[英]MySQL top 2 records per group
基本上,考慮到最后的created_datetime,我只需要為每個用戶獲取最后的2條記錄:
id | user_id | created_datetime
1 | 34 | '2015-09-10'
2 | 34 | '2015-10-11'
3 | 34 | '2015-05-23'
4 | 34 | '2015-09-13'
5 | 159 | '2015-10-01'
6 | 159 | '2015-10-02'
7 | 159 | '2015-10-03'
8 | 159 | '2015-10-06'
返回(預期輸出):
2 | 34 | '2015-10-11'
1 | 34 | '2015-09-10'
7 | 159 | '2015-10-03'
8 | 159 | '2015-10-06'
我正在嘗試這個想法:
select user_id, created_datetime,
$num := if($user_id = user_id, $num + 1, 1) as row_number,
$id := user_id as dummy
from logs group by user_id
having row_number <= 2
想法是僅保留這些前2行,並刪除所有其他行。
有任何想法嗎?
你的想法很接近。 我認為這會更好:
select u.*
from (select user_id, created_datetime,
$num := if(@user_id = user_id, @num + 1,
if(@user_id := id, 1, 1)
) as row_number
from logs cross join
(select @user_id := 0, @num := 0) params
order by user_id
) u
where row_number <= 2 ;
更改如下:
order by
而不是group by
。 where
的,而不是having
(實際上,在MySQL中having
一定的作用,但where
是比較合適)。
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