Oracle数据库练习题实例及答案分享

一、建表

1.	create table empgj(empno number(4,0) primary key,ename varchar2(10),  
2.	job varchar2(9),mgr number(4,0),hiredate date,sal number(7,2),comm  
3.	number(7,2),deptno number(2,0));  
4.	--1  
5.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
6.	(7369,'smith','clerk',7902,to_date('1980-12-17','yyyy/mm/dd'),  
7.	800.00,'',20);  
8.	commit;  
9.	--2  
10.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
11.	(7499,'allen','salesman',7698,to_date('1981-2-20','yyyy/mm/dd'),  
12.	1600.00,'300.00',20);  
13.	commit;  
14.	--3  
15.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
16.	(7521,'ward','salesman',7698,to_date('1981-2-22','yyyy/mm/dd'),  
17.	1250.00,'500.00',30);  
18.	commit;  
19.	--4  
20.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
21.	(7566,'jones','manager',7839,to_date('1981-4-2','yyyy/mm/dd'),  
22.	2975.00,'',20);  
23.	commit;  
24.	--5  
25.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
26.	(7654,'martin','salesman',7698,to_date('1981-9-28','yyyy/mm/dd'),  
27.	1250,'1400.00',30);  
28.	commit;  
29.	--6  
30.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
31.	(7698,'blake','manager',7839,to_date('1981-5-1','yyyy/mm/dd'),  
32.	2850,'',30);  
33.	commit;  
34.	--7  
35.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
36.	(7782,'clark','manager',7839,to_date('1981-6-9','yyyy/mm/dd'),  
37.	2450,'',10);  
38.	commit;  
39.	--8  
40.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
41.	(7788,'scott','analyst',7566,to_date('1987-4-19','yyyy/mm/dd'),  
42.	3000,'',20);  
43.	commit;  
44.	--9  
45.	insert into empgj(empno,ename,job,hiredate,sal,comm,deptno) values   
46.	(7839,'king','president',to_date('1981-11-17','yyyy/mm/dd'),  
47.	5000,'',10);  
48.	commit;  
49.	--10  
50.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
51.	(7844,'turner','salesman',7698,to_date('1981-9-8','yyyy/mm/dd'),  
52.	5000,'0.00',30);  
53.	commit;  
54.	--11  
55.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
56.	(7876,'adams','clerk',7788,to_date('1987-5-23','yyyy/mm/dd'),  
57.	1100,'',20);  
58.	commit;  
59.	--12  
60.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
61.	(7900,'james','clerk',7698,to_date('1981-12-3','yyyy/mm/dd'),  
62.	950,'',30);  
63.	commit;  
64.	--13  
65.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
66.	(7902,'ford','analyst',7566,to_date('1981-12-3','yyyy/mm/dd'),  
67.	3000,'',20);  
68.	commit;  
69.	--14  
70.	insert into empgj(empno,ename,job,mgr,hiredate,sal,comm,deptno) values   
71.	(7934,'miller','clerk',7782,to_date('1982-1-23','yyyy/mm/dd'),  
72.	1300,'',10);  
73.	commit;  
74.	select empno as "员工id",ename as "员工姓名", job as "职位",mgr as "员工管理者id",  
75.	hiredate as "入职日期",sal as "薪资",comm as "绩效",deptno as "员工所在部门的id"  
76.	from empgj;  
77.	  
78.	alter session set nls_date_format='yyyy/mm/dd';  
79.	  
80.	create table deptgj(deptno number(2,0),dname varchar2(14),loc varchar2(13));  
81.	--1  
82.	insert into deptgj(deptno,dname,loc) values(10,'accounting','new york');  
83.	commit;  
84.	--2  
85.	insert into deptgj(deptno,dname,loc) values(20,'research','dallas');  
86.	commit;  
87.	--3  
88.	insert into deptgj(deptno,dname,loc) values(30,'sales','chicago');  
89.	commit;  
90.	--4  
91.	insert into deptgj(deptno,dname,loc) values(30,'operations','boston');  
92.	commit; 

二、表展示

三、练习题及答案

1.取得每个部门最高薪水的人员名称

select e.ename,t.maxsal, t.deptno
 from empgj e
 join (select max(sal) as maxsal, deptno from empgj group by deptno) t 
 on t.deptno = e.deptno and e.sal= t.maxsal;

