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Java面试之SQL语句题经典案例

2024年10月30日 • Java •我要评论
一、行转列问题现有表格a，按照以下格式排列；姓名收入类型收入金额tom年奖金5wtom月工资10kjack年奖金8wjack月工资12k先需要将表格转化为：姓名月工资年奖金tom10k50kjack1

一、行转列问题

现有表格a，按照以下格式排列；

姓名收入类型收入金额
tom年奖金5w
tom月工资10k
jack年奖金8w
jack月工资12k

先需要将表格转化为：

姓名月工资年奖金
tom10k50k
jack12k80k

方法一：使用静态sql

select '姓名',
sum(case '收入类型' when '年奖金' then '收入金额' else 0 end) 年奖金,
sum(case '收入类型' when '月工资' then '收入金额' else 0 end) 月工资
from a
group by '姓名'

方法二：使用 pivot：mysql不支持

select * from
(
    select 姓名,收入类型,收入金额 from a
) test
pivot(sum(收入金额) for 收入类型 in ('月工资','年终奖')) pvt

二、准备工作：

【1】表名和字段

–1.学生表 
student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别 
–2.课程表 
course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号 
–3.教师表 
teacher(t_id,t_name) –教师编号,教师姓名 
–4.成绩表 
score(s_id,c_id,s_score) –学生编号,课程编号,分数

【2】测试数据

--建表
--学生表
create table `student`(
`s_id` varchar(20),
`s_name` varchar(20) not null default '',
`s_birth` varchar(20) not null default '',
`s_sex` varchar(10) not null default '',
primary key(`s_id`)
);
--课程表
create table `course`(
`c_id` varchar(20),
`c_name` varchar(20) not null default '',
`t_id` varchar(20) not null,
primary key(`c_id`)
);
--教师表
create table `teacher`(
`t_id` varchar(20),
`t_name` varchar(20) not null default '',
primary key(`t_id`)
);
--成绩表
create table `score`(
`s_id` varchar(20),
`c_id` varchar(20),
`s_score` int(3),
primary key(`s_id`,`c_id`)
);
--插入学生表测试数据
insert into student values('01' , '赵雷' , '1990-01-01' , '男');
insert into student values('02' , '钱电' , '1990-12-21' , '男');
insert into student values('03' , '孙风' , '1990-05-20' , '男');
insert into student values('04' , '李云' , '1990-08-06' , '男');
insert into student values('05' , '周梅' , '1991-12-01' , '女');
insert into student values('06' , '吴兰' , '1992-03-01' , '女');
insert into student values('07' , '郑竹' , '1989-07-01' , '女');
insert into student values('08' , '王菊' , '1990-01-20' , '女');
--课程表测试数据
insert into course values('01' , '语文' , '02');
insert into course values('02' , '数学' , '01');
insert into course values('03' , '英语' , '03');

--教师表测试数据
insert into teacher values('01' , '张三');
insert into teacher values('02' , '李四');
insert into teacher values('03' , '王五');

--成绩表测试数据
insert into score values('01' , '01' , 80);
insert into score values('01' , '02' , 90);
insert into score values('01' , '03' , 99);
insert into score values('02' , '01' , 70);
insert into score values('02' , '02' , 60);
insert into score values('02' , '03' , 80);
insert into score values('03' , '01' , 80);
insert into score values('03' , '02' , 80);
insert into score values('03' , '03' , 80);
insert into score values('04' , '01' , 50);
insert into score values('04' , '02' , 30);
insert into score values('04' , '03' , 20);
insert into score values('05' , '01' , 76);
insert into score values('05' , '02' , 87);
insert into score values('06' , '01' , 31);
insert into score values('06' , '03' , 34);
insert into score values('07' , '02' , 89);
insert into score values('07' , '03' , 98);

三、练习题

【1】查询"01"课程比"02"课程成绩高的学生的信息及课程分数：当对一张表中的一列数据比较时，应当将一张表拆分为两张表；

select st.*,sc.`s_score` as '语文' ,sc2.`s_score` as '数学' 
from student st 
left join score sc on st.s_id=sc.`s_id` and sc.`c_id`='01'
left join score sc2 on st.s_id=sc2.`s_id` and sc2.`c_id`='02'  
where sc.`s_score` > sc2.`s_score`;

【2】查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩：分组在 having 之前，有函数表达式时，条件判断需要使用 having，同时主要成绩需要截取为两位；

select s.`s_id`,s.`s_name`,round(avg(sc.`s_score`),2) as '平均成绩' from student s
left join score sc on s.`s_id` = sc.`s_id`
group by sc.`s_id`
having avg(sc.`s_score`) >= 60;

