Advances in database programing languages by Fançois Bancilhon; Peter Buneman (eds.)

By Fançois Bancilhon; Peter Buneman (eds.)

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End_time); 12 PUZZLE 3 THE ANESTHESIA PUZZLE The result is this view shown here for procedure #10 only and sorted by event_time for clarity. Notice that the same anesthesiologist can start more than one procedure at the same time. Events proc_id comparison_proc anest_name event_time event_type ========================================================== 10 10 Baker 08:00 +1 10 20 Baker 09:00 +1 10 10 Baker 11:00 -1 10 20 Baker 13:00 -1 Now, for each set of Events with the same proc_id id, we can compute for each event the sum of the event_types for Events that occur earlier.

Answer #1 This is really fairly straightforward, but you have to reword the query specification into the passive voice to see the answer. , a group of step_nbrs) that has certain properties in the step_status column for certain step_nbrs. Using a characteristic function in a SUM() will let us know if all the elements of the group meet the criteria; if they all do, then the count of the characteristic function is the size of the group. SELECT FROM GROUP HAVING workorder_id Projects BY workorder_id SUM(CASE WHEN step_nbr <> 0 AND step_status = ‘W’ THEN 1 WHEN step_nbr = 0 AND step_status = ‘C’ THEN 1 ELSE 0 END) = COUNT (step_nbr); or if you do not have a CASE expression, you can use some algebra: SELECT workorder_id FROM Projects AS P1 GROUP BY workorder_id HAVING SUM( ((1-ABS(SIGN(step_nbr)) * POSITION(‘W’ IN step_status)) + ((1-ABS(step_nbr)) * POSITION(‘C’ IN step_status)) ) = COUNT(step_nbr); Since this query involves only one copy of the table and makes a single pass through it, it should be as fast as possible.

UPDATE Seats SET seat_nbr = -seat_nbr WHERE seat_nbr = :my_seat; The same update statement can be used to put back into the Available list, and a “SET seat_nbr = ABS(seat_nbr)” will reset the restaurant at closing time. Answer #2 The second thought is to create a second table with a single column of occupied seating and to move numbers between the occupied and available tables. That would require a total of 1,000 rows in both tables, which is a little weird, but it leads to the next answer. PUZZLE 9 AVAILABLE SEATS 35 Answer #3 Instead, we can use a single table and create seats 0 through 1,001 (0 and 1,001 do not really exist and are never assigned to a customer.

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