Subject: [xsl] Compare current node with the last node in a sorted for-each list From: TBrouwer@xxxxxxxxxxxxxxxxxxx Date: Mon, 21 May 2001 14:20:58 +0200 |
Example: <node1> <node2 attr="name4" extra="test1"/> <other>More1</other> <x>X</x> <y>Y</y> <node2 attr="name3" extra="test2"/> <s>S</s> <other>More2</other> <node2 attr="name2" extra="test1"/> <other>More3</other> <node2 attr="name1" extra="test1"/> <z>Z</z> <other>More4</other> </node1> Is it possible to display all 'node2' in alphabetic order based on the value of the 'attr' attribute? Further, as the value of the attribute 'extra' is the same as the node before in the sorted list, then print an extra line under that node with <same /> Wanted output: <node2 attr="name1" extra="test1"/> <node2 attr="name2" extra="test1"/> <same/> <node2 attr="name3" extra="test2"/> <node2 attr="name4" extra="test1"/> Thanks in advance for any suggestions.... T. Brouwer. XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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