Question

例如，这个表格，我需要将其与id和日期分组，并使用不同的

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

但结果是错误的，例如在id 1上删除了重复内容，而不是320，它只会导致220，因为它会删除重复的amt 100。

所以我试过

SELECT SUM(DISTINCT amt) GROUP BY id, date

但我无法总结。

编辑：对不起，我忘了说结果应该是

id      amt
1       320
2       330

Answer 1

具有长版本，但查询易于理解。使用CTE的以下查询可以帮助您

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

Answer 2

尝试以下。

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

select id,amt from #tmp

Answer 3

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，您需要对按amt和id分组的不同date进行分组。接下来，您必须按id对结果进行分组，汇总部分求和的amt列（在第一步中，我们只汇总了特定日期的不同值）。

Answer 4

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 FIDDLE：http://sqlfiddle.com/#!18/2d356/14/0

Answer 5

使用CTE和Row_number的以下查询可以帮助您

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

SQL Server Sum Distinct Group by

5 个答案: