[xquery-talk] search mechanism
martin at x-hive.com
Thu Apr 6 18:26:52 PDT 2006
> Here is my query attempt which results in the following exception:
> "unexpected token: ) [at line 11, column 9]"
> let $res :=
> let $ids := distinct-values(
> let $ids1 :=
> for $id1 in //text//abstract &= 'peter'
> return xs:string($id1/ancestor::text/@id)
> let $ids2 :=
> for $id2 in //text//title &= 'java'
> return xs:string($id2/ancestor::text/@id)
> where $ids1 = $ids2
> for $doc in //text[@id = $ids]
> order by year-from-date($doc//date)
> return $doc
> return subsequence($res, 10, 20)
What is that &= syntax you're using? I've never seen that.
You always need an expression following the "return" keyword, normally a
variable from the preceding FLWOR expression. In this case however, you
need to do that a little different. let binds the whole sequence to a
variable and '=' is an existential comparison, e.g. 'one of the left
side equals one of the right side'. This means take all IDs from the
first query ('peter') and take all IDs from the second query ('java').
Then if one of the IDs matches one of the other IDs, return all.
I'd guess that what you want to say is different, e.g. put this into the
for $id1doc in //text//abstract[contains(.,'Peter')]
let $id1 := $id1doc/ancestor::text/@id/string()
for $id2doc in //text//title[contains(.,'java')]
let $id2 := $id2doc/ancestor::text/@id/string()
where $id1 = $id2
I.e. for each document that contains Peter in the abstract and each
document that contains Java in the title, if the IDs are equal return
one of them. While this looks like a nested join smart XQuery processors
can rewrite it to a hash-merge and for not so smart processors you can
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