我有三个表,其中SELECT
语句中的select列优先。
假设表A有一些列,例如:
表A:
A_ID |name
---------|-------
1 |name1
表B:
purchase |A_ID |type | market | group | rate | max | min
---------|-----|-----|--------|-------|---------|--------|---------
1 | 1 | 1 | 1 | 1 | 0.12 | 1000 | 500
1 | 1 | 2 | 1 | 1 | 0.3 | 2000 | 1500
0 | 1 | 3 | 1 | 1 | 0.2 | 5000 | 800
0 | 1 | 4 | 1 | 1 | 0.6 | 8000 | 2800
0 | 1 | 6 | 1 | 1 | 0.7 | null | 2800
表C:
purchase |A_ID |type | market | group | rate | max | min
---------|-----|-----|--------|-------|---------|--------|---------
1 | 1 | 1 | 1 | null | 0.2 | null | null
1 | 1 | 2 | 1 | null | null | 5000 | 3000
0 | 1 | 3 | 1 | null | 0.5 | 3000 | 1000
0 | 1 | 5 | 1 | null | 0.4 | 3800 | 2000
0 | 1 | 6 | 1 | null | null | null | 3000
所需结果:
purchase |A_ID |type | market | rate | max | min
---------|-----|-----|--------|---------|--------|---------
1 | 1 | 1 | 1 | 0.2 | 1000 | 500
1 | 1 | 2 | 1 | 0.3 | 5000 | 3000
0 | 1 | 3 | 1 | 0.5 | 3000 | 1000
0 | 1 | 5 | 1 | 0.4 | 3800 | 2000
0 | 1 | 4 | 1 | 0.6 | 8000 | 2800
0 | 1 | 6 | 1 | 0.7 | null | 3000
从列中获取价值的规则:
1- Table C
的优先级高于Table B
,这意味着如果两个变量在同一列中都具有值,则结果将从Table C
中提取,除非该值为null
2-结果可以是GROUP BY
上的purchase, type, market
3-结果具有FULL JOIN
,这意味着如果某行在另一侧具有等效的行,则使用优先级获取值,如果不是整行都进入结果
4-列选择值的优先级(比率|最大|最小):
rate
中的列TABLE C
有值,而没有考虑TABLE B
上的值
==>从TABLE C
rate
中的列TABLE C
是null
,但在TABLE B
中有值==>从TABLE B
中选择结果答案 0 :(得分:1)
这是使用sql server语法的,我确定您可以根据需要进行更改:
首先设置样本数据:
declare @a table(purchase int,A_ID int,[type] int,market int,[group] int,rate decimal(5,2),[max] int,[min] int)
insert @a values (1,1,1,1,1,0.12,1000,500)
,(1,1,2,1,1,0.3,2000,1500)
,(0,1,3,1,1,0.2,5000,800)
,(0,1,4,1,1,0.6,8000,2800)
,(0,1,6,1,1,0.7,null,2800)
declare @b table(purchase int,A_ID int,[type] int,market int,[group] int,rate decimal(5,2),[max] int,[min] int)
insert @b values
(1,1,1,1,null,0.2,null,null)
,(1,1,2,1,null,null,5000,3000)
,(0,1,3,1,null,0.5,3000,1000)
,(0,1,5,1,null,0.4,3800,2000)
,(0,1,6,1,null,null,null,3000)
然后查询:
select coalesce(b.purchase,a.purchase) purchase,
coalesce(b.A_ID,a.A_ID) A_ID,
coalesce(b.[type],a.[type]) [type],
coalesce(b.market,a.market) market,
coalesce(b.rate,a.rate) rate,
coalesce(b.[max],a.[max]) [max],
coalesce(b.[min],a.[min]) [min]
from @a a
full outer join @b b on b.purchase=a.purchase and b.[type]=a.[type] and b.market=a.market
order by rate
添加任何您需要的排序。
答案 1 :(得分:0)
使用Coalesce()函数始终返回列表中的第一个非空值
select coalesce(b.purchase,a.purchase) purchase,
coalesce(b.A_ID,a.A_ID) A_ID,
coalesce(b.[type],a.[type]) [type],
coalesce(b.market,a.market) market,
coalesce(b.rate,a.rate) rate,
coalesce(b.[max],a.[max]) [max],
coalesce(b.[min],a.[min]) [min]
from tableB a full outer join tableC b
on a.purchase=b.purchase and a.[type]=b.[type]