DML(Data Manipulation Language)语句 on Oracle

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DML语句(select + inser update delete)--insert
insert into tablename values(col1value,col2value···);--可以插入特定字段insert into dept2 (deptno,dname) values(60,'game2')--可以插入整张表insert into dept1 (select * from dept2)--update
update tablename set col1name=value1,col2name=value2··· where···--delete
delete from tablenamedelete from tablename where···--select
select * from tablename;select col1name,col2name from tablename;select col1name,col2name*12 from tablename;select col1name,col2name*12 "another name" from tablename;--anothername里有特殊字符，用双引号括起来,双引号可以保持别名的格式desc dual; --空表，用来算数学表达式select sysdate from dual;select colmame1||colname2 --|| 字符串连接符--''单引号括起来的是字符串，若字符串内有单引号则用两个单引号代替一个单引号--去除重复项，distinct修饰多个列时保证查询结果对这几个列的组合是不重复的select distinct colname from tablename;--value 为字符串时加''--比较运算符：<,>,不等于<>,between···and···,<=,>=;--逻辑运算符：and,or，not--其他判断    ：is null,is not null,in(value1,value2,value3)等于其中之一，not in()--比较日期时日期字符串要按标准格式写select * from tablename where colname = value;--模糊查询 like 通配符 %零个或多个 _代表一个字母 \转义字符···like '%$%%' escape '$' --自己指定转义字符为$···order by colname1 desc,colname2 asc;--desc降序的 默认升序asc;先按colmname1降序，相同的再用colmname2升序排常用sql函数：
单行函数
lower(colname) upper(cloname)substr(colname,NUM1,NUM2) --从第NUM1位开始取 NUM2位chr(ASCII)ascii('CHARACTER') ROUND(DECIMALS)-- ROUND(小数,四舍五入到第几位)to_char(colname,'$99,999.9999') --9代表一位数字，但是这位上没有数字时不显示；0也代表一位数字，该位没数字时用0补； L代表本地货币to_char(colname,'YYYY-MM-DD HH:MI:SS')to_char(colname,'YYYY-MM-DD HH24:MI:SS')to_date('1988-10-24 12:34:56','YYYY-MM-DD HH24:MI:SS')to_number('$1,250.00',"$9,999.99")nvl(colname,0) --若conname为null，则变为0组函数：
--多行函数!!!!!!!!!!!!!!!!!!!!!!!!max(colname) min(colname)avg(colname)sum(colname) count(*)--一共多少条记录count(colname)--只计算不是空值的count(distinct colname)--只计算不重复的group by colnamegroup by colname1,colname2-- 按这俩的组合分组!!!!!!!出现在select 列表中的字段如果不是出现在组函数里，那么必须存在于group by子句里!!!!!!!having  where对行操作，having对组操作，分组后再havingselect···from···where···group by··having··order by···--where后为筛选元组的条件，group by为分组条件，having后为筛选分组的条件，order by为排序条件子查询
select ename,sal from emp where sal =(select max(sal) from emp);　　--自连接（把一张表当两张表用）
select e1.ename,e2.ename from emp e1,emp e2 where e1.mgr = e2.empno;--92版select e1.ename,e2.ename from emp e1 join emp e2 on (e1.mgr = e2.empno);--99版　　--交叉连接
select ename,dname from emp,dept;--92版select ename,dname from emp cross join dept;--99版　　--等值连接
select ename,dname from emp,dept where emp.deptno = dept.deptno;--92版select ename,dname from emp join dept on (emp.deptno = dept.deptno);--99版select ename,dname from emp join dept using (deptno);--不推荐用　　--非等值连接
select ename,sal,grade--99版from emp e join salgrade son (e.sal between s.losal and s.hisal);select ename,dname,gradefrom emp e join dept d on (e.deptno=d.deptno)join salgrade son (e.sal between s.losal and s.hisal)where ename not like '_A%';　　--外连接
--左外连接：
select e1.ename,e2.ename from emp e1 left join emp e2 on (e1.mgr=e2.empno);　　--右外连接：
select ename,dname from emp e right join dept d on (e.deptno=d.deptno);　　--全外连接
select ename,dname from emp e full join dept d on (e.deptno=d.deptno);--基于Oracle示例表的子查询练习
--求部门中那些人薪水最高
select ename,sal,emp.deptnofrom emp join (select max(sal) max_sal,deptno from emp group by deptno) ton (t.max_sal = emp.sal and emp.deptno = t.deptno);　　--部门平均薪水的等级
select deptno,avg_sal,gradefrom salgrade join (select avg(sal) avg_sal,deptno from emp group by deptno) ton (t.avg_sal>salgrade.losal and t.avg_sal<salgrade.hisal);　　--部门平均的薪水等级
select deptno,avg(grade)from emp e join salgrade son (e.sal between s.losal and s.hisal)group by deptno;select deptno,avg(grade)from (select deptno,ename,sal,gradefrom emp e join salgrade son (e.sal between s.losal and s.hisal))group by deptno;　　--雇员中有哪些人是经理人
select distinct e2.enamefrom emp e1 join emp e2on (e1.mgr=e2.empno);　　--求普通员工时要注意了，not in不行，子查询里有空值，空值代表任意值
select ename from emp where empno in (select distinct mgr from emp);select ename from emp where empno not in (select distinct mgr from emp where mgr is not null)　　--不准用组函数，求薪水的最高值
