[xquery-talk] my question again

[Michael Rys]

I think it is not clear what your grouping criteria is doing. In this
case you don't need a group by to to get the result 2. Just
count(/bookstore/book[ ... predicate .... ]) gives you 2....

Best regards
Michael

> -----Original Message-----
> From: talk-bounces at xquery.com [mailto:talk-bounces at xquery.com] On
Behalf
> Of fatma helmy
> Sent: Wednesday, March 15, 2006 8:12 PM
> To: talk at xquery.com
> Subject: [xquery-talk] my question again
> 
> dear all
> i forwarded this email, i just need to know , is it
> possible to do that, if not please reply with no
> 
> Let &j= <bookstore>
> <book id="1">
> 	<title>seven years in trenton</title>
> 	<author>
> 		<firstname>joe</firstname>
> 		<lastname>bob</lastname>
> 		<degree>B.A</degree>
> 	</author>
> 	<price>12</price>
> 
> </book>
> <book>
> 	<title>history of trenton</title>
> 	<author>
> 		<firstname>mary</firstname>
> 		<lastname>bob</lastname>
> 		<degree>PhD</degree>
> 		<degree>MSC</degree>
> 	</author>
> 	<price>55</price>
> </book>
> <book>
> 	<title>trenton today, trenton tomorrow</title>
> 	<author>
> 		<firstname>toni</firstname>
> 		<lastname>bob</lastname>
> 		<degree>PhD</degree>
> 		<degree>MSC</degree>
> 	</author>
> 	<price>55</price>
> </book>
> </bookstore>
> 
> let $whrcnd = Book/author/lastname= bob and
> book/price=55
> 
> I need to select count(*) from $j where $whrcnd
> Group by <records of book>
> The result is 2
> How to transform that into xquery.
> 
> 
> 
> 
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