mysql sql面试题 sql经典习题 sql练习题
目录
-
- 读我
- 数据&建表语句(MySQL)
- student
- score
- teacher
- course
- 以下所有题目解题思路
- ---> 查询学生各科成绩并显示排名
- ---> 查询各科成绩的第2名到第3名的学生信息及该课程成绩
- ---> 查询学生总成绩并进行排名
- ---> 查询学生总成绩的前3名
- ---> 查询学生总成绩的第2名到第3名
- ---> 查询学生各科成绩及总成绩（总成绩要排名）
- ---> 查询各科成绩最高分、最低分、平均分、及格率、良好率等
- ---> 查询学生单科成绩，单科总成绩，以及每位同学的总成绩（排名），并在一张表中显示
- ---> 查询和"01"号同学学习课程完全相同的其他同学的信息
- ---> 查询和"01"号同学学习课程不完全相同的其他同学的信息
- ---> 查询没有学全所有课程的同学的信息
- ---> 查询学过"张三"老师授课的同学的信息
- ---> 查询没学过"张三"老师讲授的任一门课程的学生姓名
- ---> 查询各科成绩都大于4号同学的学号
- ---> 查询"01"课程比"02"课程成绩高的学生的信息及课程分数
- ---> 查询每一门课程都大于60分的同学
- ---> 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
- ---> 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
- ---> 查询学过编号"01"但是没有学过编号"02"的课程的同学的信息
- ---> 查询"李"姓老师的数量
- ---> 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
- ---> 查询不同老师所教课程的平均成绩（平均成绩从高到低显示，平均成绩相同时，按课程编号升序排列）
- ---> 查询所有课程成绩的第2名到第3名的学生信息及该课程成绩
- ---> 查询选修"张三"老师所授课程的学生中，成绩最高的学生信息，所学课程名称以及成绩
- ---> 查询选修"张三"老师所授课程的学生中，成绩最高的学生姓名，成绩，以及其它课程的成绩
- ---> 查询1990年出生的学生名单
- ---> 查询各学生的年龄
- ---> 查询本周过生日的学生
- ---> 查询下周过生日的学生
- --->查询12月份过生日的学生
- 总结：mysql8.x语法更为严格体现在哪些方面
- ---> 数据&建表语句(hive)
- sql 面试题-求职
读我
网上很多这样的题目，好像一共50道，但是其中有的题目的解法是错误的。题目用了一题多解（多解思路必须要掌握）
bug:score这张表当中没有成绩的同学不显示
mysql 高版本支持了
窗口函数，并且对group by 等语法要求更加严格
# 查看mysql版本
select VERSION();
# 8.0.31
数据&建表语句(MySQL)
create database test1;
use test1;
create table student
(
s_id varchar(10),
s_name varchar(10),
s_birth datetime,
s_sex varchar(10)
);
insert into student (s_id, s_name, s_birth, s_sex) values ('01', '赵雷', '1990-01-01 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('02', '钱电', '1990-12-21 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('03', '孙风', '1990-05-20 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('04', '李云', '1990-08-06 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('05', '周梅', '1991-12-01 00:00:00', '女');
insert into student (s_id, s_name, s_birth, s_sex) values ('06', '吴兰', '1992-03-01 00:00:00', '女');
insert into student (s_id, s_name, s_birth, s_sex) values ('07', '郑竹', '1989-07-01 00:00:00', '女');
insert into student (s_id, s_name, s_birth, s_sex) values ('08', '王菊', '1990-01-20 00:00:00', '女');
create table score
(
s_id varchar(10),
c_id varchar(10),
s_score decimal(18, 1)
);
insert into score (s_id, c_id, s_score) values ('01', '01', 80.0);
insert into score (s_id, c_id, s_score) values ('01', '02', 90.0);
insert into score (s_id, c_id, s_score) values ('01', '03', 99.0);
insert into score (s_id, c_id, s_score) values ('02', '01', 70.0);
insert into score (s_id, c_id, s_score) values ('02', '02', 60.0);
insert into score (s_id, c_id, s_score) values ('02', '03', 80.0);
insert into score (s_id, c_id, s_score) values ('03', '01', 80.0);
