Oracle高效的SQL语句之分析函数汇总

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[Oracle]高效的SQL语句之分析函数(一)--sum()

实际应用中我们可以通过sum()统计出组中的总计或者是累加值，具体示例如下：

1.创建演示表


create table emp


as


select * from scott.emp;




alter table emp


add constraint emp_pk


primary key(empno);




create table dept


as


select * from scott.dept;




alter table dept


add constraint dept_pk


primary key(deptno);



2. sum()语句如下：


select deptno,


ename,


sal,


　　--按照部门薪水累加（order by改变了分析函数的作用，只工作在当前行和前一行，而不是所有行）


sum(sal) over (partition by deptno order by sal) CumDeptTot,


sum(sal) over (partition by deptno) SalByDept, --统计一个部门的薪水


sum(sal) over (order by deptno,sal) CumTot, --所有雇员的薪水一行一行的累加


sum(sal) over () TotSal --统计总薪水


from emp


order by deptno, sal



3. 结果如下：


10 MILLER 1300.00 1300 8750 1300 29025


10 CLARK 2450.00 3750 8750 3750 29025


10 KING 5000.00 8750 8750 8750 29025


20 SMITH 800.00 800 10875 9550 29025


20 ADAMS 1100.00 1900 10875 10650 29025


20 JONES 2975.00 4875 10875 13625 29025


20 SCOTT 3000.00 10875 10875 19625 29025


20 FORD 3000.00 10875 10875 19625 29025


30 JAMES 950.00 950 9400 20575 29025


30 WARD 1250.00 3450 9400 23075 29025


30 MARTIN 1250.00 3450 9400 23075 29025


30 TURNER 1500.00 4950 9400 24575 29025


30 ALLEN 1600.00 6550 9400 26175 29025


30 BLAKE 2850.00 9400 9400 29025 29025

[Oracle]高效的SQL语句之分析函数(二)--max()


如果我们按照示例想得到每个部门薪水值最高的雇员的纪录，可以有四种方法实现：

先创建示例表


create table emp


as


select * from scott.emp;




alter table emp


add constraint emp_pk


primary key(empno);




create table dept


as


select * from scott.dept;




alter table dept


add constraint dept_pk


primary key(deptno);

方法1.emp中的每一行都会进行max比较，费时


select * from emp emp1 where emp1.sal=(select max(emp2.sal) from emp emp2 where emp2.deptno=emp1.deptno)

方法2.先子查询查找出max sal，然后与emp表相关联，如果逻辑复杂会产生较多代码

select * from emp emp1,(select deptno,max(sal) maxsal from emp emp2 group by emp2.deptno) emp3 where emp1.deptno=emp3.deptno and emp1.sal=emp3.maxsal

方法3.使用max分析函数


select deptno,maxsal,empno from(


select max(sal) over (partition by deptno) maxsal,emp.* from emp) emp2


where emp2.sal=emp2.maxsal

方法4.使用dense_rank分析函数，如果一个部门可能存在多笔最大薪水，就不能使用row_number()分析函数


select deptno,sal,empno from(


select emp.*,DENSE_RANK() over (partition by deptno order by sal desc) rownumber from emp) emp2


where rownumber=1

结果如下：


10 5000.00 7839


20 3000.00 7788


20 3000.00 7902


30 2850.00 7698

[Oracle]高效的SQL语句之分析函数(三)--row_number() /rank()/dense_rank()

有些时候我们希望得到指定数据中的前n列，示例如下：

得到每个部门薪水最高的三个雇员：

先创建示例表


create table emp


as


select * from scott.emp;




alter table emp


add constraint emp_pk


primary key(empno);




create table dept


as


select * from scott.dept;




alter table dept


add constraint dept_pk


primary key(deptno);



先看一下row_number() /rank()/dense_rank()三个函数之间的区别


select emp.deptno,emp.sal,emp.empno,row_number() over (partition by deptno order by sal desc) row_number, --1,2,3


rank() over (partition by deptno order by sal desc) rank, --1,1,3


dense_rank() over (partition by deptno order by sal desc) dense_rank from emp --1,1,2

结果如下:


10 5000.00 7839 1 1 1


10 2450.00 7782 2 2 2


10 1300.00 7934 3 3 3


20 3000.00 7788 1 1 1


20 3000.00 7902 2 1 1


20 2975.00 7566 3 3 2


20 1100.00 7876 4 4 3


20 800.00 7369 5 5 4


30 2850.00 7698 1 1 1


30 1600.00 7499 2 2 2

取每个部门的薪水前三位雇员：


select t.deptno,t.rank,t.sal from


(


select emp.*,row_number() over (partition by deptno order by sal desc) row_number, --1,2,3


rank() over (partition by deptno order by sal desc) rank, --1,1,3


dense_rank() over (partition by deptno order by sal desc) dense_rank from emp --1,1,2


) t


where t.rank<=3

结果如下：


10 1 5000.00


10 2 2450.00


10 3 1300.00


20 1 3000.00


20 1 3000.00


20 3 2975.00


30 1 2850.00


30 2 1600.00


30 3 1500.00

如果想输出成deptno sal1 sal2 sal3这种类型的格式
步骤一（decode）：


select t.deptno,decode(row_number,1,sal) sal1,decode(row_number,2,sal) sal2,decode(row_number,3,sal) sal3 from


(


select emp.*,row_number() over (partition by deptno order by sal desc) row_number, --1,2,3


rank() over (partition by deptno order by sal desc) rank, --1,1,3


dense_rank() over (partition by deptno order by sal desc) dense_rank from emp --1,1,2


) t


where t.rank<=3



结果如下：


10 5000


10 2450


10 1300


20 3000


20 3000


20 2975


30 2850


30 1600


30 1500

步骤二（使用聚合函数去除null,得到最终结果）：


select t.deptno,max(decode(row_number,1,sal)) sal1,max(decode(row_number,2,sal)) sal2,max(decode(row_number,3,sal)) sal3 from


(


select emp.*,row_number() over (partition by deptno order by sal desc) row_number, --1,2,3


rank() over (partition by deptno order by sal desc) rank, --1,1,3


dense_rank() over (partition by deptno order by sal desc) dense_rank from emp --1,1,2


) t


where t.rank<=3


group by t.deptno

结果如下：


10 5000 2450 1300


20 3000 3000 2975


30 2850 1600 1500

[Oracle]高效的SQL语句之分析函数(四)--lag()/lead()

有时候报表上面需要显示该笔操作的上一步骤或者下一步骤的详细信息，这个时候可以按照下面的做法：

先创建示例表：


-- Create table


create table LEAD_TABLE


(


CASEID VARCHAR2(10),


STEPID VARCHAR2(10),


ACTIONDATE DATE


)


tablespace COLM_DATA


pctfree 10


initrans 1


maxtrans 255


storage


(


initial 64K


minextents 1


maxextents unlimited


);




insert into LEAD_TABLE values('Case1','Step1',to_date('20070101','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case1','Step2',to_date('20070102','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case1','Step3',to_date('20070103','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case1','Step4',to_date('20070104','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case1','Step5',to_date('20070105','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case1','Step4',to_date('20070106','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case1','Step6',to_date('20070101','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case1','Step1',to_date('20070201','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case2','Step2',to_date('20070202','yyyy-mm-dd'));


insert into LEAD_TABLE values('Case2','Step3',to_date('20070203','yyyy-mm-dd'));


commit;



每一条记录都能连接到上/下一行的内容


select caseid,stepid,actiondate,lead(stepid) over (partition by caseid order by actiondate) nextstepid,


lead(actiondate) over (partition by caseid order by actiondate) nextactiondate,


lag(stepid) over (partition by caseid order by actiondate) prestepid,


lag(actiondate) over (partition by caseid order by actiondate) preactiondate


from lead_table

结果如下：


Case1 Step1 2007-1-1 Step2 2007-1-2


Case1 Step2 2007-1-2 Step3 2007-1-3 Step1 2007-1-1


Case1 Step3 2007-1-3 Step4 2007-1-4 Step2 2007-1-2


Case1 Step4 2007-1-4 Step5 2007-1-5 Step3 2007-1-3


Case1 Step5 2007-1-5 Step4 2007-1-6 Step4 2007-1-4


Case1 Step4 2007-1-6 Step6 2007-1-7 Step5 2007-1-5


Case1 Step6 2007-1-7 Step4 2007-1-6


Case2 Step1 2007-2-1 Step2 2007-2-2


Case2 Step2 2007-2-2 Step3 2007-2-3 Step1 2007-2-1


Case2 Step3 2007-2-3 Step2 2007-2-2

还可以进一步统计一下两者的相差天数


select caseid,stepid,actiondate,nextactiondate,nextactiondate-actiondate datebetween from (


select caseid,stepid,actiondate,lead(stepid) over (partition by caseid order by actiondate) nextstepid,


lead(actiondate) over (partition by caseid order by actiondate) nextactiondate,


lag(stepid) over (partition by caseid order by actiondate) prestepid,


lag(actiondate) over (partition by caseid order by actiondate) preactiondate


from lead_table)

结果如下：


Case1 Step1 2007-1-1 2007-1-2 1


Case1 Step2 2007-1-2 2007-1-3 1


Case1 Step3 2007-1-3 2007-1-4 1


Case1 Step4 2007-1-4 2007-1-5 1


Case1 Step5 2007-1-5 2007-1-6 1


Case1 Step4 2007-1-6 2007-1-7 1


Case1 Step6 2007-1-7


Case2 Step1 2007-2-1 2007-2-2 1


Case2 Step2 2007-2-2 2007-2-3 1


Case2 Step3 2007-2-3