Question

我已经加入了两个表，我想使用GROUP BY，但出现错误。我尝试替换所有列而不是*，但无法正常工作。

SELECT t.*

       FROM
(select O.[CUSTADDRESSID]
      ,O.[ACCOUNTNO]
      ,O.[ADDRESSTYPE]
      ,O.[ADDRESSLINE1]
      ,O.[ISCOMMUNICATION]
      ,C.[CUSTOMERID]
      ,C.[ACCOUNTNO]
      ,C.[ACCOUNTGROUPID]
      ,C.[PREPAIDACCOUNTSTATUSID]
      ,C.[PREPAIDACCOUNTSTATUSDATE]

from [ISSUER].[HISTORY].[TP_CUSTOMER_PREPAIDACCOUNTS] c
full join  [ISSUER].[PLAY].[TP_CUSTOMER_ADDRESSES] o
on c.ACCOUNTNO = o.ACCOUNTNO)  as t
group by ACCOUNTNO

我想从加入结果中删除重复的ACCOUNTNO。

Answer 1

使用t。*时，意味着您选择了所有列，但是仅按组使用了一个列，这就是为什么会出现此错误的原因，现在，如果您仅在选择中使用accountno，并使用过滤器来过滤重复的账户，则不会，那么您可以使用count聚合函数以及下面的查询

SELECT [ACCOUNTNO]
           FROM
          (select O.[CUSTADDRESSID]
          ,O.[ACCOUNTNO]
          ,O.[ADDRESSTYPE]
          ,O.[ADDRESSLINE1]
          ,O.[ISCOMMUNICATION]
          ,C.[CUSTOMERID]
          ,C.[ACCOUNTNO] as C_ACCOUNTNO
          ,C.[ACCOUNTGROUPID]
          ,C.[PREPAIDACCOUNTSTATUSID]
          ,C.[PREPAIDACCOUNTSTATUSDATE]

    from [ISSUER].[HISTORY].[TP_CUSTOMER_PREPAIDACCOUNTS] c
    full join  [ISSUER].[PLAY].[TP_CUSTOMER_ADDRESSES] o
    on c.ACCOUNTNO = o.ACCOUNTNO
   )  as t
    group by ACCOUNTNO
    having count([ACCOUNTNO])>1

但是，如果您要查询，则必须使用distinct

SELECT distinct t.*

       FROM
(select O.[CUSTADDRESSID]
      ,O.[ACCOUNTNO]
      ,O.[ADDRESSTYPE]
      ,O.[ADDRESSLINE1]
      ,O.[ISCOMMUNICATION]
      ,C.[CUSTOMERID]
      ,C.[ACCOUNTNO] as C_ACCOUNTNO
      ,C.[ACCOUNTGROUPID]
      ,C.[PREPAIDACCOUNTSTATUSID]
      ,C.[PREPAIDACCOUNTSTATUSDATE]

from [ISSUER].[HISTORY].[TP_CUSTOMER_PREPAIDACCOUNTS] c
full join  [ISSUER].[PLAY].[TP_CUSTOMER_ADDRESSES] o
on c.ACCOUNTNO = o.ACCOUNTNO)  as t

您可以了解有关group by

Answer 2

您在内部查询中两次指定了相同的ACCOUNTNO

SELECT t.*
FROM
(
select O.[CUSTADDRESSID]
      ,O.[ACCOUNTNO]             -- 1st occurance
      ,O.[ADDRESSTYPE]
      ,O.[ADDRESSLINE1]
      ,O.[ISCOMMUNICATION]
      ,C.[CUSTOMERID]
      ,C.[ACCOUNTNO]             -- second occurance
      ,C.[ACCOUNTGROUPID]
      ,C.[PREPAIDACCOUNTSTATUSID]
      ,C.[PREPAIDACCOUNTSTATUSDATE]    
from [ISSUER].[HISTORY].[TP_CUSTOMER_PREPAIDACCOUNTS] c
full join  [ISSUER].[PLAY].[TP_CUSTOMER_ADDRESSES] o
on c.ACCOUNTNO = o.ACCOUNTNO)  as t
--group by ACCOUNTNO

您可以使用ACCOUNTNO示例之一的别名

,O.[ACCOUNTNO] AS ACCNO1
,C.[ACCOUNTNO] AS ACCNO2

或者或者

因为您正在对ACCOUNTNO做FULL OUTER JOIN 您应该使用COALESCE()从TP_CUSTOMER_PREPAIDACCOUNTS或TP_CUSTOMER_ADDRESSES返回

将内部查询更改为

select  O.[CUSTADDRESSID]
       ,COALESCE(O.[ACCOUNTNO], C.[ACCOUNTNO]) AS ACCOUNTNO
       ,O.[ADDRESSTYPE]

Answer 3

您可以尝试以下查询：它将为您提供重复的帐户编号

SELECT [ACCOUNTNO],count([ACCOUNTNO])
          FROM
    (select O.[CUSTADDRESSID]
          ,O.[ACCOUNTNO]
          ,O.[ADDRESSTYPE]
          ,O.[ADDRESSLINE1]
          ,O.[ISCOMMUNICATION]
          ,C.[CUSTOMERID]
          ,C.[ACCOUNTNO] as accountno_c
          ,C.[ACCOUNTGROUPID]
          ,C.[PREPAIDACCOUNTSTATUSID]
          ,C.[PREPAIDACCOUNTSTATUSDATE]

    from [ISSUER].[HISTORY].[TP_CUSTOMER_PREPAIDACCOUNTS] c
    full join  [ISSUER].[PLAY].[TP_CUSTOMER_ADDRESSES] o
    on c.ACCOUNTNO = o.ACCOUNTNO)  as t
    group by ACCOUNTNO
    having count([ACCOUNTNO])>1

列被多次指定

3 个答案: