MYSQL错题整理（1212. 查询球队积分）
题目描述：Table: Teams
+---------------+----------+
| 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
+---------------+---------+
| 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。
参考答案：
select t.team_id,t.team_name,
sum(case when m1.host_num>m1.guest_num then 3 when m1.host_num=m1.guest_num then 1 else 0 end)num_points
from
Teams t left join
(select host_team as team,host_goals as host_num,guest_goals as guest_num from Matches
union all
select guest_team as team,guest_goals as host_num,host_goals as guest_num from Matches)m1
on t.team_id=m1.team
group by t.team_id
order by num_points desc,t.team_id asc
注意事项:
1.https://www.cnblogs.com/wangyayun/p/6133540.html 注意mysql中union和union all 的区别
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