如何在 SQL 服务器中将多行汇总为 1 行？

Question

以下是我的语法

with AA as 
(
  select distinct Store_Code, sum([Transactions]) as totaltrans
  from CRM_RETAIL_SUMMARY
  where Store_Code in ('L43','J62','L45','UA2','A35','949','C50','L44','Y45')
    and Trans_Hdr_Sale_Date BETWEEN '2018-12-1' AND '2018-12-31'
  group by Transactions, Store_Code
),
BB as 
( 
  select distinct[Store_Code] as storecode from CRM_RETAIL_SUMMARY
)
select AA.totaltrans, BB.storecode
from AA
inner join BB on AA.Store_Code = BB.storecode

然后结果显示：

storecode   totaltrans
----------------------
A35          65
A35          76
A35          48
A35          62
A35          56
UA2           5
UA2           6
UA2           8
UA2          15
UA2           9
UA2          16
949          16
949          29
949          55

但我想对每个存储代码进行如下求和，但我不知道如何：

storecode  totaltrans
---------------------
A35         307 
UA2          59

谢谢大家的回答

Answer 1

您只需store_code分组，而不是store_code transactions 。 你不需要第二个cte 。

with AA as (
  select Store_Code, sum([Transactions]) as totaltrans
  from CRM_RETAIL_SUMMARY
  where Store_Code in ('L43','J62','L45','UA2','A35','949','C50','L44','Y45')
    and Trans_Hdr_Sale_Date BETWEEN '2018-12-1' AND '2018-12-31'
  group by Store_Code
)
select * from AA;

Answer 2

你会使用group by吗？

select BB.storecode, sum(AA.totaltrans)
from AA join
     BB
     on AA.Store_Code = BB.storecode
group by BB.storecode;

当然，这是在 CTE 之后进行的。