That should work, I use the following regularly:
<xsl:if test="position() mod 2 = 1">
<xsl:attribute name="style">background-color: #DDDDDD;</xsl:attribute>
</xsl:if>
-----Original Message-----
From: Luke Ambrogio [mailto:gryzlaw(_at_)hotmail(_dot_)com]
Sent: Monday, April 21, 2003 10:09 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [xsl] <xsl:when test="position()%2=0">
first of all i wish to thank all of you guys (and gals) for providing me with
great help. now i have another query
i would like to find the position of a node whether it is odd or even, and
tried to use position() mod 2, but i don't know how to do it
can anyone tell me how to find the remainder of a division or else any ideas on
how to find whether a node is odd or even
10x, regards
luke
____________________________________________________
<http://www.incredimail.com/redir.asp?ad_id=309&lang=9> IncrediMail - Email
has finally evolved - <http://www.incredimail.com/redir.asp?ad_id=309&lang=9>
Click Here
