SQL Server按组求和

例如，在这个表中，我需要使用distinct对按id和日期分组的数据进行求和

id      amt          date
1       100          2018/06/01
1       120          2018/06/02
1       100          2018/06/03
1       100          2018/06/03
1       100          2018/06/03
2       100          2018/06/01
2       100          2018/06/01
2       100          2018/06/01
2       130          2018/06/02
2       130          2018/06/02
2       130          2018/06/02
2       130          2018/06/02
2       100          2018/06/03

首先我试过

SELECT SUM(DISTINCT amt) GROUP BY id

但是结果是错误的，它正在删除重复的，例如id 1，而不是320，它只会导致220，因为它删除了重复的amt 100

所以我试过了

SELECT SUM(DISTINCT amt) GROUP BY id, date

但我不能总结

编辑：对不起，我忘了说结果应该是

id      amt
1       320
2       330

试过下面的

SELECT id,SUM(DISTINCT amt) as amt,date #tmp from [yourTableName] GROUP BY id, date

select id,amt from #tmp

试过下面的

SELECT id,SUM(DISTINCT amt) as amt,date #tmp from [yourTableName] GROUP BY id, date

select id,amt from #tmp

---

具有长版本，但易于理解的查询。下面使用CTE的查询应该对您有所帮助

with records as
(select distinct id, date, amt from table_name)
select sum (amt), id
from records
group by 
id;

with cte
as
(      
select 1 id, 100 amt,cast('2018/06/01' as date) date
union all
select 1 id, 120 amt,cast('2018/06/02' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 100 amt,cast('2018/06/03' as date) date
)
select id,sum(amt) as sum1 from 
(
select *,ROW_NUMBER() over(partition by id,date order by id,date) s1 from cte
) b
where s1=1
group by id

---

具有长版本，但易于理解的查询。下面使用CTE的查询应该对您有所帮助

with records as
(select distinct id, date, amt from table_name)
select sum (amt), id
from records
group by 
id;

with cte
as
(      
select 1 id, 100 amt,cast('2018/06/01' as date) date
union all
select 1 id, 120 amt,cast('2018/06/02' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 100 amt,cast('2018/06/03' as date) date
)
select id,sum(amt) as sum1 from 
(
select *,ROW_NUMBER() over(partition by id,date order by id,date) s1 from cte
) b
where s1=1
group by id

试试这个：

select id, SUM(amt) from (
    select id, SUM(distinct amt) amt, [date] from @tbl
    group by id, [date]
) a group by id

首先，您需要按

id

和

date

对不同的

amt

进行分组。接下来，您必须按

id

对结果进行分组，对部分求和的

amt

列求和（在第一步中，我们只对特定日期的不同值求和）。

尝试以下操作：

select id, SUM(amt) from (
    select id, SUM(distinct amt) amt, [date] from @tbl
    group by id, [date]
) a group by id

首先，您需要按

id

和

date

对不同的

amt

进行分组。接下来，您必须按

id

对结果进行分组，对部分求和的

amt

列求和（在第一步中，我们只对特定日期的不同值求和）。

尝试以下操作

SELECT id, 
       Sum(amt) AS amt 
FROM   (SELECT DISTINCT * 
        FROM   mytable) tbl 
GROUP  BY id 

输出

+----+-----+
| id | amt |
+----+-----+
|  1 | 320 |
|  2 | 330 |
+----+-----+

SQL FIDDLE:

试试这个

SELECT id, 
       Sum(amt) AS amt 
FROM   (SELECT DISTINCT * 
        FROM   mytable) tbl 
GROUP  BY id 

输出

+----+-----+
| id | amt |
+----+-----+
|  1 | 320 |
|  2 | 330 |
+----+-----+

---

SQL FIDLE:

下面使用CTE和行号的查询应该可以帮助您

with records as
(select distinct id, date, amt from table_name)
select sum (amt), id
from records
group by 
id;

with cte
as
(      
select 1 id, 100 amt,cast('2018/06/01' as date) date
union all
select 1 id, 120 amt,cast('2018/06/02' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 100 amt,cast('2018/06/03' as date) date
)
select id,sum(amt) as sum1 from 
(
select *,ROW_NUMBER() over(partition by id,date order by id,date) s1 from cte
) b
where s1=1
group by id

---

下面使用CTE和Row_编号的查询应该对您有所帮助

with records as
(select distinct id, date, amt from table_name)
select sum (amt), id
from records
group by 
id;

with cte
as
(      
select 1 id, 100 amt,cast('2018/06/01' as date) date
union all
select 1 id, 120 amt,cast('2018/06/02' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 1 id, 100 amt,cast('2018/06/03' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 100 amt,cast('2018/06/01' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 130 amt,cast('2018/06/02' as date) date
union all
select 2 id, 100 amt,cast('2018/06/03' as date) date
)
select id,sum(amt) as sum1 from 
(
select *,ROW_NUMBER() over(partition by id,date order by id,date) s1 from cte
) b
where s1=1
group by id