怎么用分析函数找出EMP表中每个部门工资最高的员工
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