SQL聚合组由

我有一个员工表和一个leave_allocation表，它具有一对多的关系，每个员工在一段时间内都有许多假期分配。我想为每位员工获得最新的分配。SQL聚合组由

我试着查询，但日期和天值不关联到同一行

select e.employee_number, e.nme, MAX(l.date), l.days 
from employee e, leave_allocation l 
where l.employee_id = e.employee_id 
group by e.employee_number, e.nme

我怎样才能得到每名雇员的最新配置？

来源

2012-08-30 Clinton Bosch

放置MAX()日在一个子查询：

SELECT e.employee_number, e.nme, l.leavedate, la.days 
FROM employee e 
INNER JOIN 
(
    SELECT Max(date) leavedate, employee_id 
    FROM leave_allocation 
    GROUP BY employee_id 
) l 
    ON e.employee_id = l.employee_id 
INNER JOIN leave_allocation la 
    ON l.employee_id = la.employee_id 
    AND l.leavedate = la.date

我也切换到使用ANSI联接语法，而不是表之间用逗号查询。

来源

2012-08-30 16:48:07 Taryn

SELECT e.employee_number 
     ,e.nme 
     ,l.days 
    FROM employee e 
     , leave_allocation l 
     ,(SELECT employee_id 
       ,MAX(DATE) date 
      FROM leave_allocation 
     GROUP BY employee_id) m 
WHERE l.employee_id = e.employee_id 
    AND l.employee_id = m.employee_id 
    AND l.date = m.date

如果可以有多个具有相同employee_number和日期的行，则需要求和。

SELECT e.employee_number 
     ,e.nme 
     ,sum(l.days) 
    FROM employee e 
     , leave_allocation l 
     ,(SELECT employee_id 
       ,MAX(DATE) date 
      FROM leave_allocation 
     GROUP BY employee_id) m 
WHERE l.employee_id = e.employee_id 
    AND l.employee_id = m.employee_id 
    AND l.date = m.date 
GROUP BY e.employee_number 
     ,e.nme

来源

2012-08-30 16:45:57 dcp

注意，这会产生重复，如果员工ID和日期的组合不是leave_allocation独特。 –

@ Eric Petroelje - 是的，但我的假设是他们，否则，拥有2行具有相同员工ID和日期的行会有什么意义？无论如何，我提出了一个替代方案，以防万一有多行，请参阅编辑答案。 – dcp

试试这个，

SELECT e.employee_number, e.nme, c.maxDate, l.days 
FROM employee e 
      INNER JOIN leave_allocation l 
       ON l.employee_id = e.employee_id 
      INNER JOIN 
      (
       select employee_id, MAX(date) maxDate 
       from leave_allocation 
       group by employee_id 
      ) c ON c.employee_id = l.employee_ID AND 
        c.maxDate = l.date

来源

2012-08-30 16:48:50

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