Subject: Re: [xsl] sorting by comparing two nodes From: Ragulf Pickaxe <ragulf.pickaxe@xxxxxxxxx> Date: Tue, 20 Dec 2005 08:39:56 +0100 |
> I'm stuck with a sorting problem and hope to find some help here :-) > > I have the following input: > > <points>10 13,20 34,2 10,40 67</points> I am not all that clear on how you would sort these. As far as I know, even in XSLT 2.0, you can only sort on a nodeset. It seems to me that you want to sort on the comma separed list in the points element. My scetch of a solution is an XSLT 1.0, which is what I know. This may inspire you to a 2.0 solution. In a two-phase transformation, I would transform your vectors to a new XML with: <points> <point><x>10</x><y>13</y> <point><x>20</x><y>34</y> <point><x>2</x><y>10</y> <point><x>40</x><y>67</y> </points> This I would then transform to the output that you want. Your expression seems more like the javaScript way of making a sort algorithm. This is not necessary here. You will need an expression that computes the distance from a given point to the reference. The sorting function of XSLT will do the comparison between different nodes: expr:= sqrt((x-$ref/x)^2+(y-$ref/y)^2) I do not know whether these functions (sqr and sqrt) are included in XSLT 2.0. If not, you may want to look up FXSL to help you (also the case if you use 1.0). (I have also split the reference point into its X an Y parts) Then you can have a template which gives: <xsl:for-each select="point"> <xsl:sort select="expr" data-type="number"/> <!-- Put the expression here --> <xsl:value-of select="x"/><xsl:text> </xsl:text><xsl:value-of select="y"/> <xsl:if test="position()!=last()">,</xsl:if> </xsl:for-each> I hope this helps, Ragulf Pickaxe :-)
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