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 XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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