题目

数据

结果

解答

1.因为分了两场，所以把两场连接为一场

select first_player player_id,first_score score from matches #
union all
select second_player player_id,second_score score from matches

2. 连接players表，按组别和玩家 分组计算得分

select group_id,players.player_id,sum(score) score1 
from players
right join(
            select first_player player_id,first_score score from matches #
            union 
            select second_player player_id,second_score score from matches
           ) a
on a.player_id = players.player_id
group by group_id,player_id
order by group_id

3.按照组别分组，按分数排名（排名的时候，按照分数和用户id排名）！！！！！！

select group_id,player_id,score1,
       rank() over (partition by group_id order by score1 desc,player_id) rn 
from(
        select group_id,players.player_id,sum(score) score1 
        from players
        right join(
                  select first_player player_id,first_score score from matches #
                  union 
                  select second_player player_id,second_score score from matches
                   ) a
        on a.player_id = players.player_id
        group by group_id,player_id
        order by group_id
    ) b 

4.筛选出来第一名的人，（按照分数和id）

select group_id,player_id ,rn
from(
    select group_id,player_id,score1,
        rank() over (partition by group_id order by score1 desc,player_id) rn 
    from(
        select group_id,players.player_id,sum(score) score1 
        from players
        right join(
            select first_player player_id,first_score score from matches #
            union 
            select second_player player_id,second_score score from matches
        ) a
        on a.player_id = players.player_id
        group by group_id,player_id
				order by group_id
    )b 
)c
where rn = 1

标签：group,players,1194,LeetCode,player,score,优胜者,id,select
来源： https://www.cnblogs.com/Tdazheng/p/14981005.html