insert into score (s_id, c_id, s_score) values ('03', '02', 80.0);
insert into score (s_id, c_id, s_score) values ('03', '03', 80.0);
insert into score (s_id, c_id, s_score) values ('04', '01', 50.0);
insert into score (s_id, c_id, s_score) values ('04', '02', 30.0);
insert into score (s_id, c_id, s_score) values ('04', '03', 20.0);
insert into score (s_id, c_id, s_score) values ('05', '01', 76.0);
insert into score (s_id, c_id, s_score) values ('05', '02', 87.0);
insert into score (s_id, c_id, s_score) values ('06', '01', 31.0);
insert into score (s_id, c_id, s_score) values ('06', '03', 34.0);
insert into score (s_id, c_id, s_score) values ('07', '02', 89.0);
insert into score (s_id, c_id, s_score) values ('07', '03', 98.0);
create table course
(
c_id varchar(10),
c_name varchar(10),
t_id varchar(10)
);
insert into course (c_id, c_name, t_id) values ('01', '语文', '02');
insert into course (c_id, c_name, t_id) values ('02', '数学', '01');
insert into course (c_id, c_name, t_id) values ('03', '英语', '03');
create table teacher
(
t_id varchar(10),
t_name varchar(10)
);
insert into teacher (t_id, t_name) values ('01', '张三');
insert into teacher (t_id, t_name) values ('02', '李四');
insert into teacher (t_id, t_name) values ('03', '王五');
student
score
bug:score这张表当中没有成绩的不显示
（正确的做法：没有参加考试的同学成绩直接显示为0，而不是不显示），这样会导致问题。后来我在发现问题后，也不想再去修改这张表的数据了，因为太懒
teacher
course
以下所有题目解题思路
旁人务下载-解题思路文档
—> 查询学生各科成绩并显示排名
典型的分组TopN的弱化题目
select st.s_id,
st.s_name,
c.c_name,
sc.s_score,
row_number() over (partition by sc.c_id order by sc.s_score desc ) as ranking
from score sc
join student st on st.s_id = sc.s_id
join course c on sc.c_id = c.c_id;
| s_id | s_name | c_name | s_score | ranking |
|---|---|---|---|---|
| 01 | 赵雷 | 语文 | 80.0 | 1 |
| 03 | 孙风 | 语文 | 80.0 | 2 |
| 05 | 周梅 | 语文 | 76.0 | 3 |
| 02 | 钱电 | 语文 | 70.0 | 4 |
| 04 | 李云 | 语文 | 50.0 | 5 |
| 06 | 吴兰 | 语文 | 31.0 | 6 |
| 01 | 赵雷 | 数学 | 90.0 | 1 |
| 07 | 郑竹 | 数学 | 89.0 | 2 |
| 05 | 周梅 | 数学 | 87.0 | 3 |
| 03 | 孙风 | 数学 | 80.0 | 4 |
| 02 | 钱电 | 数学 | 60.0 | 5 |
| 04 | 李云 | 数学 | 30.0 | 6 |
| 01 | 赵雷 | 英语 | 99.0 | 1 |
| 07 | 郑竹 | 英语 | 98.0 | 2 |
| 03 | 孙风 | 英语 | 80.0 | 3 |
| 02 | 钱电 | 英语 | 80.0 | 4 |
| 06 | 吴兰 | 英语 | 34.0 | 5 |
| 04 | 李云 | 英语 | 20.0 | 6 |
bug:无成绩的没有排名（因为score这张表没有成绩的都不显示）
—> 查询各科成绩的第2名到第3名的学生信息及该课程成绩
select *
from (select st.s_id,
st.s_name,
c.c_name,
sc.s_score,
row_number() over (partition by sc.c_id order by sc.s_score desc ) as ranking
from score sc
join student st on st.s_id = sc.s_id
join course c on sc.c_id = c.c_id) t1
where t1.ranking >= 2
and ranking <= 3;
| s_id | s_name | c_name | s_score | ranking |
|---|---|---|---|---|
| 03 | 孙风 | 语文 | 80.0 | 2 |
| 05 | 周梅 | 语文 | 76.0 | 3 |
| 07 | 郑竹 | 数学 | 89.0 | 2 |
