Subject: [xsl] Sorting and re-ordering down a hierarchy From: "Cole, Chris" <chris.cole@xxxxxxxxxxxxxxxxxxxxx> Date: Fri, 18 Jul 2003 15:00:30 -0000 |
Hello, I'm struggling to re-organise my node tree. My tree is something like this: <input> <node> <rank>2</rank> <node> <rank>88</rank> </node> <node> <rank>7</rank> </node> <node> <rank>66</rank> </node> </node> <node> <rank>1</rank> <node> <rank>3</rank> </node> <node> <rank>2</rank> </node> <node> <rank>1</rank> </node> </node> </input> I need to reorganise the nodes so that they are organised in number order down the nodes within their tree, so the output should be like this: <output> <node> <rank>1</rank> <node> <rank>7</rank> </node> <node> <rank>66</rank> </node> <node> <rank>88</rank> </node> </node> <node> <rank>2</rank> <node> <rank>1</rank> </node> <node> <rank>2</rank> </node> <node> <rank>3</rank> </node> </node> </output> I can sort one level by using the code below, but I'm struggling to find a way to re-order the second level within the first level (and the third level later). <xsl:template match="/input"> <xsl:call-template name="ascending-numeric-sort-l1"> <xsl:with-param name="seqnuml1" select="node/rank"/> </xsl:call-template> </xsl:template> <xsl:template name="ascending-numeric-sort-l1"> <xsl:param name="seqnuml1"/> <xsl:for-each select="$seqnuml1"> <xsl:sort select="." data-type="number" order="ascending"/> <xsl:copy-of select="parent::node()"/> </xsl:for-each> </xsl:template> Can anybody offer any advice please? Thanks Chris XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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