LeetCode_sql_day18（1841.联赛信息统计）
描述
表: Teams
+----------------+---------+
| Column Name | Type |
+----------------+---------+
| team_id | int |
| team_name | varchar |
+----------------+---------+
team_id 是该表主键.
每一行都包含了一个参加联赛的队伍信息.
表: Matches
+-----------------+---------+
| Column Name | Type |
+-----------------+---------+
| home_team_id | int |
| away_team_id | int |
| home_team_goals | int |
| away_team_goals | int |
+-----------------+---------+
(home_team_id, away_team_id) 是该表主键.
每一行包含了一次比赛信息.
home_team_goals 代表主场队得球数.
away_team_goals 代表客场队得球数.
获得球数较多的队伍为胜者队伍.
写一段SQL，用来报告联赛信息. 统计数据应使用已进行的比赛来构建，其中 获胜 球队获得 三分 ，而失败球队获得 零分 。如果 打平 ，两支球队都得 一分 。
result 表的每行应包含以下信息:
team_name - Teams 表中的队伍名字matches_played - 主场与客场球队进行的比赛次数.points - 球队获得的总分数.goal_for - 球队在所有比赛中获取的总进球数goal_against - 球队在所有比赛中，他的对手球队的所有进球数goal_diff - goal_for - goal_against.按 points 降序 返回结果表。 如果两队或多队得分相同，则按 goal_diff 降序 排列。 如果仍然存在平局，则以 team_name 按字典顺序 排列它们。
查询的结果格式如下例所示。
示例 1:
输入：
Teams 表:
+---------+-----------+
| team_id | team_name |
+---------+-----------+
| 1 | Ajax |
| 4 | Dortmund |
| 6 | Arsenal |
+---------+-----------+
Matches 表:
+--------------+--------------+-----------------+-----------------+
| home_team_id | away_team_id | home_team_goals | away_team_goals |
+--------------+--------------+-----------------+-----------------+
| 1 | 4 | 0 | 1 |
| 1 | 6 | 3 | 3 |
| 4 | 1 | 5 | 2 |
| 6 | 1 | 0 | 0 |
+--------------+--------------+-----------------+-----------------+
输出：
+-----------+----------------+--------+----------+--------------+-----------+
| team_name | matches_played | points | goal_for | goal_against | goal_diff |
+-----------+----------------+--------+----------+--------------+-----------+
| Dortmund | 2 | 6 | 6 | 2 | 4 |
| Arsenal | 2 | 2 | 3 | 3 | 0 |
| Ajax | 4 | 2 | 5 | 9 | -4 |
+-----------+----------------+--------+----------+--------------+-----------+
解释：
Ajax (team_id=1) 有4场比赛: 2败2平. 总分数 = 0 + 0 + 1 + 1 = 2.
Dortmund (team_id=4) 有2场比赛: 2胜. 总分数 = 3 + 3 = 6.
Arsenal (team_id=6) 有2场比赛: 2平. 总分数 = 1 + 1 = 2.
Dortmund 是积分榜上的第一支球队. Ajax和Arsenal 有同样的分数, 但Arsenal的goal_diff高于Ajax, 所以Arsenal在表中的顺序在Ajaxzhi'qian.
数据准备
Create table If Not Exists Teams (team_id int, team_name varchar(20))
Create table If Not Exists Matches
(
home_team_id int,
away_team_id int,
home_team_goals int,
away_team_goals int
)
Truncate table Teams ;
insert into Teams (team_id, team_name)
values ('1', 'Ajax')
insert
into Teams (team_id, team_name)
values ('4', 'Dortmund')
insert into Teams (team_id, team_name)
values ('6', 'Arsenal');
Truncate table Matches;
insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals)
values ('1', '4', '0', '1')
insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals)
values ('1', '6', '3', '3')
insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals)
values ('4', '1', '5', '2')
insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals)
values ('6', '1', '0', '0');
分析
①先构造出得分情况
select *,
case
when home_team_goals > away_team_goals then 3
when home_team_goals = away_team_goals then 1
when home_team_goals < away_team_goals then 0
end as home_team_points,
case
when home_team_goals < away_team_goals then 3
when home_team_goals = away_team_goals then 1
when home_team_goals > away_team_goals then 0
end as away_team_points
from Matches
②然后分别计算球队比赛次数（主队的次数+客队的次数）、球队总得分（主队时的得分+客队时的得分）、球队总进球数（主队时的总进球数+客队时的总进球数）、对手总进球数（作为主队时对手作为客队的进球数+作为客队时对手作为主队的总进球数）
with t1 as (select *,
case
when home_team_goals > away_team_goals then 3
when home_team_goals = away_team_goals then 1
when home_team_goals < away_team_goals then 0
end as home_team_points,
case
when home_team_goals < away_team_goals then 3
when home_team_goals = away_team_goals then 1
when home_team_goals > away_team_goals then 0
end as away_team_points
from Matches)
select distinct team_name,
(select count(1) from t1 where home_team_id = Matches.home_team_id or away_team_id =Matches.home_team_id) as matches_played,
(select sum(home_team_points) from t1 where home_team_id = Matches.home_team_id) +