–>

select e.ename,t.maxsal, d.dname
 from empgj e 
join (select max(sal) as maxsal, deptno from empgj group by deptno) t 
 on t.deptno = e.deptno and e.sal= t.maxsal
 ,deptgj d
where e.deptno = d.deptno;(加上部门名称)

<+>取得sales部门中薪资最高的人名

 select ename
    from empgj
   where sal = (select max(e.sal)
                     from empgj e
                     join deptgj d
                       on e.deptno = d.deptno
                     where d.dname = 'sales');

2.哪些人的薪水在部门的平均薪水之上

select e.ename, e.sal, t.avgsal
  from empgj e
  join (select avg(sal) as avgsal, deptno from empgj group by deptno) t 

  on t.deptno = e.deptno and e.sal > t.avgsal;

3.不准用组函数(max),取得最高薪水

select sal from (select sal from empgj order by sal desc) where rownum = 1;

4.取得平均薪水最高的部门的部门编号

select deptno
  from (select avg(sal), deptno
          from empgj
         group by deptno
         order by avg(sal) desc)
 where rownum = 1;
(ps:这种方法只能查询最高平均薪资只有一个部门的情况,如果两个部门平均薪资一样且都是最高,执行此查询则会漏查,希望有能解决此问题的朋友不吝赐教)

5.取得平均薪水最高的部门的部门名称

(1)select dname
  from deptgj
 where deptno = (select deptno
                   from (select avg(sal), deptno
                           from empgj
                          group by deptno
                          order by avg(sal) desc)
                  where rownum = 1);
(2)select t2.avgsal, t2.deptno, d.dname
  from (select t.avgsal, t.deptno
          from (select avg(sal) as avgsal, deptno
                  from empgj
                 group by deptno
                 order by avgsal desc) t
         where rownum = 1) t2
  join deptgj d on t2.deptno = d.deptno;

6.取得比普通员工(员工代码没有在mgr字段上出现的)的最高薪水还要低的经理人姓名

select ename
  from empgj
 where sal <
       (select max(sal)
          from empgj
         where empno not in
               (select distinct mgr from empgj where mgr is not null))
   and job = 'manager';
ps:这里注意not in的用法,其后不能跟有null的数据,注意去除;一般使用not exists

7.取得薪水最高的前五名员工

select ename, sal
  from (select ename, sal from empgj order by sal desc)
 where rownum < 6;

8.取得薪水最高的第六到第十名员工

select ename, sal
  from (select ename, sal, rownum rn
          from (select ename, sal from empgj order by sal desc))
 where rn >= 6
   and rn <= 10;
ps:有人可能会想,我直接查询rownum大于5小于11不就使查询语句简单些吗,比如:
select ename,sal from
(select ename, sal from empgj order by sal desc)
where rownum >= 6 and rownum <=10;
但是这样尝试的查询并没有结果,这是因为不能对rownum使用>(大于1的数值)、>=(大于或等于1的数值)、=(大于或等于1的数值),否则无结果
这是因为:

(1)rownum是伪列,必须要要有返回结果后,每条返回记录就会对应产生一个rownum数值;

(2)返回结果记录的rownum是从1开始排序的,因此第一条始终是1;

这样,当查询到第一条记录时,该记录的rownum为1,但条件要求rownum>1,因此不符合,继续查询下一条;因为前面没有符合要求的记录,因此下一条记录过来后,其rownum还是为1,如此循环,就不会产生结果

按照我所写查询语句是将rownum伪列加入到新表中成为实际列,就可进行某个范围的查询。rownum的其他用法可参考此篇文章进行深入了解点击打开链接

9.取得最后入职的5名员工

select ename
  from (select ename, hiredate from empgj order by hiredate desc)
 where rownum <= 5;

10.列出所有员工及直接上级的姓名

(1)select a.ename, b.ename from empgj a left join empgj b on a.mgr = b.empno;
(2)select t.ename, e.ename
  from (select ename, mgr from empgj) t
  left join empgj e on e.empno = t.mgr;

sql语句的书写很灵活,有时一个条件会有多种写法的查询语句,除了及时复盘,更应经常训练,才会运用熟练

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