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_id away_team_id home_team_goals away_team_goals home_team_points away_team_points team_id team_name 1 4 0 1 0 3 4 Dortmund 1 6 3 3 1 1 1 Ajax 4 1 5 2 3 0 6 Arsenal 6 1 0 0 1 1 分别求出各队作为
主队和客队时的分数、球数结果 team_name matches_played points goal_for goal_against 结果(最终) Dortmund 2 6 6 2 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 Arsenal 2 2 3 3 Ajax 4 2 5 9 在此基础上求出goal_diff 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
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