SELECT SUM(nums) FROM demo GROUP BY name

表名：Sum_Test

Id Name Count Type

1 A 12 1
2 B 12 1
3 C 12 1
4 D 12 1
5 B 12 2
6 B 12 1
7 A 12 2
8 A 12 1
9 C 12 1
10 B 12 1

结果： Type 1 ： 84
Type 2 ： 36

求type 为1和2的Count的总和，如果Name即有Type为1也有Type为2的，则该Count算到Type为2的里面，如： A

Type即有1也有2则A的求和时将A的Count值全部算作为Type=2里面

方法一：

现将Type为2的Name的Type值都改为2

UPDATE Sum_Test SET Type=2 WHERE Name IN (SELECT Name FROM Sum_Test SHERE Type=2)

然后再求和

SELECT SUM(Count) FROM Sum_Test GROUP BY Type

方法二：

select sum([count]) from
(
select [count],[Type]='2' from sum_test where [name] in (select [name] from sum_test where type=2)
UNION ALL
select [count],[Type]='1' from sum_test where [name] not in (select [name] from sum_test where type=2)
) t
group by [type]

这两种方法虽然可以得到想要的结果，但是有很大的弊端

方法三：

select (select sum([count]) from sum_test) - (select sum(t.counts) from
(select []name, sum([count]) counts from sum_test
where [name] in(select [name] from sum_test where [type] = 2)
group by [name]) t) type1, (select sum(t.counts) from
(select [name], sum([count]) counts from sum_test
where [name] in(select [name] from sum_test where [type] = 2)
group by [name]) t) type2

select * from (select name, nums from demo) pivot (sum(nums) for name in (select namefrom demo gruop by name));

你试试需要这种效果

content不太对

这不是 select name ,sum(nums) from demo group by name?

select sum(nums) as name from demo
GROUP BY name

2楼正解，不过楼主是DBA吗？基础知识不太牢固

select name,totalnum from (select name as name,sum(nums) as totalnum from demo group by name) pivot (totalnum for name in ('aa','bb','cc','dd'))

你这个就是行变列
使用case 就好了

select
sum(case when 'aa' then 值 else 0 end)aa,
...
from 表名

这样也可以得