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RE: <xsl:when test="position()%2=0">
Subject: RE: <xsl:when test="position()%2=0">
From: "Craig Kattner" <CKattner@xxxxxxxxxxxxxx>
Date: Mon, 21 Apr 2003 10:36:57 -0500
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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>
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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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