LeetCode：Database 70.查询球队积分

Posted 2021-06-26 Xiao Miao

要求：写出一条SQL语句以查询每个队的 team_id，team_name 和 num_points。结果根据 num_points 降序排序，如果有两队积分相同，那么这两队按 team_id 升序排序。

积分规则如下：

赢一场得三分；
平一场得一分；
输一场不得分。

Table: Teams的结构

+---------------+----------+
| Column Name   | Type     |
+---------------+----------+
| team_id       | int      |
| team_name     | varchar  |
+---------------+----------+
此表的主键是 team_id，表中的每一行都代表一支独立足球队。

Table: Matches的结构

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| match_id      | int     |
| host_team     | int     |
| guest_team    | int     | 
| host_goals    | int     |
| guest_goals   | int     |
+---------------+---------+
此表的主键是 match_id，表中的每一行都代表一场已结束的比赛，比赛的主客队分别由它们自己的 id 表示，他们的进球由 host_goals 和 guest_goals 分别表示。

Teams 表:

+-----------+--------------+
| team_id   | team_name    |
+-----------+--------------+
| 10        | Leetcode FC  |
| 20        | NewYork FC   |
| 30        | Atlanta FC   |
| 40        | Chicago FC   |
| 50        | Toronto FC   |
+-----------+--------------+

Matches 表:

+------------+--------------+---------------+-------------+--------------+
| match_id   | host_team    | guest_team    | host_goals  | guest_goals  |
+------------+--------------+---------------+-------------+--------------+
| 1          | 10           | 20            | 3           | 0            |
| 2          | 30           | 10            | 2           | 2            |
| 3          | 10           | 50            | 5           | 1            |
| 4          | 20           | 30            | 1           | 0            |
| 5          | 50           | 30            | 1           | 0            |
+------------+--------------+---------------+-------------+--------------+

Result Table:

+------------+--------------+---------------+
| team_id    | team_name    | num_points    |
+------------+--------------+---------------+
| 10         | Leetcode FC  | 7             |
| 20         | NewYork FC   | 3             |
| 50         | Toronto FC   | 3             |
| 30         | Atlanta FC   | 1             |
| 40         | Chicago FC   | 0             |
+------------+--------------+---------------+

SQL语句：
注：也可使用union all 将主客队team_id当做一列，成绩当做一列，通过Group by分组求和，下面是同一个team_id的在主队成绩与在客队的成绩求和（将成绩分成两列）

with a as(
select host_team,guest_team,
case when host_goals>guest_goals then 3 when host_goals=guest_goals then 1 else 0 end as p1,
case when host_goals<guest_goals then 3 when host_goals=guest_goals then 1 else 0 end as p2 
from matches)

select b.team_id,b.team_name,ifnull(c.p3,0)+ifnull(d.p4,0) as num_points
from teams b
left join
(select host_team,sum(p1) as p3
from a
group by host_team
)c 
on b.team_id=c.host_team
left join
(select guest_team,sum(p2) as p4
from a
group by guest_team)d
on b.team_id=d.guest_team
order by num_points desc,b.team_id asc;