oracle求同比,环比函数(LAG与LEAD)的详解

数据库  /  管理员 发布于 6年前   780

Lag和Lead函数可以在一次查询中取出同一字段的前N行的数据和后N行的值。

这种操作可以使用对相同表的表连接来实现，不过使用LAG和LEAD有更高的效率。

代码如下:

CREATE TABLE salaryByMonth
(
 employeeNo varchar2(20),
 yearMonth varchar2(6),
 salary number
) ;

insert into SALARYBYMONTH (EMPLOYEENO, YEARMONTH, SALARY)
values (1, '200805', 500);
insert into SALARYBYMONTH (EMPLOYEENO, YEARMONTH, SALARY)
values (1, '200802', 150);
insert into SALARYBYMONTH (EMPLOYEENO, YEARMONTH, SALARY)
values (1, '200803', 200);
insert into SALARYBYMONTH (EMPLOYEENO, YEARMONTH, SALARY)
values (1, '200804', 300);
insert into SALARYBYMONTH (EMPLOYEENO, YEARMONTH, SALARY)
values (1, '200708', 100);
commit;

SELECT EMPLOYEENO
      ,YEARMONTH
      ,SALARY
      ,MIN(SALARY) KEEP(DENSE_RANK FIRST ORDER BY YEARMONTH) OVER(PARTITION BY EMPLOYEENO) FIRST_SALARY -- 基比分析 salary/first_salary
      ,LAG(SALARY, 1, 0) OVER(PARTITION BY EMPLOYEENO ORDER BY YEARMONTH) AS PREV_SAL -- 环比分析，与上个月份进行比较
      ,LAG(SALARY, 12, 0) OVER(PARTITION BY EMPLOYEENO ORDER BY YEARMONTH) AS PREV_12_SAL -- 同比分析，与上个年度相同月份进行比较   
      ,SUM(SALARY) OVER(PARTITION BY EMPLOYEENO, SUBSTR(YEARMONTH, 1, 4) ORDER BY YEARMONTH RANGE UNBOUNDED PRECEDING) LJ --累计值
  FROM SALARYBYMONTH
 ORDER BY EMPLOYEENO
         ,YEARMONTH