RE: [xsl] <xsl:sort> question.

Subject: RE: [xsl] <xsl:sort> question.
From: "Michael Kay" <michael.h.kay@xxxxxxxxxxxx>
Date: Fri, 19 Apr 2002 18:20:02 +0100
>
>    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!!
>
> **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>
>
<xsl:for-each select="."> selects a single node and then sorts it, which
isn't a very useful thing to do.

Michael Kay
Software AG
home: Michael.H.Kay@xxxxxxxxxxxx
work: Michael.Kay@xxxxxxxxxxxxxx


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