sql 练习1

本人搜集的sql 练习,已经在oracle pl/sql develper下测试通过,分享之

 

create table DEPT
(
DEPTNO NUMBER(2) not null,
DNAME VARCHAR2(14),
LOC    VARCHAR2(13)
)
tablespace USERS;
alter table DEPT add constraint PK_DEPT primary key (DEPTNO);
insert into DEPT (DEPTNO, DNAME, LOC)
values (10, 'ACCOUNTING', 'NEW YORK');
insert into DEPT (DEPTNO, DNAME, LOC)
values (20, 'RESEARCH', 'DALLAS');
insert into DEPT (DEPTNO, DNAME, LOC)
values (30, 'SALES', 'CHICAGO');
insert into DEPT (DEPTNO, DNAME, LOC)
values (40, 'OPERATIONS', 'BOSTON');
commit;
create table EMP
(
EMPNO    NUMBER(4) not null,
ENAME    VARCHAR2(10),
JOB      VARCHAR2(9),
MGR      NUMBER(4),
HIREDATE DATE,
SAL      NUMBER(7,2),
COMM     NUMBER(7,2),
DEPTNO   NUMBER(2)
)
tablespace USERS;
alter table EMP add constraint PK_EMP primary key (EMPNO);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7369, 'SMITH', 'CLERK', 7902, to_date('17-12-1980', 'dd-mm-yyyy'), 800, null, 20);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7499, 'ALLEN', 'SALESMAN', 7698, to_date('20-02-1981', 'dd-mm-yyyy'), 1600, 300, 30);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7521, 'WARD', 'SALESMAN', 7698, to_date('22-02-1981', 'dd-mm-yyyy'), 1250, 500, 30);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7566, 'JONES', 'MANAGER', 7839, to_date('02-04-1981', 'dd-mm-yyyy'), 2975, null, 20);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7654, 'MARTIN', 'SALESMAN', 7698, to_date('28-09-1981', 'dd-mm-yyyy'), 1250, 1400, 30);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7698, 'BLAKE', 'MANAGER', 7839, to_date('01-05-1981', 'dd-mm-yyyy'), 2850, null, 30);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7782, 'CLARK', 'MANAGER', 7839, to_date('09-06-1981', 'dd-mm-yyyy'), 2450, null, 10);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7788, 'SCOTT', 'ANALYST', 7566, to_date('19-04-1987', 'dd-mm-yyyy'), 3000, null, 20);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7839, 'KING', 'PRESIDENT', null, to_date('17-11-1981', 'dd-mm-yyyy'), 5000, null, 10);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7844, 'TURNER', 'SALESMAN', 7698, to_date('08-09-1981', 'dd-mm-yyyy'), 1500, 0, 30);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7876, 'ADAMS', 'CLERK', 7788, to_date('23-05-1987', 'dd-mm-yyyy'), 1100, null, 20);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7900, 'JAMES', 'CLERK', 7698, to_date('03-12-1981', 'dd-mm-yyyy'), 950, null, 30);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7902, 'FORD', 'ANALYST', 7566, to_date('03-12-1981', 'dd-mm-yyyy'), 3000, null, 20);
insert into EMP (EMPNO, ENAME, JOB, MGR, HIREDATE, SAL, COMM, DEPTNO)
values (7934, 'MILLER', 'CLERK', 7782, to_date('23-01-1982', 'dd-mm-yyyy'), 1300, null, 10);
commit;

/*
1．列出至少有一个员工的所有部门。
2．列出薪金比“SMITH”多的所有员工。
3．列出所有员工的姓名及其直接上级的姓名。
4．列出受雇日期早于其直接上级的所有员工。
5．列出最低薪金大于1500的各种工作。
6．列出在每个部门工作的员工数量、平均工资和平均服务期限。
*/
SELECT DNAME FROM EMP E, DEPT D WHERE E.DEPTNO=D.DEPTNO GROUP BY DNAME HAVING COUNT(EMPNO)>=1

SELECT * FROM EMP E WHERE E.SAL>(SELECT SAL FROM EMP WHERE EMP.ENAME='SMITH')

select A.ENAME, B.ENAME from EMP A, EMP B WHERE A.MGR=B.EMPNO

select A.ENAME from EMP A, EMP B WHERE A.MGR=B.EMPNO AND A.HIREDATE<B.HIREDATE

SELECT JOB,MIN(SAL) MSAL FROM EMP GOURP BY (JOB) HAVING MIN(SAL)>1500;

select job,min(sal) msal from emp group by job having min(sal)>1500;

SELECT DEPTNO, COUNT(ENAME), AVG(SAL), AVG(HIREDATE) FROM EMP GOURP BY DEPTNO

select deptno,count(*), trunc(avg(sal+nvl(comm,0))) avgsal, trunc(avg(sysdate-hiredate)) avgday from emp group by deptno;

/*1.         取出emp表中comm最小的所有记录。*/
select * from emp where comm=(select min(comm) from emp);

select * from emp where rownum=1 order by comm;

/*2.         显示所有职工的empname以及其对应的deptname；如果emp表中的deptno在dept表中找不到，则deptname输出为空。*/

select e.ename,d.dname from emp e left join dept d on e.deptno=d.deptno;
/*3.         把emp表中的所有comm设置为comm=comm.+100;如果comm为null，则comm=1。*/

update emp set comm=case
when comm is not NULL then comm+100
when comm is null then 1
end