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񡜧employee

񡜧salary


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select a.*,b.* from emplyee a 
inner join salary b on b.employee_id = a.employee
where max(b.month)= b.month

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SQL code:
select *
from (
select t.*,
row_number() over(partition by employee_id order by month desc) rn
from salary t
)
where rn = 1

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select sum(salary) from salary
group by month
having month = 3

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SQL code:
SELECT * from SALARY WHERE (EMPLOYEE_ID,MONTH) IN
(SELECT EMPLOYEE_ID,MAX(MONTH) from SALARY GROUP BY EMPLOYEE_ID)

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select a.*,b.* from emplyee a  
inner join salary b on b.employee_id = a.id ---¿´´í×ֶΡ£
where max(b.month)= b.month

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ÒýÓÃ

select a.*,b.* from emplyee a
inner join salary b on b.employee_id = a.id ---¿´´í×ֶΡ£
where max(b.month)= b.month

ÓÐÎÊÌâ 

´Ë´¦²»ÔÊÐí´¦Àí max(b.month)= b.month

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SQL code:

With t As
(Select 1 eid,1 mon,10