:output method="html" version="1.0" encoding="ISO-8859-1" indent="yes"
omit-xml-declaration="yes"/>
<xsl:strip-space elements="*"/>
<xsl:template match="/">
<xsl:for-each
select="sn_servicioproyecto/sn_masterinformacion/paginas/sn_paginas">
<xsl:_value-of select="pagina_id"/>
</xsl:for-each>

i can print all the pagina id   values.
but i have one problem, if only there is one sn_paginas/pagina_id i dont
want to print it, only i want to print them if there are more than one node.
how can i make it?
can you help me?
thanks




<Prev in Thread] Current Thread [Next in Thread>