MySQL 中每个 ID 的前 N 行

Question

我有一个 table 我的用户分数是这样的：

id | kills
----------
2  | 1
1  | 1
1  | 5
1  | 3
2  | 4
2  | 5
3  | 5

我想获得每个玩家的前两行超过 2 次击杀。所以结果应该是这样的

id | kills
----------
1  | 5
1  | 3
2  | 4
2  | 5
3  | 5

我试过了，但没用：

SELECT * 
FROM user_stats us 
WHERE 
(
    SELECT COUNT(*) 
    FROM user_stats f 
    WHERE f.id=us.id AND f.kills > 2
) <= 2;

Answer 1

select * from user_stats
where (id,kills) in (select id, max(kills) from user_stats where kills > 2 group by id
                    union
                    select id, min(kills) from user_stats where kills > 2 group by id)

Answer 2

我怀疑您只想要杀死数大于 2 的用户的两个最大值。如果是这样，请使用变量：

select us.*
from (select us.*,
             (@rn := if(@i = id, @rn + 1,
                        if(@i := id, 1, 1)
                       )
             ) as seqnum
      from user_stats us cross join
           (select @rn := 0, @i := -1) params
      where us.kills > 2
      order by us.id, kills desc
     ) us
where seqnum <= 2;

Answer 3

试试这个。我来自 Oracle，其中 rownum 是所选行的计数。这应该有相同的效果。

select @rownum:=@rownum+1, us.*
from user_stats us , (select @rownum := 0) r
where id in (
    select id from user_stats f 
    group by id 
    having count(*) > 2
)
and @rownum < 3;

Answer 4

基于 vkp 的响应。当 id 超过 1 次击杀时取最小值和最大值

select id, max(kills) 
  from user_stats 
  group by id 
  having count(kills) > 2 
union 
select id, min(kills) 
  from user_stats 
  group by id 
  having count(kills) > 2 
order by id

MySQL 中每个 ID 的前 N 行

first N row of each id in MySQL

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MySQL 中每个 ID 的前 N ​​行

first N row of each id in MySQL

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MySQL 中每个 ID 的前 N 行