Version 6 (modified by 2 years ago) ( diff ) | ,
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Напредни извештаи од базата
Листа на извештаи
- Патникот може да побара историјат за одредено време кој ќе се состои од тоа колку километри поминал со секој возач и колку вкупно платил.
SQL прашалници
SELECT d.driver_id, SUM(d.km_travelled) as km_travelled_with_driver, SUM(pay.total_sum_payed) as total_price_paid, SUM(d.km_travelled)/SUM(pay.total_sum_payed) as price_per_km FROM request r JOIN drive d ON r.request_id = d.request_id JOIN payment pay ON pay.passenger_id = r.passenger_id WHERE r.passenger_id = '77d9c766-e0c1-4e53-86d0-bafc3101a1ac' AND d.start_time between now() and now() - interval '2 months' GROUP BY d.driver_id ORDER BY price_per_km ASC
SELECT dr.driver_id, dr.email, (CASE WHEN num_grades > 2 THEN AVG(de.grade) ELSE NULL END) as driver_grade, COUNT(*) as number_of_drives, SUM(p.total_sum_payed) as total_money_made, COUNT(DISTINCT de.request_id) as number_of_different_requests, COUNT(DISTINCT r.passenger_id) as number_of_different_passengers, (SUM(p.total_sum_payed))/COUNT(DISTINCT de.request_id) as average_money_per_request, SUM(d.km_travelled) as total_km_travelled FROM driver dr JOIN drive de ON dr.driver_id = de.driver_id JOIN payment p ON de.drive_id = p.drive_id LEFT JOIN request r ON de.request_id = r.request_id GROUP BY dr.driver_id, dr.email ORDER BY driver_grade DESC }}
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