| 05 | 周梅 | 数学 | 87.0 | 3 |
| 07 | 郑竹 | 英语 | 98.0 | 2 |
| 02 | 钱电 | 英语 | 80.0 | 3 |
—> 查询学生总成绩并进行排名
select st.s_id,
st.s_name,
sum(sc.s_score) as sumScore,
row_number() over (order by sum(sc.s_score) desc ) as ranking
from score sc
join student st on st.s_id = sc.s_id
join course c on sc.c_id = c.c_id
group by st.s_id, st.s_name;
| s_id | s_name | sumscore | ranking |
|---|---|---|---|
| 01 | 赵雷 | 269.0 | 1 |
| 03 | 孙风 | 240.0 | 2 |
| 02 | 钱电 | 210.0 | 3 |
| 07 | 郑竹 | 187.0 | 4 |
| 05 | 周梅 | 163.0 | 5 |
| 04 | 李云 | 100.0 | 6 |
| 06 | 吴兰 | 65.0 | 7 |
bug:score这张表当中没有成绩的都不显示，导致王菊没有显示出来
—> 查询学生总成绩的前3名
会了下边的题目，求TopN就非常简单
—> 查询学生总成绩的第2名到第3名
这里的方法2有局限性
-- 方法1
select *
from (select st.s_id,
st.s_name,
sum(sc.s_score) as sumScore,
row_number() over (order by sum(sc.s_score) desc ) as ranking
from score sc
join student st on st.s_id = sc.s_id
join course c on sc.c_id = c.c_id
group by st.s_id, st.s_name) t1
where t1.ranking >= 2
and t1.ranking <= 3;
-- 方法2
select st.s_id,
st.s_name,
sum(sc.s_score) as sumScore,
row_number() over (order by sum(sc.s_score) desc ) as ranking
from score sc
join student st on st.s_id = sc.s_id
join course c on sc.c_id = c.c_id
group by st.s_id, st.s_name
limit 1,2;
| s_id | s_name | sumscore | ranking |
|---|---|---|---|
| 03 | 孙风 | 240.0 | 2 |
| 02 | 钱电 | 210.0 | 3 |
—> 查询学生各科成绩及总成绩（总成绩要排名）
这里的2种解法都很重要
解法1
select st.s_id,
st.s_name,
sum(case when c.c_name = '语文' then sc.s_score else 0 end) as chinese,
sum(case when c.c_name = '数学' then sc.s_score else 0 end) as math,
sum(case when c.c_name = '英语' then sc.s_score else 0 end) as english,
sum(sc.s_score) as sumScore,
row_number() over (order by sum(sc.s_score) desc ) as ranking
from score sc
join course c on sc.c_id = c.c_id
join student st on st.s_id = sc.s_id
group by st.s_id, st.s_name;
# case when 可以用 if 来简化
sum(case when c.c_name = '语文' then sc.s_score else 0 end)<=>sum(if(c.c_name = '语文', sc.s_score, 0))
| s_id | s_name | chinese | math | english | sumScore | ranking |
|---|---|---|---|---|---|---|
| 01 | 赵雷 | 80.0 | 90.0 | 99.0 | 269.0 | 1 |
| 03 | 孙风 | 80.0 | 80.0 | 80.0 | 240.0 | 2 |
| 02 | 钱电 | 70.0 | 60.0 | 80.0 | 210.0 | 3 |
| 07 | 郑竹 | 0.0 | 89.0 | 98.0 | 187.0 | 4 |
| 05 | 周梅 | 76.0 | 87.0 | 0.0 | 163.0 | 5 |
| 04 | 李云 | 50.0 | 30.0 | 20.0 | 100.0 | 6 |
| 06 | 吴兰 | 31.0 | 0.0 | 34.0 | 65.0 | 7 |
解法2
select t0.s_id,
t0.s_name,
ifnull(t1.chinese, 0),
ifnull(t2.math, 0),
ifnull(t3.english, 0),
t0.sumScore,
t0.ranking
from (select st.s_id,
st.s_name,
sum(sc.s_score) as sumScore,
row_number() over (order by sum(sc.s_score) desc ) as ranking
from score sc
join student st on sc.s_id = st.s_id
group by st.s_id, st.s_name) t0
left join (select sc.s_id,
sc.s_score as chinese
from score sc
join course c on sc.c_id = c.c_id and c.c_name = '语文') t1 on t0.s_id = t1.s_id
left join (select sc.s_id,
sc.s_score as math
from score sc
join course c on sc.c_id = c.c_id and c.c_name = '数学') t2 on t0.s_id = t2.s_id
left join (select sc.s_id,
sc.s_score as english
from score sc
join course c on sc.c_id = c.c_id and c.c_name = '英语') t3
on t0.s_id = t3.s_id;
—> 查询各科成绩最高分、最低分、平均分、及格率、良好率等
格式要求：显示课程ID，课程name，最高分，最低分，平均分，及格率，良好率
评级标准：及格为>=60，良好为：[60，80]
select c.c_id,
c.c_name,
max(sc.s_score) as maxScore,
min(sc.s_score) as minScore,
round(avg(sc.s_score), 2) as avgScore,
# count+if
# count时，必须用null，不能用0，因为0所在也被计数，null则不被计数
round(count(if(sc.s_score >= 60, 1, null)) / count(sc.s_id), 2) as passRate,
# sum+case when
round(sum(case when sc.s_score >= 60 and sc.s_score <= 80 then 1 else 0 end) / count(sc.s_id), 2) as goodRate
from score sc
join course c on sc.c_id = c.c_id
group by c.c_id, c.c_name;
| c_id | c_name | maxScore | minScore | avgScore | passRate | goodRate |
|---|---|---|---|---|---|---|
| 01 | 语文 | 80.0 | 31.0 | 64.50 | 0.67 | 0.67 |
| 02 | 数学 | 90.0 | 30.0 | 72.67 | 0.83 | 0.33 |
| 03 | 英语 | 99.0 | 20.0 | 68.50 | 0.67 | 0.33 |
—> 查询学生单科成绩，单科总成绩，以及每位同学的总成绩（排名），并在一张表中显示
select st.s_id,
st.s_name,
sum(case when c.c_name = '语文' then sc.s_score else 0 end) as chinese,
sum(case when c.c_name = '数学' then sc.s_score else 0 end) as math,
sum(case when c.c_name = '英语' then sc.s_score else 0 end) as english,
sum(sc.s_score) as sumScore,
row_number() over (order by sum(sc.s_score) desc ) as ranking
from score sc
join course c on sc.c_id = c.c_id
join student st on st.s_id = sc.s_id
group by st.s_id, st.s_name
union all
select '单科总成绩', # 将整个表看成一组
'',
sum(t1.chinese),
sum(t1.math),
sum(t1.english),
'',
''
from (select st.s_id,
st.s_name,
sum(case when c.c_name = '语文' then sc.s_score else 0 end) as chinese,
sum(case when c.c_name = '数学' then sc.s_score else 0 end) as math,
sum(case when c.c_name = '英语' then sc.s_score else 0 end) as english
from score sc
join course c on sc.c_id = c.c_id
join student st on st.s_id = sc.s_id
group by st.s_id, st.s_name) t1;
| s_id | s_name | chinese | math | english | sumScore | ranking |
|---|---|---|---|---|---|---|
| 01 | 赵雷 | 80.0 | 90.0 | 99.0 | 269.0 | 1 |
| 03 | 孙风 | 80.0 | 80.0 | 80.0 | 240.0 | 2 |
| 02 | 钱电 | 70.0 | 60.0 | 80.0 | 210.0 | 3 |
| 07 | 郑竹 | 0.0 | 89.0 | 98.0 | 187.0 | 4 |
| 05 | 周梅 | 76.0 | 87.0 | 0.0 | 163.0 | 5 |
| 04 | 李云 | 50.0 | 30.0 | 20.0 | 100.0 | 6 |
| 06 | 吴兰 | 31.0 | 0.0 | 34.0 | 65.0 | 7 |
| 单科总成绩 | 387.0 | 436.0 | 411.0 |
—> 查询和"01"号同学学习课程完全相同的其他同学的信息
select *
from student
where s_id in (select s_id
from score
where s_id != '01'
group by s_id
having group_concat(distinct c_id order by c_id separator '|') =