【3】查询所有同学的学生编号、学生姓名、选课总数、所有课程的总成绩；

select s.`s_id`,s.`s_name`,count(sc.`c_id`) as '选课总数',sum(case when sc.`s_score` is null then 0 else sc.`s_score` end) as '总成绩' from student s 
left join score sc on s.`s_id` = sc.`s_id`
group by sc.`s_id`

【4】查询学过 “张三” 老师授课的同学的信息；

select s.* from student s 
left join score sc on s.`s_id` = sc.`s_id`
left join course c on sc.`c_id` = c.`c_id`
left join teacher t on t.`t_id` = c.`t_id`
where t.`t_name` = "张三"

【5】查询没学过"张三"老师授课的同学的信息；

 select st.* from student st where st.s_id not in(
  select sc.s_id from score sc where sc.c_id in (select c.c_id from course c left join teacher t on t.t_id=c.t_id where  t.t_name="张三")
  )

【6】查询学过编号为"01"并且也学过编号为"02"的课程的同学的信息

select s.* from student s
inner join score sc on s.`s_id` = sc.`s_id` 
inner join score sc1 on s.`s_id` = sc1.`s_id`
where sc.`c_id`='01' and sc1.`c_id`='02'

--方式二
select a.* 
from
    student a,
    score b,
    score c
where
    a.s_id = b.s_id
    and a.s_id = c.s_id
    and b.c_id = '01'
    and c.c_id = '02';

【7】查询至少有一门课与学号为"01"的同学所学相同的同学的信息

select distinct s.* from student s 
left join score c on s.`s_id` = c.`s_id`
where c.`c_id` in (
    select sc.`c_id` from student s
    left join score sc on s.`s_id` = sc.`s_id`
    where s.`s_id`='01'
);

【8】查询和"01"号的同学学习的课程完全相同的其他同学的信息

select distinct s.* from student s 
left join score c on s.`s_id` = c.`s_id`
group by s.`s_id`
having count(c.`c_id`) = (
    select count(sc.`c_id`) from student s
    left join score sc on s.`s_id` = sc.`s_id`
    where s.`s_id`='01'
);

【9】查询没学过"张三"老师讲授的任一门课程的学生姓名

select s.`s_name` from student s
where s.`s_id` not in(
    select sc.`s_id` from score sc
    left join course c on sc.`c_id` = c.`c_id`
    left join teacher t on t.`t_id` = c.`t_id`
    where t.`t_name`="张三"
)

【10】查询两门及其以上不及格课程的同学的学号，姓名及其平均成绩

select s.`s_id`,s.`s_name`,avg(sc.`s_score`) from student s 
inner join score sc on s.`s_id` = sc.`s_id`
where s.`s_id` in (
    select sc.`s_id` from score sc 
    where sc.`s_score`<60
    group by sc.`s_id`
    having count(1)>=2
)
group by s.`s_id`

【11】按平均成绩从高到低显示所有学生的所有课程的成绩以及平均成绩：这里要注意 where 和 on 的区别：on 条件是在生成临时表时使用的条件，它不管on中的条件是否为真，都会返回左（右）边表中的记录。(返回左（右）表全部记录)。此时可能会出现与右表不匹配的记录即为空的记录。即使on后边的条件不为真也会返回左（右）表中的记录。where 条件是在临时表生成好后，再对临时表进行过滤的条件。

select s.`s_id`,s.`s_name`,sc.`s_score` as "语文" ,sc1.`s_score` as "数学",sc2.`s_score` as "英语",avg(sc3.`s_score`) "平均分" from student s 
left join score sc on s.`s_id` = sc.`s_id` and sc.`c_id` = "01"
left join score sc1 on s.`s_id` = sc1.`s_id` and sc1.`c_id` = "02"
left join score sc2 on s.`s_id` = sc2.`s_id` and sc2.`c_id` = "03"
left join score sc3 on s.`s_id` = sc3.`s_id`  
group by s.`s_id`
order by avg(sc3.`s_score`) desc

