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RE: [xsl] <xsl:when test="position()%2=0">

Subject: RE: [xsl] <xsl:when test="position()%2=0">
From: "Craig Kattner" <CKattner@xxxxxxxxxxxxxx>
Date: Mon, 21 Apr 2003 10:36:57 -0500
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@xxxxxxxxxxx]
Sent: Monday, April 21, 2003 10:09 AM
To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
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
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