Solving a relational query: part 4

As Paul Rubin pointed out the SQL in the last post was broken – here’s some that, I think (please do not let it be wrong a second time!), by relying on SQL’s coercion of a tuple to a scalar, would work (Paul proposed an even more ambitions way of doing this in a single join):

SELECT DISTINCT U.NAME, U.EMAIL
FROM USER U, BID B, LOT L
WHERE U.USER_ID = B.BIDDER_ID AND
B.AMOUNT > 100 AND B.LOT_ID = L.LOT_ID
AND (SELECT COUNT(*) FROM
BID, LOT WHERE
BID.LOT_ID = B.LOT_ID AND BID.BIDDER_ID = U.USER_ID
AND BID.LOT_ID = LOT.LOT_ID) > 1

Still no clue as to how to do this with relational algebra though.

Related articles

• Solving a relational query: part 2 (cartesianproduct.wordpress.com)
• Solving a relational query: part 1 (cartesianproduct.wordpress.com)
• Relational division again (cartesianproduct.wordpress.com)

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