select * from emp;
select * from emp where empno>8000;
select empno,ename,sal,comm from emp where sal<comm;
select deptno,count(*) from emp group by deptno;
select deptno,sum(sal) total_sal from emp where job='MANAGER' group by deptno;
select job,count(distinct deptno) from emp where mgr is not null group by job order by job;
select job,count(distinct deptno) from emp where mgr is not null group by job order by count(distinct deptno) desc,job;
select job, count(distinct deptno) uniq_deptno from emp where mgr is not null group by job order by uniq_deptno desc, job;
select deptno, to_char(hiredate,'yyyy'),count(*) from emp group by deptno,to_char(hiredate,'yyyy');
排序是很耗资源的,所以是最后执行,大家应该记住这一点。
下面看一则需求。
![](http://images0.cnblogs.com/blog/695132/201411/271003084186904.png)