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【每日SQL打卡】DAY 20丨查询球队积分【难度中等】
2022-07-29 11:43:00 【不吃西红柿】
难度中等
SQL架构
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 |
+------------+--------------+---------------+边栏推荐
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