Question

我想在CASE语句中应用计算，但结果不是我预期的，我不知道如何应用查询来过滤日期（取决于月份）。我的查询有什么问题？请帮忙......谢谢

这是我的查询

SELECT DISTINCT a.SERVICE_CODE, 
       CASE WHEN
           b.SERVICE_CODE IS NULL THEN a.AMT ELSE  SUM (a.AMT) - SUM(b.AMT) END AS TOTAL_AMT
       FROM Table a
 LEFT OUTER JOIN Table b ON (a.SERVICE_CODE = b.SERVICE_CODE)
 WHERE  b.SERVICE_CODE >='3000' AND c.SERVICE_CODE >='3000'

表a

Invoice_No               date          Service_code     Amt
001                     1/7/2014       6000             300
002                     1/8/2014       6003             700
003                     5/8/2014       6003             100
004                     10/8/2014      6005             1000

表b

Credit_No                 date          Service_code     Amt
c100                     1/7/2014       6000             300
c200                     13/8/2014      6003             700

期望的输出

Service_code              Total_Amt
6003                      100
6005                      1000

谢谢

Answer 1

试试这个

在MYSQL

select t1.SERVICE_CODE,(t1.amt- case when  c.amt is null then 0 else c.amt end) 
as AMT  from t1 c right outer join (select t.Invoice_No,SERVICE_CODE,
sum(amt) as amt from t group by SERVICE_CODE) as t1  
on c.SERVICE_CODE=t1.SERVICE_CODE 
inner join tablea a on a.invoice_no=t1.Invoice_No

Fiddle

Answer 2

试试这个：

select 
bcode, 
case 
    when ccode is null then bamt 
    else bamt - camt 
end as amt
from
(select service_code bcode, sum(amt) bamt
 from b
 where service_code >= 3000
 group by service_code) b
left join
(select service_code ccode, sum(amt) camt
 from c
 where service_code >= 3000
group by service_code) c on b.bcode = c.ccode
order by bcode

根据示例数据，表b可能包含表service_code可能不会c的值。因此，请使用左外部联接来链接b和c。此外，由于invoice_no不需要显示在输出中，因此您根本不需要加入a。

Demo

Sql Case Column 1和Column相同的值然后应用计算

2 个答案: