SQL DB2 与连接操作符相关联的子句或合并语句中的On子句无效。

Question

所以我试图运行一个查询，发现它产生了一个我不清楚的错误。

    select a.employer_id, a.customer_id, s.service, count(a.account_num) as "TOTAL_COUNT" 
        from dhs01.msa_account  a
left outer join dhs01.alt_services s
    on a.customer_id = s.customer_id
        and s.service = 'TH'
        and a.employer_id not in (select s2.employer_id from dhs01.alt_services s2 where s2.service = 'TE')   -<<<<----- THIS IS THE BAD LINE
where 
        a.ACCT_MAINT_FEE = 'E'
        AND a.STAT = 'O'
        AND a.OPEN_CONFIRM_DATE < (CURRENT_DATE - (DAY(CURRENT_DATE)-1) DAYS) ---first of the current 

    group by a.employer_id, a.customer_id, s.service;

查询非常接近工作，但出于某种原因，子选择查询会抛出一个错误（如果我去掉子选择，我得到的结果基本上是正确的，除了没有排除那些我想排除的结果）。 在db2中，我一般都会看到subselects在on子句中的工作，实际上也有类似的例子。

Answer 1

试试。

from dhs01.msa_account  a
left join table
(
select s.*
from dhs01.alt_services s 
where s.customer_id = a.customer_id and s.service = 'TH'
and not exists
(
select 1
from dhs01.alt_services s2 
where s2.service = 'TE' and s2.employer_id = a.employer_id
)
) s on 1=1

而不是:

from dhs01.msa_account  a
left outer join dhs01.alt_services s
    on a.customer_id = s.customer_id
        and s.service = 'TH'
        and a.employer_id not in (select s2.employer_id from dhs01.alt_services s2 where s2.service = 'TE')   -<<<<----- THIS IS THE BAD LINE

Answer 2

你的查询似乎是完全正确的，但你仍然可以使用下面的查询。

SELECT          a.employer_id, 
                a.customer_id, 
                s.service, 
                COUNT(a.account_num) AS "TOTAL_COUNT" 
FROM            dhs01.msa_account a 
LEFT OUTER JOIN dhs01.alt_services s 
ON              a.customer_id = s.customer_id 
AND             s.service = 'TH' 
WHERE           a.acct_maint_fee = 'E' 
AND             a.stat = 'O' 
AND             a.open_confirm_date < (CURRENT_DATE - (DAY(CURRENT_DATE)-1) DAYS) --first of the current 
AND NOT EXISTS  (SELECT NULL
                 FROM   dhs01.alt_services s2 
                 WHERE  s2.service = 'TE'
                 AND    a.employer_id = s2.employer_id)
GROUP BY        a.employer_id, 
                a.customer_id, 
                s.service;

Answer 3

如果你想过滤，你可以尝试以下查询 On Clause:

select a.employer_id, a.customer_id, s.service, count(a.account_num) as "TOTAL_COUNT" 
    from dhs01.msa_account  a
left outer join dhs01.alt_services s
on a.customer_id = s.customer_id
    and s.service = 'TH'
    and a.employer_id <> (select distinct s2.employer_id from dhs01.alt_services s2 where s2.service = 'TE' and a.employer_id=s2.employer_id)   
where 
    a.ACCT_MAINT_FEE = 'E'
    AND a.STAT = 'O'
    AND a.OPEN_CONFIRM_DATE < (CURRENT_DATE - (DAY(CURRENT_DATE)-1) DAYS) ---first of the current 

group by a.employer_id, a.customer_id, s.service;

如果你想排除所有的 employer_id，并且有 service = 'TE' 在执行时间较短的情况下，你应该尝试以下查询：-。

select a.employer_id, a.customer_id, s.service, count(a.account_num) as "TOTAL_COUNT" 
    from dhs01.msa_account  a
left outer join dhs01.alt_services s
on a.customer_id = s.customer_id
    and s.service = 'TH'   
where 
    a.ACCT_MAINT_FEE = 'E'
    AND a.STAT = 'O'
    AND a.OPEN_CONFIRM_DATE < (CURRENT_DATE - (DAY(CURRENT_DATE)-1) DAYS) ---first of the current 
AND not exists(select 1 from dhs01.alt_services s2 where s2.service = 'TE' and a.employer_id=s2.employer_id)

group by a.employer_id, a.customer_id, s.service;