Question

我将这些表格包含以下列：

员工24（EMPLOYEEID，FIRSTNAME，LASTNAME，GENDER，JOBROLES）;
Leave25（EMPLOYEEID，LEAVEID，LEAVETYPE，STARTDATE，ENDDATE）;

我想写一个查询来查找已经获得最大休假天数的员工：

select emp.*, L.Startdate - L.Enddate 
from leave25 l, employee24 emp
where emp.employeeid = l.employeeid 
  and l.startdate - l.enddate = (select max(startdate - enddate) from  leave25)

Answer 1

你可以使用以下嵌套SQL和DENSE_RANK（）函数（包括tie（相同的值））：

SELECT EMPLOYEEID, FIRSTNAME, LASTNAME, LEAVE_DAY
  FROM
(
SELECT L.EMPLOYEEID, FIRSTNAME, LASTNAME, 
       (ENDDATE - STARTDATE + 1) LEAVE_DAY,
       DENSE_RANK() OVER (ORDER BY (ENDDATE - STARTDATE + 1) DESC) DR
  FROM EMPLOYEE24 E JOIN LEAVE25 L ON ( E.employeeid = L.employeeid )
 GROUP BY L.EMPLOYEEID, FIRSTNAME, LASTNAME, ENDDATE - STARTDATE + 1                        
)            
WHERE DR = 1
ORDER BY 1;

SQL Fiddle Demo

修改（根据您的评论＆＃34; 查询过去1个月内获得最多离职数量的员工＆＃ 34;）：

SELECT EMPLOYEEID, FIRSTNAME, LASTNAME,
              LEAVE_DAY
  FROM            
(              
  SELECT EMPLOYEEID, FIRSTNAME, LASTNAME,
         LEAVE_DAY, DENSE_RANK() OVER (ORDER BY LEAVE_DAY DESC) DR 
    FROM
  (
    SELECT L.EMPLOYEEID, FIRSTNAME, LASTNAME,
         DECODE(SIGN(ENDDATE - ADD_MONTHS(TRUNC(SYSDATE),-1)+1),-1,0,
                     ENDDATE - ADD_MONTHS(TRUNC(SYSDATE),-1)+1) -
         DECODE(SIGN(STARTDATE - ADD_MONTHS(TRUNC(SYSDATE),-1)+1),1,
                    (STARTDATE - ADD_MONTHS(TRUNC(SYSDATE),-1)),0)
         as LEAVE_DAY
      FROM EMPLOYEE24 E JOIN LEAVE25 L ON ( E.employeeid = L.employeeid )
  ) 
)
WHERE DR = 1
ORDER BY 1;

SQL Fiddle Demo 2

Answer 2

试试这个：

Select Top 1  e.EMPLOYEEID,e.Name,SUM(DATEDIFF(d,StartDate,EndDate)+1) as LeaveTaken 
from Leave25 l
INNER JOIN Employee24 e on l.EMPLOYEEID=e.EMPLOYEEID
Group By e.EMPLOYEEID,e.Name
Order by LeaveTaken desc

查询具有最大叶数的员工

2 个答案: