Question

如何选择wach user_id的行等于每个user_id的选择计数？

我的示例表：

mp3_id  |  user_id
--------------------
120     |  840
123     |  840
126     |  840
128     |  455
130     |  840
131     |  840
132     |  840
135     |  840
144     |  840
158     |  840
159     |  455
161     |  455
169     |  455
180     |  840
181     |  455
184     |  455
186     |  455
189     |  455

我的简单查询：

select mp3_id where user_id IN (840,455) limit 8

retuen：

mp3_id  |  user_id
--------------------
120     |  840
123     |  840
126     |  840
128     |  455
130     |  840
131     |  840
132     |  840
135     |  840

但我想选择这个：

mp3_id  |  user_id
--------------------
120     |  840
123     |  840
126     |  840
130     |  840
128     |  455
159     |  455
161     |  455
169     |  455

我希望每个user_id返回相等的行数。如何？感谢

Answer 1

您可以使用UNION执行此操作：

select mp3_id where user_id = 840 limit 4
union all
select mp3_id where user_id = 455 limit 4

Answer 2

尝试此查询：

select
yt.mp3_id,
e.user_id
from
(
    select distinct user_id from your_table
) e
join your_table yt on true
where yt.mp3_id in (select tt.mp3_id from your_table tt where tt.user_id = e.user_id order by tt.mp3_id limit 4)

和相同的查询但有条件

select
yt.mp3_id,
e.user_id
from
(
    select distinct user_id from your_table where user_id in (840,455)
) e
join your_table yt on true
where yt.mp3_id in (select tt.mp3_id from your_table tt where tt.user_id = e.user_id order by tt.mp3_id limit 4)

Answer 3

SELECT x.* 
  FROM my_table x 
  JOIN my_table y 
    ON y.user_id = x.user_id 
   AND y.mp3_id <= x.mp3_id 
 GROUP 
    BY x.mp3_id HAVING COUNT(*) <= 4 
 ORDER  
    BY user_id DESC
     , mp3_id;

或更快

SELECT mp3_id, user_id FROM
 (
 SELECT x.*, CASE WHEN @prev = user_id THEN @i:=@i+1 ELSE @i:=1 END i, @prev:=user_id FROM my_table x, (SELECT @prev:=null,@i:=1) vars ORDER BY user_id DESC, mp3_id
 ) a
 WHERE i<=4;

mysql选择in（）中的每个id返回等行数

3 个答案: