RE: [xsl] how to convert from list to grid ?

["Michael Kay" <mike@xxxxxxxxxxxx>]

In 2.0, with <list> as the context node, it's something like this:

<xsl:variable name="list" select="."/>
<table>
<xsl:for-each select="1 to max(item/@x)">
  <xsl:variable name="x" select="."/>
  <tr>
  <xsl:for-each select="1 to max(item/@y)">
    <xsl:variable name="y" select="."/>
    <td>
       <xsl:value-of select="count($list/item[@x=$x and @y=$y])"/>
    </td>
  </xsl:for-each>
  </tr>
</xsl:for-each>
</table>

In 1.0, you can replace the for-each loops with a recursion, or with
something like

<xsl:for-each select="//node()[position() &lt;= $max-x]">

Michael Kay
http://www.saxonica.com/


> -----Original Message-----
> From: Bru, Pierre [mailto:Pierre.Bru@xxxxxxxxxxxx] 
> Sent: 30 October 2005 22:48
> To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx
> Subject: [xsl] how to convert from list to grid ?
> 
> hello,
>  
> I've been racking my brains for a couple of weeks now but I 
> can not find
> a way to do the folowing: I have a list reprenting the type 
> and integer
> coordinate of various items on a square grid like this one:
>  
> <list>
>   <item id="1" type="foo" x="1" y="7" />
>   <item id="2" type="bar" x="8" y="2" />
>   <item id="3" type="foo" x="8" y="2" />
>   <item id="4" type="bar" x="8" y="3" />
>   <item id="5" type="bar" x="8" y="3" />
>   <item id="6" type="bar" x="8" y="3" />
>   ...
> </list>
>  
> - "id" are unique
> - each cell may contain any number of item (including 0).
> - there are as many lines as needed for each cell (see [8,2] or [8,3])
> - the cell are in no particular order for now but this can be 
> changed as
> needed
>  
> I would like to transform this list into a square grid (table?) with :
> - if the cell contains no item, an empry cell
> - otherwise, the count of item(s) of each type (one value for 
> each type
> present on the cell)
>  
> is this feasable with XSLT ?
>  
> TIA,
> Pierre.