Subject: RE: Strange Sort... From: "Spychalski, Frank" <frank.spychalski@xxxxxxx> Date: Wed, 19 Jul 2000 22:26:42 +0200 |
Hello, I think the easiest way is chaining 2 XSLTs, the first copiest the y values from children to attributes and the second XSLT sorts the results. But I'm curious if one of the xslt-gurus will come up with a solution that works with one xslt. bye Frank - Work is the curse of the drinking class. (Oscar Wilde) Frank Spychalski frank@xxxxxxxxxxxxx ->-----Original Message----- ->From: Brian Young [mailto:Brian.Young@xxxxxxx] ->Sent: Wednesday, July 19, 2000 11:51 AM ->To: 'xsl-list@xxxxxxxxxxxxxxxx' ->Subject: Strange Sort... -> -> ->Hello, -> ->I have the following XML hierarchy: -> -><bucket> -> <object id="first" x="35" y="10" /> -> <object id="second" x="15" y="58" /> -> <subbucket> -> <object> -> <id>third</id> -> <x>28</x> -> <y>145</y> -> </object> -> </subbucket> -> <object id="fourth" x="65" y="27" /> -></bucket> -> ->Ok, enough with the funny looks. I know that the use of id, ->x, and y should be consistent as either children or an ->attribute. ; ). But, this is what I'm dealing with now, ->until I can get it changed. -> ->I'd like to be able to sort all four objects based on the ->value of their "y"s, regardless of whether they are a child ->or an attribute. Is that possible? Doing two sorts, with ->one being the major and the other the minor will not work. I ->think it has to be in one xsl:sort. -> ->Any help is greatly appreciated. -> ->Thanks, -> Brian Young XSL-List info and archive: http://www.mulberrytech.com/xsl/xsl-list
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