sql 语句练习与答案

作者:计算机教程

1学生表student S#学号,sname姓名,difdate日期,班级grade 2课程表 course c#课程号 ,名字cname 3成绩单score s#学号 c#课程号 成绩score --1统计每个班级有多少人 select grade,count(sname) from ze_student group by grade; --2、2007级的各学生的平均成绩 没有成绩的为0; select a.sname,(select avg(nvl(b.score,0)) from ze_score b where b.s#=a.s#) from ze_student a where grade=2007; --3 每科 平均成绩和最高成绩 最低成绩 2007级 保留2位小数点 四舍五入 select b.c#,avg(b.score),max(b.score),min(nvl(b.score,0)) from ze_student a,ze_score b where b.s# = a.s# and a.grade =2007 group by b.c#; --4 给2007级 数学加5分 update ze_score set score=nvl(score,0) 5 where s# in (select s# from ze_student where grade=2007) and c# =(select c# from ze_course where cname='数学'); --5 90分以上的为优秀 90到85为良好,60分 不及格 各人平均成绩 select s#, c, case when c=90 then '优秀' when c90 and c=60 then '及格' else '不及格' end as jige from (select s#,avg(nvl(score,0)) as c from ze_score group by s# ) order by jige desc; --6 求同月出生的 人数 select to_char(difdate,'mm') as 月份,count(s#) as 出生人数 from ze_student group by to_char(difdate,'mm'); --7 各科的及格率和平均成绩 截取 保留2位 --及格率 select c#,avg(nvl(score,0))as 平均成绩,sum(nvl(score,0))as 总成绩, count(s#) as 各科人数, trunc(sum( case when nvl(score,0)60 then '1' else '0' end)/count(s#),2) as 及格率 from ze_score group by c#; --每人的及格率 select s#, avg(nvl(score,0))as 平均成绩,sum(nvl(score,0))as 总成绩, count(c#) as 总科目, sum( case when nvl(score,0)60 then 1 else 0 end )/count(c#) as 及格率 from ze_score group by s#; --8删除 姓名是张三 的大学语文 成绩 select * from ze_score where s# in (select s# from ze_student where sname in '张三') and c#=(select c# from ze_course where cname ='大学语文'); --9 将数学替换成高等数学 update ze_course set cname='高等数学'where cname like '%数学%'; --10 格式化 ,显示 将学号修改成S开头 不足12位补0; --查询 select concat('S',lpad(s#,11,0)) as s# from ze_score ; select concat('S',lpad(s#,11,0)) as s# from ze_student ; --格式化 update ze_score set s#= concat('S',lpad(s#,9,0)); update ze_student set s#= concat('S',lpad(s#,9,0)); 四个足球队 select a.name,b.name from qiu a,qiu b where a.nameb.name; commit rollback 服务器类型 服务器协议 全局数据库名称 服务器IP地址 服务器端口号 用户名和密码

 

作为一枚Java后端开发者,数据库知识必不可少,对数据库的掌握熟悉度的考察也是对这个人是否有扎实基本功的考察。特别对于初级开发者,面试可能不会去问框架相关知识,但是绝对不会不去考察数据库知识,这里收集一些常见类型的SQL语句,无论对于平常开发还是准备面试,都会有助益。

基本表结构:

        student(sno,sname,sage,ssex)学生表
        course(cno,cname,tno) 课程表
        sc(sno,cno,score) 成绩表

*        teacher(tno,tname) 教师表*

* *

101,查询课程1的成绩比课程2的成绩高的所有学生的学号
select a.sno from
(select sno,score from sc where cno=1) a,
(select sno,score from sc where cno=2) b
where a.score>b.score and a.sno=b.sno

 

102,查询平均成绩大于60分的同学的学号和平均成绩*
select a.sno as "学号", avg(a.score) as "平均成绩" 
from
(select sno,score from sc) a 
group by sno having avg(a.score)>60
*

* *

103,查询所有同学的学号、姓名、选课数、总成绩*
select a.sno as 学号, b.sname as 姓名,
count(a.cno) as 选课数, sum(a.score) as 总成绩
from sc a, student b
where a.sno = b.sno
group by a.sno, b.sname
*

或者:

*selectstudent.sno as 学号, student.sname as 姓名,
 count(sc.cno) as 选课数, sum(score) as 总成绩
from student left Outer join sc on student.sno = sc.sno
group by student.sno, sname

