Subject: Re: [xsl] 10 records per page From: "J.Pietschmann" <j3322ptm@xxxxxxxx> Date: Mon, 03 Feb 2003 23:09:51 +0100 |
I have common question. The question is how to show 10 records at a time and also show navigation links i.e page1 page2 which will show the other records. I have checked the example at http://www.dpawson.co.uk/xsl/sect2/sect21.html but can't see how it make it work in mine. I am very new to xsl!
This is mine: (It shows the first 10 records)!
<xsl:for-each select="feedbackDetailsTag[position() mod 10 = 1]"> <tr> <td width="11%" height="18" bgcolor="#FFFFFF" class="bold">
... You need two nested for-each, the outer one selecting the groups, the inner one selecting the elements within the group. You've only allpied the outer. Try <xsl:for-each select="feedbackDetailsTag[position() mod 10 = 1]"> <xsl:for-each select=".|following-sibling::*[position() < 10]"> <tr> <td width="11%" height="18" bgcolor="#FFFFFF" class="bold"> ...
BTW please don't start a new thread by replying to some other message.
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