我正在寻找UPDATE(计算SET值)现有表格的解决方案,其中包含前一个5的第2和第3大数字的总和。
ID | Price | Sum |
| | (2nd + 3rd largest of the previous 5)|
-------------------------------------------------------
1 | 6 | |
2 | 1 | |
3 | 8 | |
4 | 3 | |
5 | 5 | |
6 | 9 | should be 11 |
7 | 1 | should be 13 |
8 | 6 | should be 13 |
9 | 6 | should be 11 |
10 | 9 | should be 12 |
11 | 2 | should be 15 |
12 | 4 | should be 12 |
在EXCEL中,它可以通过:= LARGE(范围,2)+ LARGE(范围,3),其中范围始终指向最后5个数字。
我知道MYSQL的函数有GREATEST(value1,value2,...)和LEAST(value1,value2,...),但是这个函数只返回GREATEST或LEAST值。
如果我需要IGNORE第一个最大数字并且只加上第二和第三大数字,我该如何使这个挑战发挥作用?
这个想法围绕着这个原则:
UPDATE table
SET SUM =
GREATEST(2nd max price) where ID between ID-5 AND ID-1
+
GREATEST(3rd max price) where ID between ID-5 AND ID-1
答案 0 :(得分:1)
你可以试试这个。我希望这就是你想要的。
update yourtable,
(
select id,sum(number) as sum from(
select id,number,
case id when @id then @rownum := @rownum+1 else @rownum := 1 and @id:= id end as r
from(
select t.id,t1.number
from yourtable t
join(
select id,number from yourtable order by number desc
) t1 on t1.id between (t.id - 5) and (t.id - 1)
where t.id > 5
order by t.id asc, t1.number desc
) t2
join (select @rownum:=0, @id:=0) as x
)as t3 where r in(2,3) -- 2nd max + 3rd max.
group by id
)tab
set yourtable.sum = tab.sum
where yourtable.id = tab.id
此查询update
列SUM
与您(2nd Greater + 3rd Greater) from 5 previous ID
匹配。但是,如果您只想在没有更新的情况下查看结果,只需删除UPDATE
语句。
p.s:Number
表示您桌面上的price
。