LeetCode-1212. 查询球队积分（中等）

Table: Teams

+---------------+---------+
| 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。

积分规则如下：

赢一场得三分；
平一场得一分；
输一场不得分。
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。

查询结果格式如下：

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

Matches table:
+------------+--------------+---------------+-------------+--------------+
| 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 |
+------------+--------------+---------------+

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/team-scores-in-football-tournament
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

审题：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。

根据进球数，计算比分，最后求和。

思考：首先根据结果对数据加工，算出主队客队的本场比赛积分，然后求所有球队id的比分和。

解题：

方法一：思路：求id，队名，分数。

首先通过union all将比赛结果表，将对两队的结果，联合起来。如果五场比赛，相当于10个球队，将比赛表左右两边对调，就可以是五场比赛的十支球队结果都包含了。

然后通过teams表left join 使得左边肯定多于右边，判断通过on后边的条件对比，右边分组里的每一行通过对比判断积分，求和。

最后按照条件排序即可。

select t.team_id,t.team_name,
sum(case when m.host_goals > m.guest_goals then 3 
      when m.host_goals = m.guest_goals then 1
      else 0 end) num_points
from teams t 
left join
(select host_team, guest_team, host_goals, guest_goals
from matches
union all
select guest_team host_team, host_team guest_team, guest_goals host_goals, host_goals guest_goals
from matches) m
on m.host_team = t.team_id
group by t.team_id
order by num_points desc, t.team_id asc

-- 练习
-- left join 
-- LEFT JOIN 关键字从左表（table1）返回所有的行，即使右表（table2）中没有匹配。如果右表中没有匹-- -- 配，则结果为 NULL。

select t.team_id, t.team_name,
sum(case when m.host_goals>m.guest_goals then 3 when m.host_goals = m.guest_goals then 1 else 0 end) num_points from teams t left join
(select host_team, guest_team, host_goals, guest_goals
from matches
union all
select guest_team host_team, host_team guest_team, guest_goals host_goals, host_goals guest_goals
from matches) m
on m.host_team = t.team_id
group by t.team_id
order by num_points desc,t.team_id asc;

方法二：

select t.team_id,t.team_name,sum(
    CASE WHEN t.team_id= m.host_team and m.host_goals>m.guest_goals THEN 3
         WHEN m.host_goals = m.guest_goals THEN 1
         WHEN t.team_id = m.guest_team and m.guest_goals>m.host_goals THEN 3 ELSE 0 END
    ) as num_points 
from Teams t 
left join Matches m 
on t.team_id=m.host_team or t.team_id=m.guest_team 
group by t.team_id 
order by num_points desc ,team_id asc

知识点：

UNION 只会选取不同的值。请使用 UNION ALL 来选取重复的值！

红楼终究一场梦

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LeetCode-1212. 查询球队积分（中等）

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