我想帮助mysql查询引用上面的单元格 例如,在下表中:
primary key(id) day count percentage change
1 monday 1 0
2 tuesday 2 (1-0)*100%=100%
3 wednesday 5 (2-1)*100%=100%
4 thursday 9 (5-2)*100%=300%
5 friday 27 (9-5)*100%=400%
百分比更改结果基于计数列前两天的结果。有没有办法合并主键(id)来引用“上面”的单元格?
答案 0 :(得分:2)
没有一个简单的方法。你需要使用自我加入。
select t.*,
(coalesce(t_1.count, 0) - coalesce(t_2.count, 0)) * 100.0
from t left outer join
t t_1
on t.id = t_1.id + 1 left outer join
t t_2
on t.id = t_2.id + 2
left outer join
确保所有原始行都保留,即使没有前面的ID。
这是有效的,因为ID是顺序的。如果id不是顺序的,并且计数单调递增,则可以使用相关子查询执行此操作:
select t.*,
(coalesce((select max(`count`) as val
from table1 t_1
where t_1.`count` < t.`count`
), 0)
) -
(coalesce((select max(`count`)
from table1 t_2
where t_2.`count` < (select max(`count`) from table1 t_1 where t_1.`count` < t.`count`)
), 0)
)
from table1 t
注意:如果一行中两个值相同,则无法正常工作。为此,您需要使用id:
select t.*,
(coalesce((select max(`count`) as val
from table1 t_1
where t_1.`id` < t.`id`
), 0)
) -
(coalesce((select max(`count`)
from table1 t_2
where t_2.`id` < (select max(`id`) from table1 t_1 where t_1.`id` < t.`id`)
), 0)
)
from table1 t
如果计数没有增加,那么你必须得到id,并再次加入值。真有趣!这是代码:
select t.*,
coalesce(t1.count, 0) - coalesce(t2.count, 0)
from (select t.*,
(select max(`id`) as id1 from table1 t_1 where t_1.`id` < t.`id`
) as id1,
(select max(`count`) from table1 t_2
where t_2.`id` < (select max(`id`) from table1 t_1 where t_1.`id` < t.`id`)
) id2
from table1 t
) t left outer join
table1 t1
on t.id1 = t1.id left outer join
table1 t2
on t.id2 = t2.id