select distinct e1sal from(select distinct e1.sal e1sal,e2.sal e2salfrom emp e1 left join emp e2on (e1.sal<e2.sal))where e2sal is null;select distinct e1.sal from emp e1 left join emp e2 on (e1.sal<e2.sal) where e2.sal is null;select distinct sal from empwhere sal not in (select distinct e1.sal e1salfrom emp e1 join emp e2on (e1.sal<e2.sal));　　--求平均薪水最高的部门的部门编号
select deptno,avgsal from (select deptno,avg(sal) avgsal from emp group by deptno)where avgsal=(select max(avgsal) from (select deptno,avg(sal) avgsal from emp group by deptno));　　--组函数嵌套，最多两层
select deptno,avgsal from (select deptno,avg(sal) avgsal from emp group by deptno)where avgsal=(select max(avg(sal)) from emp group by deptno);　　--求平均薪水最高的部门的部门名称
select dname from dept where deptno =(select deptno from (select deptno,avg(sal) avgsal from emp group by deptno)where avgsal=(select max(avgsal) from (select deptno,avg(sal) avgsal from emp group by deptno)))　　--求平均薪水的等级最低的部门的部门名称
select dname from dept where deptno =(select deptno from ( select deptno,grade from(select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade son (t.avg_sal between s.losal and hisal))where grade =(select min(grade) from(select deptno,grade from(select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade son (t.avg_sal between s.losal and hisal))) )select d.deptno,dname,avg_sal,grade from (select deptno,avg_sal,grade from(select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade son (t.avg_sal between s.losal and hisal)) t join dept d on (t.deptno=d.deptno)where grade =(select min(grade) from(select deptno,grade from(select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade son (t.avg_sal between s.losal and hisal)))select dname from(select t.deptno,dname,grade from (select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade son (t.avg_sal between s.losal and s.hisal)join dept d on (t.deptno=d.deptno))where grade=(select min(grade) from(select t.deptno,dname,grade from (select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade son (t.avg_sal between s.losal and s.hisal)join dept d on (t.deptno=d.deptno)))　　--创建视图
grant create table,create view to scott;create view V$_dept_sal_info as(select t.deptno,dname,grade from( (select deptno,avg(sal) avg_sal from emp group by deptno) t join salgrade son (t.avg_sal between s.losal and s.hisal)join dept d on (t.deptno=d.deptno)))select dname from V$_dept_sal_info where grade=(select min(grade) from V$_dept_sal_info2)　　--求部门经理人中平均薪水最低的部门名称(思考题)
select dname from dept where deptno=(select deptno from (select deptno,avg(sal) avg_sal from (select empno,ename,deptno,sal from emp where empno in (select e2.empno from emp e1 join emp e2 on e1.mgr=e2.empno))group by deptno)where avg_sal=(select min(avg(sal)) from (select empno,ename,deptno,sal from emp where empno in (select e2.empno from emp e1 join emp e2 on e1.mgr=e2.empno))group by deptno))　　--求比普通员工的最高薪水还要高的经理人名称
select ename from emp whereempno in(select distinct mgr from emp where mgr is not null)and sal > (select max(sal) from emp where empno not in(select distinct mgr from emp where mgr is not null))　　--求薪水最高的前五名雇员
--结果集中存在伪字段rownum，行号，只能和< <=一块用
select ename from(select ename,sal from emp order by sal desc）where rownum < 6　　--求薪水最高的第六到第十名雇员
select ename,sal,r from(select ename,sal,rownum r from(select ename,sal from emp order by sal desc))where r>=6 and r<=10　　--练习：求最后入职的五名员工
select ename,to_char(hiredate,'YYYY_MM_DD HH24:MI:SS') from(select ename,hiredate from emp order by hiredate desc)where rownum<=5　　--<的话直接用就行了
--面试题：比较效率
  select * from emp where deptno = 10 and ename like '%A%';select * from emp where ename like '%A%' and deptno = 10;　　--有3个表S,C,SC S(SNO,SNAME) C(CNO,CNAME,CTEACHER) SC(SNO,CNO,SCGRADE)
--找出没选过“黎明”老师的所有学生姓名
select disctinct s.sno,s.sname from sjoin sc on (s.sno=sc.sno)join c on(c.cno=sc.cno)where cteacher<>'黎明'　　--列出两门以上（含两门）不及格学生姓名及平均成绩
select sname,avg_g from s join(select sno,avg(scgrade) avg_g from sc where scgrade<60 group by sno having count(cno)>=2) t on (s.sno=t.sno)　　--既学过1号课程又学过2号课程的所有学生的姓名

select sname from s where sno in(select sno from sc where cno=2 and sno in (select distinct sno from sc where cno=1))