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

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Subject: RE: <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: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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Current Thread
<xsl:when test="position()%2=0"> Luke Ambrogio - Mon, 21 Apr 2003 11:05:23 -0400 (EDT) Andrew Watt - Mon, 21 Apr 2003 11:51:56 -0400 (EDT) <Possible follow-ups> Craig Kattner - Mon, 21 Apr 2003 11:32:06 -0400 (EDT) <=

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