sqlplus sys/tiger as sysdba;
alter user scott account unlock;
用户已更改
切换用户：conn scott/tiger as sysdba;
修改密码：alter user scott indentified by tiger;
1.select语句：
select sysdate from dual;
select ename,sal*12+comm from emp;(结果是错的，如果comm为空，那么sal*12+comm整个为空)
select ename ,sal*12 year_sal from emp;
select ename||sal from emp;
select ename || ‘dhfaf‘ from emp;
select ename || ‘jfkjdk‘‘kdjfkla‘ from emp;
distinct:
select distinct deptno from emp; //重复的编号去除掉
select distinct deptno,job from emp; //查询的结果会出现重复的编号和job，但deptno+job不重复
2.where:
select ename,sal,comm from emp where comm is null;
select ename,sal,comm from emp where comm is not null;
select ename,sal,comm from emp where sal in (800,1500,2000);
select ename,sal,comm from emp where sal in (‘Simth‘,‘abc‘,‘ABC‘);
如果想查询大于2007年2月25日以后入职的员工信息
首先select sysdate from emp;
查询结果是25-2月-07
必须得按这个字符串特定的格式来写：
select ename,sal,hiredate from emp where hiredate >‘25-2月-07‘;
select ename,sal,from emp where deptno =10 and/or sal>1000;
where ename,sal from emp where sal not in (800,1500); //取反
select ename from emp where ename like ‘%ALL%‘; //%表示0个或多个字母
select enaem from emp where ename like ‘_A%‘; //_表示占位符
如果名字里面本身就有%：
select ename from emp where ename like ‘%$%%‘ escape ‘$‘; //指定$为转义字符
3.order by
select * from dept order by deptno desc; //降序排列
select empno,ename from emp order by empno asc; //升序排列
select empno,ename from emp where empno <>10 order by empno asc;//先过滤再排序
按照两个字段排序：
select ename,sal,deptno from emp order by deptno asc,ename desc; //先按deptno升序排列再在相同deptno中ename用降序排列
4.函数lower
select lower(ename) from emp;
查询员工名字不管大小写，第二个字母是a的名字：
select ename from emp where lower(ename) like ‘_a%‘;
或者：select ename from emp where ename like ‘_a%‘ or ename like ‘_A%‘;
5.函数substr 截取
select substr(ename,1,3) from emp; //从第一个字符开始截取，长度为3
6.函数chr 把而刺客码转换成字符
select chr(65) from dual; //查询结果为A
7.函数ascii 把字符转换成ASCII
select ascii(‘A‘) from dual; //查询结果为65
8.函数round 四舍五入
select round(23.652) from dual; //24
select round(23.652,2) from dual; //23.65
select round(23.652,1) from dual; //23.7
select round(23.652,-1) from dual; //20
*9.函数to_char 把数字/日期转换成相关字符串并且有特定的格式控制
select to_char(sal,‘$99,999.9999‘) from emp;
select to_char(sal,‘L99,999.9999‘) from emp; //人民币
select to_char(hiredate,‘YYYY-MM-DD HH24:MI:SS‘) from emp;
*10.to_date 把特定格式的字符串转换成日期类型
select ename,hiredate from emp where hiredate >to_date(‘1981-2-20 12:34:56‘,‘YYYY-MM-DD HH24:MI:SS‘);
*11.to_number 把特定格式的数字转换成相关类型
select sal from emp where sal > to_number(‘$1.250.00‘,‘$9,999.99‘);
*12.nvl 处理空值的
select ename,sal*12 + nvl(comm,0) from emp;
*13.组函数：输入好多条记录，输出只有一条
max(sal) min(sal) avg(sal) sum(sal)
select to_char(avg(sal),‘99999.99999‘) from emp;
select round(avg(sal),2) from emp;
count(*) //表示表中有多少条记录(不是空值的有多少个)
select count(*) from emp where deptno=10;
select count(distinct deptno) from emp; //部门编号有多少个
*14.group by
哪个部门的平均薪水是多少：
select deptno,avg(sal) from emp group by deptno;
按照多个字段进行分组：
select deptno,job,max(sal) from emp group by deptno,job;
输出薪水最高的员工的名字：
错误的是：select ename，max(sal) from emp;
真确的是：select ename from emp where sal=(select max(sal) from emp);
输出每个部门挣钱最多人的名字
错误的是：select ename,max(sal) from emp group by deptno;
真确的是：select deptno,max(sal) from emp group by deptno; //deptno在分组里面确确实实只有一个
*15.having 是对分组进行限制
where语句是对单条的记录进行过滤
把平均薪水大于2000的部门取出来
select avg(sal）,deptno from emp group by deptno having avg(sal)>2000;
顺序：select * from emp
where sal>1000
group by deptno
having
order by
例子：薪水大于1200的雇员按照部门编号进行分组，分组后的平均薪水必须大于1500，查询分组后的平均工资，按工资的倒叙排列
select avg(sal) from emp where sal>1200 group by deptno having avg(sal)>1500 order by avg(sal) desc;
*16.子查询：在一个select语句里面套了个另外的select语句
套的select语句可以出现在where里面，也可以出现在from语句后面
求哪些人的工资位于所有平均工资之上
select ename,sal from emp where sal>(select avg(sal) from emp);
按照部门进行分组后，每个部门里面挣钱最多的员工信息
select ename,sal,deptno from emp
where sal=(select max(sal) from emp group by deptno);//错误的
select ename,sal,deptno from emp
where sal in (select max(sal) from emp group by deptno);//错误的
select ename,sal from emp
join (select max(sal) max_sal,deptno from emp group by deptno ) t
on (emp.sal=t.max_sal and emp.deptno=t.deptno); //正确的
*17.自连接：其实是在同一张表内进行两次查询.为同一张表其不同别名，当成两张表来用
找出员工对应的经理的名字（分析：员工名字和经理名字都在emp表的ename列中）
select t1.ename,t2.ename from emp t1 ,emp t2 where t1.mgr=t2.empno;
*18.
等值连接：找出每位员工的部门名称
select ename,dname from emp join dept on (emp.deptno=dept.deptno);
求部门平均薪水的等级
select deptno,avg_sal,grade from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
join salgrade s on (t.avg_sal between s.losal and s.hisal);
求部门平均的薪水等级
select deptno ,avg(grade) from
(select deptno,ename,grade from emp join salgrade s on (emp.sal between s.losal and s.hisal)) t
group by deptno;
雇员中有哪些人是经理
select ename from emp where empno in (select distinct mgr from emp);
面试题：
1.不用组函数，求出薪水的最高值（自连接）
左边的一张表中的薪水小于右边一张表的薪水，有一条或多条是连接不上右边，就是左边的最大值没有小于右边表的薪水
select distinct sal from emp where sal not in
(select distinct e1.sal from emp e1 join emp e2 on (e1.sal<e2.sal));
2.平均薪水最高的部门的部门编号
select deptno,avg_sal from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
where avg_sal=
(select max(avg_sal) from t);
3.平均薪水最高的部门的部门
select dname from dept where deptno =
(
select deptno,avg_sal from
(select avg(sal) avg_sal,deptno from emp group by deptno) t
where avg_sal=
(select max(avg_sal) from t)
);