Subject: RE: [xsl] how to convert from list to grid ? From: "Michael Kay" <mike@xxxxxxxxxxxx> Date: Sun, 30 Oct 2005 23:48:16 -0000 |
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() <= $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.
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