【12】查询各科成绩最高分、最低分和平均分：以如下形式显示：课程id，课程 name，最高分，最低分，平均分，及格率，中等率，优良率，优秀率（及格为>=60，中等为：70-80，优良为：80-90，优秀为：>=90）

select c.`c_id`,c.`c_name`,max(s.`s_score`) "最高分",min(s.`s_score`) "最低分",avg(s.`s_score`) "平均分", 
((select count(1) from score sc where sc.`c_id` = c.`c_id` and sc.`s_score` >= 60)/(select count(1) from score sc where sc.c_id = c.c_id)) "及格率",
((select count(1) from score sc where sc.`c_id` = c.`c_id` and 80 >= sc.`s_score` and sc.`s_score` >= 70)/(select count(1) from score sc where sc.c_id = c.c_id)) "中等率",
((select count(1) from score sc where sc.`c_id` = c.`c_id` and 90 >= sc.`s_score` and sc.`s_score` >= 80)/(select count(1) from score sc where sc.c_id = c.c_id)) "优良率",
((select count(1) from score sc where sc.`c_id` = c.`c_id` and sc.`s_score` >= 90)/(select count(1) from score sc where sc.c_id = c.c_id)) "优秀率"
from course c
left join score s on c.`c_id` = s.`c_id`
group by c.`c_id`;

【13】查询所有课程的成绩第2名到第3名的学生信息及该课程成绩：union：对两个结果集进行并集操作，不包括重复行，同时进行默认规则的排序；union all：对两个结果集进行并集操作，包括重复行，不进行排序；注意 limit下标是从0开始的。

(select s.*,c.`c_name`,sc.`s_score` "成绩" from student s
left join score sc on s.`s_id` = sc.`s_id` and sc.`c_id`="01"
left join course c on sc.`c_id` = c.`c_id`
order by sc.`s_score` desc
limit 1,2)
union all
(select s.*,c.`c_name`,sc.`s_score` "成绩" from student s
left join score sc on s.`s_id` = sc.`s_id` and sc.`c_id`="02"
left join course c on sc.`c_id` = c.`c_id` 
order by sc.`s_score` desc
limit 1,2)
union all
(select s.*,c.`c_name`,sc.`s_score` "成绩" from student s
left join score sc on s.`s_id` = sc.`s_id` and sc.`c_id`="03"
left join course c on sc.`c_id` = c.`c_id` 
order by sc.`s_score` desc
limit 1,2)

【14】查询学生平均成绩及其名次：重点是名次的获取，通过变量 @i 进行递增获取。

set @i=0;
select test.*,@i:=@i+1 "名次" from(
select s.`s_name`,round(avg(sc.`s_score`),2) "平均成绩" from score sc
left join student s on s.`s_id` = sc.`s_id`
group by sc.`s_id`
order by avg(sc.`s_score`) desc) test;

【15】查询不同课程成绩相同的学生的学生编号、课程编号、学生成绩：思路就是先查询一条数据，然后与表中的数据比较相同的成绩，且科目号不相同的数据行，如果大于1则返回当前行即可。逐行比较；

select st.s_id,st.s_name,sc.c_id,sc.s_score from student st 
left join score sc on sc.s_id=st.s_id
left join course c on c.c_id=sc.c_id
where (
select count(1) from student st2 
left join score sc2 on sc2.s_id=st2.s_id
left join course c2 on c2.c_id=sc2.c_id
where sc.s_score=sc2.s_score and c.c_id!=c2.c_id 
)>=1

【16】 查询每门功成绩最好的前两名

select a.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="01"
order by sc.s_score desc limit 0,2) a
union all
select b.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="02"
order by sc.s_score desc limit 0,2) b
union all
select c.* from (select st.s_id,st.s_name,c.c_name,sc.s_score from student st
left join score sc on sc.s_id=st.s_id
inner join course c on c.c_id=sc.c_id and c.c_id="03"
order by sc.s_score desc limit 0,2) c

方式二

select a.s_id,a.c_id,a.s_score from score a
where (select count(1) from score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 order by a.c_id

【17】查询本周过生日的学生：此处可能有问题,week函数取的为当前年的第几周,2017-12-12是第50周而2018-12-12是第49周,可以取月份,day,星期几(%w)， 再判断本周是否会持续到下一个月进行判断,太麻烦。

select st.* from student st 
where week(now())=week(date_format(st.s_birth,'%y%m%d'))

【18】查询下周过生日的学生

select st.* from student st 
where week(now())+1=week(date_format(st.s_birth,'%y%m%d'))

【19】查询本月过生日的学生

select st.* from student st 
where month(now())=month(date_format(st.s_birth,'%y%m%d'))

【20】查询下月过生日的学生： 注意，如果当前月为12月时，用month(now())+1为13而不是1，可用 timestampadd() 函数或 mod 取模

select st.* from student st 
where month(timestampadd(month,1,now()))=month(date_format(st.s_birth,'%y%m%d'))

方法二：

select st.* from student st where (month(now()) + 1) mod 12 = month(date_format(st.s_birth,'%y%m%d'))

总结 

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