(select sum(away_team_points) from t1 where away_team_id = Matches.home_team_id) as points,
(select sum(home_team_goals) from t1 where home_team_id = Matches.home_team_id) +
(select sum(away_team_goals) from t1 where away_team_id = Matches.home_team_id) as goal_for,
(select sum(away_team_goals) from t1 where home_team_id = Matches.home_team_id) +
(select sum(home_team_goals) from t1 where away_team_id = Matches.home_team_id) as goal_against
from matches , teams where matches.home_team_id = teams.team_id
union
select distinct team_name,
(select count(1) from t1 where away_team_id = Matches.away_team_id or home_team_id =Matches.away_team_id) as matches_played,
(select ifnull(sum(home_team_points),0 ) from t1 where home_team_id = Matches.away_team_id) +
(select ifnull(sum(away_team_points),0) from t1 where away_team_id = Matches.away_team_id) as points,
(select ifnull(sum(home_team_goals),0) from t1 where home_team_id = Matches.away_team_id) +
(select ifnull(sum(away_team_goals),0) from t1 where away_team_id = Matches.away_team_id) as goal_for,
(select ifnull(sum(away_team_goals),0) from t1 where home_team_id = Matches.away_team_id) +
(select ifnull(sum(home_team_goals),0) from t1 where away_team_id = Matches.away_team_id) as goal_against
from matches , teams where matches.away_team_id = teams.team_id
③基于上述结果 求goal_diff并且按照题目要求排序
select team_name,
matches_played,
points,
goal_for,
goal_against
,(goal_for-goal_against) as goal_diff from t2
order by points desc,goal_diff desc,team_name desc;
图解：
输入home_team_idaway_team_idhome_team_goalsaway_team_goalshome_team_pointsaway_team_pointsteam_idteam_name1401034Dortmund1633111Ajax4152306Arsenal610011分别求出各队作为
主队和客队时的分数、球数结果team_namematches_playedpointsgoal_forgoal_against结果(最终)Dortmund2662主队的+客队的主队的+客队的主队的+客队的主队的+客队的主队的+客队的Arsenal2233Ajax4259在此基础上求出goal_diffteam_namematches_playedpointsgoal_forgoal_againstgoal_diffDortmund26624Arsenal22330Ajax4259-4
代码
with t1 as (select *,
case
when home_team_goals > away_team_goals then 3
when home_team_goals = away_team_goals then 1
when home_team_goals < away_team_goals then 0
end as home_team_points,
case
when home_team_goals < away_team_goals then 3
when home_team_goals = away_team_goals then 1
when home_team_goals > away_team_goals then 0
end as away_team_points
from Matches)
, t2 as (
select home_team_id,
(select count(1) from t1 where home_team_id = Matches.home_team_id or away_team_id =Matches.home_team_id) as matches_played,
(select sum(home_team_points) from t1 where home_team_id = Matches.home_team_id) +
(select sum(away_team_points) from t1 where away_team_id = Matches.home_team_id) as points,
(select sum(home_team_goals) from t1 where home_team_id = Matches.home_team_id) +
(select sum(away_team_goals) from t1 where away_team_id = Matches.home_team_id) as goal_for,
(select sum(away_team_goals) from t1 where home_team_id = Matches.home_team_id) +
(select sum(home_team_goals) from t1 where away_team_id = Matches.home_team_id) as goal_against
# goal_for-goal_against as goal_diff
from matches
union all
(select away_team_id,
(select count(1) from t1 where away_team_id = Matches.away_team_id or home_team_id =Matches.away_team_id) as matches_played,
(select sum(away_team_points) from t1 where away_team_id = Matches.away_team_id) +
(select sum(home_team_points) from t1 where home_team_id = Matches.away_team_id) as points,
(select sum(home_team_goals) from t1 where home_team_id = Matches.away_team_id) +
(select sum(away_team_goals) from t1 where away_team_id = Matches.away_team_id) as goal_for,
(select sum(away_team_goals) from t1 where home_team_id = Matches.away_team_id) +
(select sum(home_team_goals) from t1 where away_team_id = Matches.away_team_id) as goal_against
from Matches)
)
select distinct (select team_name from teams where team_id=t2.home_team_id)team_name,
matches_played,
points,
goal_for,
goal_against
,(goal_for-goal_against) as goal_diff from t2
order by points desc,goal_diff desc,team_name;
总结
最后要考虑到有的球队只有客队场 所以使用union 既要关联到主队id又要关联到客队id