*104,查询姓“张”的老师的个数

selectcount(distinct(tname)) from teacher where tname like '张%‘
或者:
select tname as "姓名", count(distinct(tname)) as "人数" 
from teacher 
where tname like'张%'
group by tname

 

105,查询没学过“张三”老师课的同学的学号、姓名
select student.sno,student.sname from student
where sno not in (select distinct(sc.sno) from sc,course,teacher
where sc.cno=course.cno and teacher.tno=course.tno and teacher.tname='张三')

 

106,查询同时学过课程1和课程2的同学的学号、姓名
select sno, sname from student
where sno in (select sno from sc where sc.cno = 1)
and sno in (select sno from sc where sc.cno = 2)
或者:

selectc.sno, c.sname from
(select sno from sc where sc.cno = 1) a,
(select sno from sc where sc.cno = 2) b,
student c
where a.sno = b.sno and a.sno = c.sno
或者:

select student.sno,student.sname from student,sc where student.sno=sc.sno and sc.cno=1
and exists( select * from sc as sc_2 where sc_2.sno=sc.sno and sc_2.cno=2)

 

107,查询学过“李四”老师所教所有课程的所有同学的学号、姓名
select a.sno, a.sname from student a, sc b
where a.sno = b.sno and b.cno in
(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四')

或者:

select a.sno, a.sname from student a, sc b,
(select c.cno from course c, teacher d where c.tno = d.tno and d.tname = '李四') e
where a.sno = b.sno and b.cno = e.cno

 

108,查询课程编号1的成绩比课程编号2的成绩高的所有同学的学号、姓名
select a.sno, a.sname from student a,
(select sno, score from sc where cno = 1) b,
(select sno, score from sc where cno = 2) c
where b.score > c.score and b.sno = c.sno and a.sno = b.sno

 

109,查询所有课程成绩小于60分的同学的学号、姓名
select sno,sname from student
where sno not in (select distinct sno from sc where score > 60)

 

110,查询至少有一门课程与学号为1的同学所学课程相同的同学的学号和姓名
select distinct a.sno, a.sname
from student a, sc b
where a.sno <> 1 and a.sno=b.sno and
b.cno in (select cno from sc where sno = 1)

或者:

select s.sno,s.sname 
from student s,
(select sc.sno 
from sc
where sc.cno in (select sc1.cno from sc sc1 where sc1.sno=1)and sc.sno<>1
group by sc.sno)r1
where r1.sno=s.sno

 

Java知音公众号整理一些各大公司常用的面试笔试题,供大家在每天闲暇之余学习其中几道题目,日积月累,等到出去面试时,一切都水到渠成,面试时就自然会游刃有余了。

图片 1

 

作为一枚Java后端开发者,数据库知识必不可少,对数据库的掌握熟悉度的考察也是对这个人是否有扎实基本功的考察。特别对于初级开发者,面试可能不会去问框架相关知识,但是绝对不会不去考察数据库知识,这里收集一些常见类型的SQL语句,无论对于平常开发还是准备面试,都会有助益。

基本表结构:

        student(sno,sname,sage,ssex)学生表
        course(cno,cname,tno) 课程表
        sc(sno,cno,score) 成绩表

*        teacher(tno,tname) 教师表*

 

111、把“sc”表中“王五”所教课的成绩都更改为此课程的平均成绩
update sc set score = (select avg(sc_2.score) from sc sc_2 wheresc_2.cno=sc.cno)
from course,teacher where course.cno=sc.cno and course.tno=teacher.tno andteacher.tname='王五'

112、查询和编号为2的同学学习的课程完全相同的其他同学学号和姓名
这一题分两步查:

1,

select sno
from sc
where sno <> 2
group by sno
having sum(cno) = (select sum(cno) from sc where sno = 2)

2,
select b.sno, b.sname
from sc a, student b
where b.sno <> 2 and a.sno = b.sno
group by b.sno, b.sname
having sum(cno) = (select sum(cno) from sc where sno = 2)

113、删除学习“王五”老师课的sc表记录
delete sc from course, teacher
where course.cno = sc.cno and course.tno = teacher.tno and tname = '王五'

