Re: Grouping by a piece of a date

["Steve Muench" <smuench@xxxxxxxxxxxxx>]

| given xml with date attributes, I want to be able to group them by day.
| Using the Muenchian method I can group by the entire value, but I can't
| figure out how to group by only a piece of it.  For example, given the
| following XML:
| 
| <xml xmlns:s="uuid:BDC6E3F0-6DA3-11d1-A2A3-00AA00C14882"
| xmlns:dt="uuid:C2F41010-65B3-11d1-A29F-00AA00C14882"
| xmlns:rs="urn:schemas-microsoft-com:rowset" xmlns:z="#RowsetSchema">
:
| <rs:data>
| <z:row iID="1" dCreated="1900-01-01T01:00:00"/>
| <z:row iID="2" dCreated="1900-01-02T01:00:00"/>
| <z:row iID="3" dCreated="1900-01-02T02:00:00"/>
| <z:row iID="4" dCreated="1900-01-04T01:00:00"/>
| </rs:data>
| </xml>

If I've not misunderstood what you're going for, then
you should be able to use the substring-before() function
in the <xsl:key> declaration like this:

<xsl:key name="foo"
         match="/xml/rs:data/z:row"
         using="substring-before(@dCreated,'T')"/>


This way, an attempt to lookup:

     key("foo","1900-01-02")

should find the <z:row> elements:

 <z:row iID="2" dCreated="1900-01-02T01:00:00"/>
 <z:row iID="3" dCreated="1900-01-02T02:00:00"/>

Because both of these have the substring-before their
"dCreated" attribute value of "1900-01-02".

______________________________________________________________
Steve Muench, Lead XML Evangelist & Consulting Product Manager
BC4J & XSQL Servlet Development Teams, Oracle Rep to XSL WG
Author "Building Oracle XML Applications", O'Reilly
http://www.oreilly.com/catalog/orxmlapp/





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