(select group_concat(distinct c_id order by c_id separator '|')
from score
where s_id = '01'));
-- 方法2 join
select *
from student
where s_id in (select s_id
from (select s_id,
group_concat(distinct c_id order by c_id separator '|') as a
from score
where s_id != '01'
group by s_id) t1
join (select group_concat(distinct c_id order by c_id separator '|') as b
from score
where s_id = '01') t2
on t1.a = t2.b);
-- 方法3 join
select st.*
from (select s_id,
group_concat(distinct c_id order by c_id separator '|') as a
from score
where s_id != '01'
group by s_id) t1
join (select group_concat(distinct c_id order by c_id separator '|') as b
from score
where s_id = '01') t2
on t1.a = t2.b
join student st on st.s_id = t1.s_id;
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 02 | 钱电 | 1990-12-21 00:00:00 | 男 |
| 03 | 孙风 | 1990-05-20 00:00:00 | 男 |
| 04 | 李云 | 1990-08-06 00:00:00 | 男 |
以下必须是HQL语法规则
select *
from student
join (select s_id
from (select s_id,
concat_ws('|', collect_set(c_id)) as num1
from score
where s_id <> '01'
group by s_id) t1
join(select concat_ws('|', collect_set(c_id)) as num2
from score
where s_id = '01') t2
on t1.num1 = t2.num2) t3
on student.s_id = t3.s_id;
—> 查询和"01"号同学学习课程不完全相同的其他同学的信息
-- 方法1
select *
from student
where s_id in (select s_id
from (select s_id,
group_concat(distinct c_id order by c_id separator '|') as a
from score
where s_id != '01'
group by s_id) t1
left join (select group_concat(distinct c_id order by c_id separator '|') as b
from score
where s_id = '01') t2
on t1.a = t2.b
where t2.b is null);
-- 方法2（适用于hive）
select student.s_id, student.s_name
from student
join (select *
from (select s_id, concat_ws('|', collect_set(c_id)) as num1
from score
group by s_id
having s_id <> '01') t1 -- <>等价于!=
left join (select concat_ws('|', collect_set(c_id)) as num2
from score
where s_id = '01'
) t2 on t1.num1 = t2.num2
where t2.num2 is null) t3
on t3.s_id = student.s_id;
-- answer3
可以使用上面提到过的 group_concat
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 5 | 周梅 | 1991-12-01 00:00:00 | 女 |
| 6 | 吴兰 | 1992-03-01 00:00:00 | 女 |
| 7 | 郑竹 | 1989-07-01 00:00:00 | 女 |
| 8 | 王菊 | 1990-01-20 00:00:00 | 女 |
—> 查询没有学全所有课程的同学的信息
别处有引用，勿动！！！
-- answer 2
select *
from student
where s_id in (select s_id
from score
group by s_id
having count(distinct c_id) < (select count(distinct c_id) from course));
-- answer 3
select student.*
from student
where s_id in (select s_id
from (select s_id,
count(distinct c_id) as num1
from score
group by s_id) t1
left join(select count(distinct c_id) as num2
from course) t2
on t1.num1 = t2.num2
where t2.num2 is null);
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 05 | 周梅 | 1991-12-01 | 女 |
| 06 | 吴兰 | 1992-03-01 | 女 |
| 07 | 郑竹 | 1989-07-01 | 女 |
| 08 | 王菊 | 1990-01-20 | 女 |
hive中如果使用concat_ws(‘|’, collect_set(c_id)，需要注意的事项
select concat_ws('|', collect_set(c_id)) as num2
from course;
| num2 |
|---|
| 03|01|02 |
select s_id, concat_ws('|', collect_set(c_id)) as num1
from score
group by s_id;
| s_id | num1 |
|---|---|
| 01 | 01|02|03 |
| 02 | 01|02|03 |
| 03 | 01|02|03 |
| 04 | 01|02|03 |
| 05 | 01|02 |
| 06 | 01|03 |
| 07 | 02|03 |
select *
from (select s_id, concat_ws('|', collect_set(c_id)) as num1
from score
group by s_id) t1
join (select concat_ws('|', collect_set(c_id)) as num2
from course) t2
on t1.num1 = t2.num2;
总结：num1的数据集和num2的数据集最好来自同一张表
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 02 | 钱电 | 1990-12-21 00:00:00 | 男 |
| 03 | 孙风 | 1990-05-20 00:00:00 | 男 |
| 04 | 李云 | 1990-08-06 00:00:00 | 男 |
| 05 | 周梅 | 1991-12-01 00:00:00 | 女 |
| 06 | 吴兰 | 1992-03-01 00:00:00 | 女 |
| 07 | 郑竹 | 1989-07-01 00:00:00 | 女 |
—> 查询学过"张三"老师授课的同学的信息
直接关联查询
-- 直接关联查询
select student.*,teacher.t_name
from student
join score on student.s_id = score.s_id
join course on course.c_id = score.c_id
join teacher on course.t_id = teacher.t_id and t_name = '张三';
| s_id | s_name | s_birth | s_sex | t_name |
|---|---|---|---|---|
| 01 | 赵雷 | 1990-01-01 00:00:00 | 男 | 张三 |
| 02 | 钱电 | 1990-12-21 00:00:00 | 男 | 张三 |
| 03 | 孙风 | 1990-05-20 00:00:00 | 男 | 张三 |
| 04 | 李云 | 1990-08-06 00:00:00 | 男 | 张三 |
| 05 | 周梅 | 1991-12-01 00:00:00 | 女 | 张三 |
| 07 | 郑竹 | 1989-07-01 00:00:00 | 女 | 张三 |
—> 查询没学过"张三"老师讲授的任一门课程的学生姓名
select student.*
from student
where s_id not in (select s_id
from score
join course on score.c_id = course.c_id
join teacher on teacher.t_id = course.t_id and t_name = '张三');
select student.*
from student
left join (select s_id
from score
join course on score.c_id = course.c_id
join teacher on teacher.t_id = course.t_id and t_name = '张三') t1
on student.s_id = t1.s_id
where t1.s_id is null;
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 06 | 吴兰 | 1992-03-01 00:00:00 | 女 |
| 08 | 王菊 | 2017-01-20 00:00:00 | 女 |
—> 查询各科成绩都大于4号同学的学号
select t1.*
from (select s_id,
sum(case when score.c_id = '01' then score.s_score else 0 end) as chinese,
sum(case when score.c_id = '02' then score.s_score else 0 end) as math,
sum(case when score.c_id = '03' then score.s_score else 0 end) as english
from score
group by s_id) t1
join (select sum(case when score.c_id = '01' then score.s_score else 0 end) as chinese,
sum(case when score.c_id = '02' then score.s_score else 0 end) as math,
sum(case when score.c_id = '03' then score.s_score else 0 end) as english
from score
where score.s_id = '04') t2
where t1.chinese > t2.chinese
and t1.math > t2.math
and t1.english > t2.english;
| s_id | chinese | math | english |
|---|---|---|---|
| 4 | 50.0 | 30.0 | 20.0 |
| s_id | chinese | math | english |
|---|---|---|---|
| 1 | 80.0 | 90.0 | 99.0 |
| 2 | 70.0 | 60.0 | 80.0 |
| 3 | 80.0 | 80.0 | 80.0 |
—> 查询"01"课程比"02"课程成绩高的学生的信息及课程分数
方法1
select student.s_id,
student.s_name,
student.s_sex,
sum(if(c_id = '01', s_score, 0)) as c01_score,
sum(if(c_id = '02', s_score, 0)) as c02_score
from score
join student on score.s_id = student.s_id
group by student.s_id, student.s_name, student.s_sex
having c01_score > c02_score;
| s_id | s_name | s_sex | c01_score | c02_score |
|---|---|---|---|---|
| 02 | 钱电 | 男 | 70.0 | 60.0 |
| 04 | 李云 | 男 | 50.0 | 30.0 |
| 06 | 吴兰 | 女 | 31.0 | 0.0 |
方法2
select student.s_id,
student.s_name,
t3.c01_score,
t3.c02_score
from student
join (select t1.s_id, t1.c01_score, t2.c02_score
from (select s_id, s_score as c01_score
from score
where c_id = '01') t1
join
(select s_id, s_score as c02_score
from score
where c_id = '02') t2
on t1.s_id = t2.s_id
where t1.c01_score > t2.c02_score) t3
on student.s_id = t3.s_id;
| s_id | s_name | s_sex | c01_score | c02_score |
|---|---|---|---|---|
| 02 | 钱电 | 男 | 70.0 | 60.0 |
| 04 | 李云 | 男 | 50.0 | 30.0 |
寄语：2种方法都很重要
—> 查询每一门课程都大于60分的同学
select s_id,
s_name
from student
where s_id in (select s_id
from score
group by s_id
having min(s_score) > 60
);
| s_id | s_name |
|---|---|
| 01 | 赵雷 |
| 03 | 孙风 |
| 05 | 周梅 |
| 07 | 郑竹 |
—> 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
数据&建表
create table score_cross
(
s_id string,
c_id string,
s_score int
)
insert overwrite table score_cross
values ('01', '01', 80),
('01', '02', 100),
('01', '03', 100),
('02', '02', 80),
('03', '02', 80);
| s_id | c_id | s_score |
|---|---|---|
| 01 | 01 | 80 |
| 01 | 02 | 100 |
| 01 | 03 | 100 |
| 02 | 02 | 80 |
| 03 | 02 | 80 |
无任何连接条件，产生了笛卡尔集合
select *
from score_cross s1, score_cross s2
where s1.c_id<>s2.c_id and s1.s_score=s2.s_score
| s_id | c_id | s_score | s_id | c_id | s_score |
|---|---|---|---|---|---|
| 01 | 01 | 80 | 03 | 02 | 80 |
| 01 | 01 | 80 | 02 | 02 | 80 |
| 01 | 02 | 100 | 01 | 03 | 100 |
| 01 | 03 | 100 | 01 | 02 | 100 |
| 02 | 02 | 80 | 01 | 01 | 80 |
| 03 | 02 | 80 | 01 | 01 | 80 |
去重
select distinct s1.*
from score_cross s1, score_cross s2
where s1.c_id<>s2.c_id and s1.s_score=s2.s_score
| s_id | c_id | s_score |
|---|---|---|
| 02 | 02 | 80 |
| 01 | 01 | 80 |
| 01 | 02 | 100 |
| 01 | 03 | 100 |
| 03 | 02 | 80 |
—> 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 st.*
from (select s_id from score where c_id = '01') t1
join (select s_id from score where c_id = '02') t2
on t1.s_id = t2.s_id
join student st on st.s_id = t1.s_id;
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 01 | 赵雷 | 1990-01-01 00:00:00 | 男 |
| 02 | 钱电 | 1990-12-21 00:00:00 | 男 |
