SQL简单语句总结习题

Posted 2020-06-30 小猩

创建一个表记员工个人信息：

 

              --创建一个表
create table  plspl_company_info(
              empno number(5) not null, 
              ename varchar2(10) not null,
              job varchar2(10),
              manager number(5),
              hiredate date,
              sal  number(5),
              comm  number(5),
              deptno number(3)
              );
            
insert into plspl_company_info values (7369, ‘SMITH‘, ‘CLERK‘, 7902, date ‘1980-12-17‘, 800, NULL, 10);
insert into plspl_company_info values (7293, ‘ALLEN‘, ‘SALESMAN‘, 7689, date ‘1981-03-27‘, 1867, NULL, 20);
insert into plspl_company_info values (7562, ‘JAMES‘, ‘SALESMAN‘, 7689, date ‘1981-09-11‘, 1796, NULL, 20);
insert into plspl_company_info values (7936, ‘JONES‘, ‘ANAYST‘, 7656, date ‘1980-09-01‘, 3250, NULL, 30);
insert into plspl_company_info values (7688, ‘WEST‘, ‘MANAGER‘, 7839, date ‘1981-02-28‘, 2985, 900, 40);
insert into plspl_company_info values (7499, ‘PAUL‘, ‘MANAGER‘, 7839, date ‘1980-03-26‘, 3600, NULL, 40);
insert into plspl_company_info values (7778, ‘FORD‘, ‘CLERK‘, 7902, date ‘1987-04-17‘, 960, NULL, 10);
insert into plspl_company_info values (7289, ‘ADAMS‘, ‘SALESMAN‘, 7689, date ‘1980-01-09‘, 1956, NULL, 20);
insert into plspl_company_info values (7531, ‘MATIN‘, ‘SALESMAN‘, 7689, date ‘1980-05-14‘, 1906, NULL, 20);
insert into plspl_company_info values (7916, ‘KING‘, ‘ANAYST‘, 7656, date ‘1982-06-04‘, 2864, NULL, 30);
insert into plspl_company_info values (7365, ‘BLKAE‘, ‘CLERK‘, 7902, date ‘1981-09-11‘, 1200, 1100, 10);
insert into plspl_company_info values (7784, ‘CHRIS‘, ‘CLERK‘, 7902, date ‘1981-07-16‘, 1376, NULL, 10);

select * from plspl_company_info;


执行结果：

SQL>

Table created

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

1 row inserted

EMPNO ENAME JOB MANAGER HIREDATE SAL COMM DEPTNO
------ ---------- ---------- ------- ----------- ------ ------ ------
7369 SMITH CLERK 7902 1980/12/17 800 10
7293 ALLEN SALESMAN 7689 1981/3/27 1867 20
7562 JAMES SALESMAN 7689 1981/9/11 1796 20
7936 JONES ANAYST 7656 1980/9/1 3250 30
7688 WEST MANAGER 7839 1981/2/28 2985 900 40
7499 PAUL MANAGER 7839 1980/3/26 3600 40
7778 FORD CLERK 7902 1987/4/17 960 10
7289 ADAMS SALESMAN 7689 1980/1/9 1956 20
7531 MATIN SALESMAN 7689 1980/5/14 1906 20
7916 KING ANAYST 7656 1982/6/4 2864 30
7365 BLKAE CLERK 7902 1981/9/11 1200 1100 10
7784 CHRIS CLERK 7902 1981/7/16 1376 10

12 rows selected

SQL>

 

习题：

1，选出部门30里的所有员工信息

select * from plspl_company_info where deptno = 30;

 2，列出所有办事员(CLERK)的姓名，编号和部门编号

select ename, empno , deptno from  plspl_company_info where job = ‘CLERK‘;

3，找出薪金高于佣金60%的员工

select * from plspl_company_info where comm > sal*0.6 ;

4，找出部门10的所有经理（MANAGER）和部门20的所有办事员（CLERK）

select * from plspl_company_info where (deptno = 10 and job = ‘MANAGER‘) or (deptno = 20 and job = ‘CLERK‘) ;

5，找出部门10的所有经理（MANAGER）和部门20的所有办事员（CLERK），以及既不是经理有不是办事员，但薪金大于或等于2000的所有员工的详细资料

select * 
from plspl_company_info 
where (deptno = 10 and job = ‘MANAGER‘) or (deptno = 20 and job = ‘CLERK‘) or ((job not in(‘MANAGER‘, ‘CLERK‘)) and sal >= 2000) ;

6，找出收取佣金的员工的不同工作

select distinct job from plspl_company_info where comm is not null ;

7，找出不收取佣金或者收取佣金低于100的原

select * from plspl_company_info where (comm is  null) or comm < 100 ;

8，显示不带“R”的员工姓名

select * from plspl_company_info where ename not like ‘%A%‘;

9，显示姓名字段的任何位置包含‘A‘的所有员工的姓名，显示结果按照基本工资由高到低；如果工资相同，则按照雇佣年限由早到晚排序；如果雇佣时间相同，则按照职位排序

select * from plspl_company_info where ename like ‘%A%‘ order by sal DESC, hiredate, job;