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Subject:
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xsl select statement
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Posted By:
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pat123
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Post Date:
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4/19/2006 6:43:59 AM
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Hi,
I am using this select statement to return unique clients from my xml file and put them into a drop down list.
//portfolio/campaign[@client_name][not(@client_name=preceding-sibling::campaign/@client_name)]
This line works perfect, but I also need another select statement to return unique creative types from my xml file.
Tried this:
//portfolio/campaign/creative[@client_type][not(@client_type=preceding-sibling::creative/@client_type)]
This is the struture of my xml
<portfolio>
<campaign client_name="client1">
<creative creative_type = "banner"></creative>
<creative creative_type = "island"></creative>
<creative creative_type = "banner"></creative>
</campaign>
<campaign client_name="client2">
<creative creative_type = "banner"></creative>
<creative creative_type = "island"></creative>
<creative creative_type = "banner"></creative>
</campaign>
<campaign client_name="client3">
<creative creative_type = "banner"></creative>
<creative creative_type = "island"></creative>
<creative creative_type = "banner"></creative>
</campaign>
</portfolio>
The select statement for creative types should return only two results:
banner
island
But mine returns:
banner
island
banner
island
banner
island
It returns unique types for creatives with the same parent but ignores other campaigns.
I hope this is clear. I need a select statement that will return the proper result as shown above. Can anyone help?
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Reply By:
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mhkay
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Reply Date:
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4/19/2006 7:15:24 AM
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In 2.0 of course you can simply use the distinct-values() function.
In 1.0, try
portfolio/campaign/creative/@creative_type[not(. = ../preceding::creative/@creative-type)]
or for better performance, use Muenchian grouping.
Michael Kay
http://www.saxonica.com/
Author, XSLT Programmer's Reference and XPath 2.0 Programmer's Reference
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Reply By:
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pat123
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Reply Date:
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4/19/2006 8:17:36 AM
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Thanks, that worked perfect.
portfolio/campaign/creative/@creative_type[not(. = ../preceding::creative/@creative_type)]
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