Subject: Re: [xsl] Changing table output from horizontal to vertical From: "Mark" <mark@xxxxxxxxxxxx> Date: Sun, 18 Sep 2011 11:08:54 -0700 |
I'll try it. Go raibh maith agat Mark
Hi, I am using an XSLT template to produce an XHTML table.
Currently, the stylesheet fills the table in a horizontal direction; I would like to fill the table in a vertical direction.
Instead of thinking of filling a table, think of it as a mapping from the input tree to the output grid... Each output tr will start with the nth input item and then have the n + NRth item, then n + NR * 2, and so on, where NR is the number of rows (rounded up to the nearest integer if there's a partial row).
You can work out the total number of rows (I'd store it in an XSLT variable for clarity) e.g. in the template for List you could do <xsl:template match="List"> <xsl:variable name="rowcount" select="(ncount(Item[@lang eq $langcode]) + 2) div 3" as="xs:integer" /> (if you are still in 1998 and XSLT 1, leave of the "as" attribute and use = instead of eq)
<!--* now you can map the Items to table cells. *--> <xsl:for-each select="Item[@lang eq $langcode][position() <= $rowcount]"> <xsl:apply-templates select="." /> </xsl:for-each> </xsl:template>
In this example $langcode is a global parameter to the stylesheet, but to handle both languages make it a tunnel parameter, or in XSLT 1 make it a literal parameter and add it as an argument to the List template, or if List generates the tables, you could instead do, <xsl:for-each select="distinct-values(Item/@lang)"> generate a table in here </xsl:for-each> perhaps with an xsl:sort instruction in there :)
-- Liam Quin - XML Activity Lead, W3C, http://www.w3.org/People/Quin/ Pictures from old books: http://fromoldbooks.org/
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