114、向sc表中插入一些记录,这些记录要求符合以下条件:
将没有课程3成绩同学的该成绩补齐, 其成绩取所有学生的课程2的平均成绩

insert sc select sno, 3, (select avg(score) from sc where cno = 2)
from student
where sno not in (select sno from sc where cno = 3)

115、按平平均分从高到低显示所有学生的如下统计报表:
-- 学号,企业管理,马克思,UML,数据库,物理,课程数,平均分

select sno as 学号
,max(case when cno = 1 then score end) AS 企业管理
,max(case when cno = 2 then score end) AS 马克思
,max(case when cno = 3 then score end) AS UML
,max(case when cno = 4 then score end) AS 数据库
,max(case when cno = 5 then score end) AS 物理
,count(cno) AS 课程数
,avg(score) AS 平均分
FROM sc
GROUP by sno
ORDER by avg(score) DESC

116、查询各科成绩最高分和最低分:

以如下形式显示:课程号,最高分,最低分
*select cno as 课程号, max(score) as 最高分, min(score) 最低分
from sc group by cno

select  course.cno as '课程号'
,MAX(score) as '最高分'
,MIN(score) as '最低分'
from sc,course
where sc.cno=course.cno
group by course.cno*

117、按各科平均成绩从低到高和及格率的百分数从高到低顺序
SELECT t.cno AS 课程号,
max(course.cname)AS 课程名,
isnull(AVG(score),0) AS 平均成绩,
100
SUM(CASE WHEN isnull(score,0)>=60 THEN 1 ELSE 0 END)/count(1) AS 及格率
FROM sc t, course
where t.cno = course.cno
GROUP BY t.cno
ORDER BY 及格率 desc*

118、查询如下课程平均成绩和及格率的百分数(用"1行"显示): 

企业管理(001),马克思(002),UML (003),数据库(004) 
select 
avg(case when cno = 1 then score end) as 平均分1,
avg(case when cno = 2 then score end) as 平均分2,
avg(case when cno = 3 then score end) as 平均分3,
avg(case when cno = 4 then score end) as 平均分4,
100
sum(case when cno = 1 and score > 60 then 1 else 0 end) / sum(casewhen cno = 1 then 1 else 0 end) as 及格率1,
100 * sum(case when cno = 2 and score > 60 then 1 else 0 end) / sum(casewhen cno = 2 then 1 else 0 end) as 及格率2,
100 * sum(case when cno = 3 and score > 60 then 1 else 0 end) / sum(casewhen cno = 3 then 1 else 0 end) as 及格率3,
100 * sum(case when cno = 4 and score > 60 then 1 else 0 end) / sum(casewhen cno = 4 then 1 else 0 end) as 及格率4
from sc*

119、查询不同老师所教不同课程平均分, 从高到低显示
select max(c.tname) as 教师, max(b.cname) 课程, avg(a.score) 平均分
from sc a, course b, teacher c
where a.cno = b.cno and b.tno = c.tno
group by a.cno
order by 平均分 desc

或者:
select r.tname as '教师',r.rname as '课程' , AVG(score) as '平均分'
from sc,
(select  t.tname,c.cno as rcso,c.cname as rname
from teacher t ,course c
where t.tno=c.tno)r
where sc.cno=r.rcso
group by sc.cno,r.tname,r.rname 
order by AVG(score) desc

120、查询如下课程成绩均在第3名到第6名之间的学生的成绩:
-- [学生ID],[学生姓名],企业管理,马克思,UML,数据库,平均成绩

select top 6 max(a.sno) 学号, max(b.sname) 姓名,
max(case when cno = 1 then score end) as 企业管理,
max(case when cno = 2 then score end) as 马克思,
max(case when cno = 3 then score end) as UML,
max(case when cno = 4 then score end) as 数据库,
avg(score) as 平均分
from sc a, student b
where a.sno not in 

(select top 2 sno from sc where cno = 1 order by score desc)
  and a.sno not in (select top 2 sno from sc where cno = 2 order by scoredesc)
  and a.sno not in (select top 2 sno from sc where cno = 3 order by scoredesc)
  and a.sno not in (select top 2 sno from sc where cno = 4 order by scoredesc)
  and a.sno = b.sno
group by a.sno

 

 

* *

Java知音公众号整理一些各大公司常用的面试笔试题,供大家在每天闲暇之余学习其中几道题目,日积月累,等到出去面试时,一切都水到渠成,面试时就自然会游刃有余了。

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