This is the mail archive of the
xsl-list@mulberrytech.com
mailing list .
sort problem
- To: XSL-List at mulberrytech dot com
- Subject: sort problem
- From: Claudio Lo Piccolo <lopi-out at katamail dot com>
- Date: 24 Jul 2000 13:51:16 +0200
- Reply-To: xsl-list at mulberrytech dot com
Hi, I'm a very novice for both xsl and this list.
I'm given the following xml:
<parade>
<personaggio punteggio="14">
<nome>
Carolina di Monaco
</nome>
<tendenza>
salita
</tendenza>
</personaggio>
<personaggio punteggio="10">
<nome>
Luca Cordero di Montezemolo
</nome>
<tendenza>
salita
</tendenza>
</personaggio>
</parade>
I can list the items with the following xsl:
<xsl:stylesheet xmlns:xsl="http://www.w3.org/TR/WD-xsl">
<xsl:template match="/">
<TABLE WIDTH="316" CELLSPACING="0" CELLPADDING="0" BORDER="0">
<tr>
<td align="center" valign="top">Posizione</td>
<td align="center" valign="top">Personaggio</td>
<td valign="top">Apparizioni</td>
<td valign="top">Tendenza</td>
</tr>
<xsl:apply-templates select="parade/personaggio" />
</TABLE>
</xsl:template>
<xsl:template match="personaggio">
<TR>
<td align="center"></td>
<TD><xsl:value-of select="nome" /></TD>
<TD align="center"><xsl:value-of select="@punteggio" /></TD>
<TD align="center"><xsl:value-of select="tendenza" /></TD>
</TR>
</xsl:template>
</xsl:stylesheet>
but I don't know how to sort the items by the field punteggio
Thanks for your attention
___________________________________________________
Se vuoi un indirizzo di posta elettronica gratuito,
iscriviti a http://www.katamail.com
XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list