如何用分析函数找出EMP表中每个部门工资最高的员工

EMP表是Oracle测试账户SCOTT中的一张雇员表，首先，我们来看看emp表的数据

SQL> select * from emp;

EMPNO ENAME      JOB              MGR HIREDATE         SAL       COMM     DEPTNO
----- ---------- --------- ---------- --------- ---------- ---------- ----------
 7369 SMITH      CLERK           7902 17-DEC-80        800                    20
 7499 ALLEN      SALESMAN        7698 20-FEB-81       1600       300          30
 7521 WARD       SALESMAN        7698 22-FEB-81       1250       500          30
 7566 JONES      MANAGER         7839 02-APR-81       2975                    20
 7654 MARTIN     SALESMAN        7698 28-SEP-81       1250      1400          30
 7698 BLAKE      MANAGER         7839 01-MAY-81       2850                    30
 7782 CLARK      MANAGER         7839 09-JUN-81       2450                    10
 7788 SCOTT      ANALYST         7566 19-APR-87       3000                    20
 7839 KING       PRESIDENT            17-NOV-81       5000                    10
 7844 TURNER     SALESMAN        7698 08-SEP-81       1500         0          30
 7876 ADAMS      CLERK           7788 23-MAY-87       1100                    20
 7900 JAMES      CLERK           7698 03-DEC-81        950                    30
 7902 FORD       ANALYST         7566 03-DEC-81       3000                    20
 7934 MILLER     CLERK           7782 23-JAN-82       1300                    10

14 rows selected.

其中，empno是员工编号，同时也是该表的主键，ename是员工姓名，sal是员工工资，deptno是员工部门。

如何找出每个部门的最高工资的员工信息呢？

常用的方法是关联查询，SQL语句如下：

select emp.deptno,ename,sal
from emp,
(select deptno,max(sal)maxsal from emp group by deptno) t
where emp.deptno=t.deptno and emp.sal=t.maxsal;

结果如下：

    DEPTNO ENAME             SAL
---------- ---------- ----------
        30 BLAKE            2850
        20 SCOTT            3000
        10 KING             5000
        20 FORD             3000

下面我们来看看执行计划：

Execution Plan
----------------------------------------------------------
Plan hash value: 269884559

-----------------------------------------------------------------------------
| Id  | Operation            | Name   | Rows   | Bytes  | Cost (%CPU) | Time     |
-----------------------------------------------------------------------------
|   0 | SELECT STATEMENT     |        |      3 |    117 |      7  (15)| 00:00:01 |
|*  1 |  HASH JOIN           |        |      3 |    117 |      7  (15)| 00:00:01 |
|   2 |   VIEW               |        |      3 |     78 |      4  (25)| 00:00:01 |
|   3 |    HASH GROUP BY     |        |      3 |     21 |      4  (25)| 00:00:01 |
|   4 |     TABLE ACCESS FULL| EMP    |     14 |     98 |      3   (0)| 00:00:01 |
|   5 |   TABLE ACCESS FULL  | EMP    |     14 |    182 |      3   (0)| 00:00:01 |
-----------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - access("EMP"."DEPTNO"="T"."DEPTNO" AND "EMP"."SAL"="T"."MAXSAL")

Statistics
----------------------------------------------------------
      0  recursive calls
      0  db block gets
     13  consistent gets
      0  physical reads
      0  redo size
    625  bytes sent via SQL*Net to client
    419  bytes received via SQL*Net from client
      2  SQL*Net roundtrips to/from client
      0  sorts (memory)
      0  sorts (disk)
      4  rows processed

不难看出，该查询针对同一个表走了两次全盘扫描，成本为7，逻辑读为13。

如何对上述查询进行优化呢？在这里，用到分析函数LAST_VALUE，LAST_VALUE返回排序集中的最后一个值。

SELECT deptno,ename,sal,
       LAST_VALUE(sal)
       OVER(PARTITION BY deptno
            ORDER BY sal
            ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)maxsal
FROM emp;

输出结果如下：

    DEPTNO ENAME             SAL     MAXSAL
---------- ---------- ---------- ----------
        10 MILLER           1300       5000
        10 CLARK            2450       5000
        10 KING             5000       5000
        20 SMITH             800       3000
        20 ADAMS            1100       3000
        20 JONES            2975       3000
        20 SCOTT            3000       3000
        20 FORD             3000       3000
        30 JAMES             950       2850
        30 MARTIN           1250       2850
        30 WARD             1250       2850
        30 TURNER           1500       2850
        30 ALLEN            1600       2850
        30 BLAKE            2850       2850

14 rows selected.

