[XSL-LIST Mailing List Archive Home] [By Thread] [By Date] [Recent Entries] [Reply To This Message] RE: Grouping problem?
Whoops! My solution had an error which meant it also didn't work with Dimitre's example. My previous example didn't correctly handle the case where the sum of elements so far was an exact multiple of 10. Recursion can be tricky! :-) Here it is corrected: <xsl:template match="root"> <xsl:copy> <xsl:call-template name="group-ele"> <xsl:with-param name="ele-list" select="ele"/> </xsl:call-template> </xsl:copy> </xsl:template> <xsl:template name="group-ele"> <xsl:param name="ele-list" select="/.."/> <xsl:param name="count" select="0"/> <xsl:if test="$ele-list"> <xsl:if test="$count = 10"> <br/> </xsl:if> <xsl:variable name="first-ele" select="$ele-list[1]"/> <xsl:variable name="new-count" select="$count mod 10 + $first-ele/@sum"/> <xsl:if test="$new-count >= 10"> <br/> </xsl:if> <xsl:copy-of select="$first-ele"/> <xsl:call-template name="group-ele"> <xsl:with-param name="ele-list" select="$ele-list[position()>1]"/> <xsl:with-param name="count" select="$new-count"/> </xsl:call-template> </xsl:if> </xsl:template> Cheers Con > -----Original Message----- > From: owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx > [mailto:owner-xsl-list@xxxxxxxxxxxxxxxxxxxxxx]On Behalf Of Conal Tuohy > Sent: Wednesday, 23 April 2003 09:26 > To: xsl-list@xxxxxxxxxxxxxxxxxxxxxx > Subject: RE: Grouping problem? > > > Lars Huttar wrote: > > > I like it! Interesting way to form a group. > > > > It might be slow for large source documents, maybe order(N*N) > > where N is the number of ele elements (because for each element you > > have to sum all preceding elements); but I can't see a way > > around that... unless you want to recursively loop through > > the elements, > > keeping a running total. > > Like this: > > <xsl:template match="root"> > <xsl:copy> > <xsl:call-template name="group-ele"> > <xsl:with-param name="ele-list" select="ele"/> > </xsl:call-template> > </xsl:copy> > </xsl:template> > > <xsl:template name="group-ele"> > <xsl:param name="ele-list" select="/.."/> > <xsl:param name="count" select="0"/> > <xsl:if test="$ele-list"> > <xsl:variable name="first-ele" select="$ele-list[1]"/> > <xsl:variable name="new-count" select="$count + > $first-ele/@sum"/> > <xsl:if test="$new-count > 10"> > <br/> > </xsl:if> > <xsl:copy-of select="$first-ele"/> > <xsl:call-template name="group-ele"> > <xsl:with-param name="ele-list" > select="$ele-list[position()>1]"/> > <xsl:with-param name="count" > select="$new-count mod 10"/> > </xsl:call-template> > </xsl:if> > </xsl:template> > > As you say, Lars, this approach would probably be a lot > quicker for large > documents. > > Cheers! > > Con > > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list > > XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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