7

我设计了以下mysql数据库表，

ticket(id, code, cust_name);
passenger(id, ticket_id, name, age, gender, fare);
service(id, passenger_id, item, cost);

一张票可以有很多乘客，每个乘客可以购买多项服务。我想要的是获得每张门票费用的总和。

我试过下面的sql，

SELECT 
    SUM(fare) as total_fare,
    (SELECT SUM(cost) as total_cost FROM services WHERE passenger.id = services.passenger_id) as total_service_cost
FROM
    ticket
JOIN passenger ON passenger.ticket_id = ticket.id

虽然，结果将乘客票价的总和作为 total_fare 但对于服务成本，它仅将第一位乘客的总服务成本相加并返回。

我认为我需要更多嵌套的查询，需要帮助，如果可能的话，我怎样才能得到结果，因为总和总和乘客票价和服务成本总和。

好的，这里是要澄清的 sql 插入语句，

INSERT INTO `ticket` (`id`, `code`, `cust_name`) VALUES
(1, 'TK01', 'Dipendra Gurung');

INSERT INTO `passenger` (`id`, `ticket_id`, `name`, `age`, `gender`, `fare`) VALUES
(1, 1, 'John', '28', 'M', 120),
(2, 1, 'Kelly', '25', 'F', 120);

INSERT INTO `services` (`id`, `passenger_id`, `item`, `cost`) VALUES
(1, 1, 'S1', 30),
(2, 1, 'S2', 50),
(3, 2, 'S3', 50);

我想得到票“TK01”的总费用（包括总票价和服务总和）。sql 必须返回总票价为 120+120 = 240，总服务为 30+50+50 = 130。

谢谢！:)

4

6 回答 6

18

首先，在您当前的表模式中，您无法区分以不同票证出售给同一乘客的服务。因此，您无法正确计算total_cost每张票。你必须ticket_id在你的service桌子上。

现在，如果您有一个ticket_idinservice表，那么具有相关子查询的解决方案可能看起来像

SELECT t.*,
       (SELECT SUM(fare) 
          FROM passenger
         WHERE ticket_id = t.id) total_fare,
       (SELECT SUM(cost) 
          FROM service
         WHERE ticket_id = t.id) total_cost
  FROM ticket t

或与JOINs

SELECT t.id, 
       p.fare total_fare,
       s.cost total_cost
FROM ticket t LEFT JOIN 
(
  SELECT ticket_id, SUM(fare) fare
    FROM passenger
   GROUP BY ticket_id
) p 
  ON t.id = p.ticket_id LEFT JOIN 
(
  SELECT ticket_id, SUM(cost) cost
    FROM service
   GROUP BY ticket_id
) s
  ON t.id = s.ticket_id

注意：这两个查询都考虑到乘客每张票可以有多个服务或根本没有服务的事实。

现在使用您当前的架构

SELECT t.*,
       (SELECT SUM(fare) 
          FROM passenger
         WHERE ticket_id = t.id) total_fare,
       (SELECT SUM(cost) 
          FROM service s JOIN passenger p
            ON s.passenger_id = p.id
         WHERE p.ticket_id = t.id) total_cost
  FROM ticket t

和

SELECT t.id, 
       p.fare total_fare,
       s.cost total_cost
FROM ticket t LEFT JOIN 
(
  SELECT ticket_id, SUM(fare) fare
    FROM passenger
   GROUP BY ticket_id
) p 
  ON t.id = p.ticket_id LEFT JOIN 
(
  SELECT p.ticket_id, SUM(cost) cost
    FROM service s  JOIN passenger p
      ON s.passenger_id = p.id
   GROUP BY p.ticket_id
) s
  ON t.id = s.ticket_id

只是为了得到每张票的总和

SELECT t.*,
       (SELECT SUM(fare) 
          FROM passenger
         WHERE ticket_id = t.id) +
       (SELECT SUM(cost) 
          FROM service s JOIN passenger p
            ON s.passenger_id = p.id
         WHERE p.ticket_id = t.id) grand_total
  FROM ticket t

或者

SELECT t.id, 
       p.fare + s.cost grand_total
FROM ticket t LEFT JOIN 
(
  SELECT ticket_id, SUM(fare) fare
    FROM passenger
   GROUP BY ticket_id
) p 
  ON t.id = p.ticket_id LEFT JOIN 
(
  SELECT p.ticket_id, SUM(cost) cost
    FROM service s  JOIN passenger p
      ON s.passenger_id = p.id
   GROUP BY p.ticket_id
) s
  ON t.id = s.ticket_id
于 2013-08-13T06:29:00.317 回答
3

您可以将三个表连接在一起，然后您可以直接进行 SUM 而无需子选择。如果您希望每张票使用它，则需要使用GROUP BY按ticket.id 分组。

就像是：

SELECT t.id, SUM(p.fare) AS total_far, SUM(s.cost) AS total_cost
FROM 
    ticket t, passenger p, service s
WHERE t.id = p.ticket_id AND s.passenger_id = p.id 
GROUP BY t.id;
于 2013-08-13T06:10:58.333 回答
2 
SELECT 
    SUM(fare) as total_fare,
    SUM(cost) as total_service_cost
FROM
    ticket
left join passenger ON ticket.id = passenger.ticket_id
left join service ON passenger.id = service.passenger_id
于 2013-08-13T06:19:20.097 回答
1

这个怎么样？

with fares as(
select p.id id, p.ticket_id ticket_id, sum(coalesce(s.cost,0)) cost
  from passenger p left outer join service s
       on p.id = s.passenger_id
 group by p.id, p.ticket_id)
select q.ticket_id, sum(f.cost), sum(q.fare), sum(f.cost + q.fare)
  from fares f inner join passenger q
       on f.id = q.id
 group by q.ticket_id;
于 2013-08-13T06:55:39.900 回答
0

那这个呢？

SELECT 
    SUM(fare) as total_fare,
    SUM(cost) as total_cost as total_service_cost
FROM
    ticket
JOIN passenger ON passenger.ticket_id = ticket.id
JOIN service ON passenger.id = service.passenger_id

如果您需要为每张票求和，请添加GROUP BY ticket.id

于 2013-08-13T06:11:20.153 回答
0 
select
    t.code,
    sum(p.fare) as total_fare,
    sum(s.cost) as total_cost
from ticket as t
    inner join passenger as p on p.ticket_id = t.id
    inner join service as s on s.passenger_id = p.id
group by t.code
于 2013-08-13T06:15:40.770 回答