Sql 使用group by时出现无效列名错误

acctkey列上出现无效列错误。你能告诉我我做错了什么吗

select COUNT(acctkey) as Num, case
            when source_sys_cd in ('TRS','CLS') then source_sys_cd + '|'+[acct_num]                 
            when source_sys_cd = 'Hogan CIS' and [acct_type_cd] = 'DDA' then 'DDA' + '|'+ [acct_id]
            when source_sys_cd = 'Hogan CIS' and [acct_type_cd] != 'DDA' then 'TDA' +'|'+ [acct_id]
            when source_sys_cd = 'CLN' then source_sys_cd + '|'+ [acct_num]+ [acct_id]
            when source_sys_cd = 'RCC' then source_sys_cd + '|'+ [acct_id]
            when source_sys_cd = 'ITF' then source_sys_cd + '|'+ [acct_id]+ [acct_num]
            when source_sys_cd = 'SEC' then source_sys_cd + '|'+ [acct_id]
            else source_sys_cd + '|'+ [acct_num]
            end  as acctkey
            from mtb..STAGING_CUST_ACCT
            group by source_sys_cd
            ,acct_id
            ,acct_num
            ,acctkey
            ,acct_type_cd

---

在您的情况下，您可以通过以下方式从组中删除

acctkey

，从技术上解决此问题：

        group by source_sys_cd, acct_id, acct_num, acct_type_cd

所有这些列都包含在

acctkey

的定义中（如果我没有遗漏任何内容的话）。但是，我怀疑这是您真正想要的，因为这可能会为

acctkey

生成重复项

相反，我认为你想要：

with t as (
        select t.*,
           (case
            when source_sys_cd in ('TRS','CLS') then source_sys_cd + '|'+[acct_num]                 
            when source_sys_cd = 'Hogan CIS' and [acct_type_cd] = 'DDA' then 'DDA' + '|'+ [acct_id]
            when source_sys_cd = 'Hogan CIS' and [acct_type_cd] != 'DDA' then 'TDA' +'|'+ [acct_id]
            when source_sys_cd = 'CLN' then source_sys_cd + '|'+ [acct_num]+ [acct_id]
            when source_sys_cd = 'RCC' then source_sys_cd + '|'+ [acct_id]
            when source_sys_cd = 'ITF' then source_sys_cd + '|'+ [acct_id]+ [acct_num]
            when source_sys_cd = 'SEC' then source_sys_cd + '|'+ [acct_id]
            else source_sys_cd + '|'+ [acct_num]
            end)  as acctkey
        from mtb..STAGING_CUST_ACCT t
       )
select count(*), acctkey
from t
group by acctkey;

---

您不能在sql Server中的group by中使用别名可能重复