不难看出，sal等于maxsal的行即为每个部门最高工资的员工,下面用嵌套子查询得到目标结果。

SELECT deptno,ename,sal FROM (
       SELECT deptno,ename,sal,
              LAST_VALUE(sal)
              OVER(PARTITION BY deptno
                   ORDER BY sal
                   ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)maxsal
       FROM emp) WHERE sal=maxsal;

输出结果如下：

    DEPTNO ENAME             SAL
---------- ---------- ----------
        10 KING             5000
        20 SCOTT            3000
        20 FORD             3000
        30 BLAKE            2850

下面我们来看看该语句的执行计划：

Execution Plan
----------------------------------------------------------
Plan hash value: 4130734685

----------------------------------------------------------------------------
| Id  | Operation            | Name  | Rows  | Bytes | Cost (%CPU)| Time       |
----------------------------------------------------------------------------
|   0 | SELECT STATEMENT     |       |    14 |   644 |     4  (25)| 00:00:01 |
|*  1 |  VIEW                |       |    14 |   644 |     4  (25)| 00:00:01 |
|   2 |   WINDOW SORT        |       |    14 |   182 |     4  (25)| 00:00:01 |
|   3 |    TABLE ACCESS FULL |  EMP  |    14 |   182 |     3   (0)| 00:00:01 |
----------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - filter("SAL"="MAXSAL")

Statistics
----------------------------------------------------------
      0  recursive calls
      0  db block gets
      6  consistent gets
      0  physical reads
      0  redo size
    619  bytes sent via SQL*Net to client
    419  bytes received via SQL*Net from client
      2  SQL*Net roundtrips to/from client
      1  sorts (memory)
      0  sorts (disk)
      4  rows processed

可见，引入了分析函数以后，成本和逻辑读都减少了一半。

通过查询的结果，我们可以看出，20号部门有两个人的工资最高，有时候，我们只想得到一个人的信息，如何实现呢？

在这里我们会用到分析函数LAG，具体SQL如下：

SELECT deptno,ename,sal,LAG(sal)OVER(ORDER BY deptno) presal FROM (
       SELECT deptno,ename,sal,
              LAST_VALUE(sal)
              OVER(PARTITION BY deptno
              ORDER BY sal
              ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)maxsal
       FROM emp) WHERE sal=maxsal;

输出结果如下：

    DEPTNO ENAME             SAL     PRESAL
---------- ---------- ---------- ----------
        10 KING             5000
        20 SCOTT            3000       5000
        20 FORD             3000       3000
        30 BLAKE            2850       3000

剔除sal等于presal的行

SELECT deptno,ename,sal FROM (
       SELECT deptno,ename,sal,LAG(sal)OVER(ORDER BY deptno) presal FROM (
              SELECT deptno,ename,sal,
                     LAST_VALUE(sal)
                     OVER(PARTITION BY deptno
                     ORDER BY sal
                     ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING)maxsal
               FROM emp)
       WHERE sal=maxsal) WHERE sal <> presal or presal is null;

输出结果如下：

    DEPTNO ENAME             SAL
---------- ---------- ----------
        10 KING             5000
        20 SCOTT            3000
        30 BLAKE            2850

总结：

在实际生产环境中，此类应用还是蛮多的，譬如如何查询每个时段耗时最大的工单。当然，通过上述演示，我们也看出了group by函数的局限性。

关于LAST_VALUE和LAG函数的具体应用及说明，可参考Oracle官方文档：

1. LAST_VALUE

2. LAG