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服务器之家 - 数据库 - Oracle - oracle求同比,环比函数(LAG与LEAD)的详解

oracle求同比,环比函数(LAG与LEAD)的详解

2019-11-24 18:40oracle教程网 Oracle

本篇文章是对oracle求同比,环比函数(LAG与LEAD)进行了详细的分析介绍,需要的朋友参考下

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。

复制代码代码如下:


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

 

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