| 03 | 孙风 | 1990-05-20 00:00:00 | 男 |
| 04 | 李云 | 1990-08-06 00:00:00 | 男 |
| 05 | 周梅 | 1991-12-01 00:00:00 | 女 |
另一解题思路：用 in
select student.*
from student
where s_id in (select s_id from score where c_id = '01' and s_id in (select s_id from score where c_id = '02'));
—> 查询学过编号"01"但是没有学过编号"02"的课程的同学的信息
select st.*
from (select s_id from score where c_id = '01') t1
left join (select s_id from score where c_id = '02') t2 on t1.s_id = t2.s_id
left join student st on t1.s_id = st.s_id
where t2.s_id is null;
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 06 | 吴兰 | 1992-03-01 00:00:00 | 女 |
另一解题思路：用 not in
不管是mysql中，还是hive中，都不建议使用 not in
select st.*
from score sc
join student st on sc.s_id = st.s_id
where sc.c_id = '01'
and sc.s_id not in (select s_id
from score
where c_id = '02');
—> 查询"李"姓老师的数量
select count(1)
from teacher
where t_name like "李%";
res：1
—> 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
错误做法：
select s_id, round(avg(s_score), 2) as avgscore
from score
where s_score < 60
group by s_id
having count(s_id) >= 2
| s_id | avgscore |
|---|---|
| 04 | 33.33 |
| 06 | 32.50 |
以上方法是错误的，它只把两门及其以上不及格课程的同学的学号给筛选出来了，但是平均成绩的计算方式是错误的
原表中6号同学只有2门成绩，并且都不及格
添加6号同学的2号课程为85分
执行上述语句
正确做法：
select st.s_id,
st.s_name,
round(avg(sc.s_score), 2) as avgScore
from score sc
join (select s_id
from score
where s_score < 60
group by s_id
having count(1) >= 2) t1 on sc.s_id = t1.s_id
join student st on st.s_id = t1.s_id
group by st.s_id, st.s_name;
| s_id | s_name | avgscore |
|---|---|---|
| 04 | 李云 | 33.33333 |
| 06 | 吴兰 | 50.00000 |
—> 查询不同老师所教课程的平均成绩（平均成绩从高到低显示，平均成绩相同时，按课程编号升序排列）
order by 中被排序的字段，must be either aggregated, or mentioned in GROUP BY clause
具体解释：
avgScore是被聚合函数应用的字段
c_id 是被分组的字段
select t.t_name,
c.c_name,
round(avg(sc.s_score), 2) as avgScore
from score sc
join course c on sc.c_id = c.c_id
join teacher t on c.t_id = t.t_id
group by t.t_name, c.c_name, c.c_id
order by avgScore desc, c.c_id asc;
| c_name | t_name | avgscore |
|---|---|---|
| 数学 | 张三 | 72.67 |
| 英语 | 王五 | 68.50 |
| 语文 | 李四 | 64.50 |
—> 查询所有课程成绩的第2名到第3名的学生信息及该课程成绩
上边已经出现过这样类型的题目了，但是这里又出现了，出现的目的是为了告知，数据排序时，如果选择join的方式不对，那么可能无序。例如你把 left join 换成 join 可以试试
select t1.s_id,
c_name,
t1.s_score,
t1.ranking
from (select s_id,
c_id,
s_score,
row_number() over (partition by c_id order by s_score desc ) as ranking
from score) t1
left join student st on st.s_id = t1.s_id
left join course c on c.c_id = t1.c_id
where t1.ranking >= 2
and t1.ranking <= 3;
| s_id | c_name | s_score | ranking |
|---|---|---|---|
| 03 | 语文 | 80.0 | 2 |
| 05 | 语文 | 76.0 | 3 |
| 07 | 数学 | 89.0 | 2 |
| 05 | 数学 | 87.0 | 3 |
| 07 | 英语 | 98.0 | 2 |
| 03 | 英语 | 80.0 | 3 |
—> 查询选修"张三"老师所授课程的学生中，成绩最高的学生信息，所学课程名称以及成绩
mysql8.x系列语法更为严格，以下语句会报错： sql_mode=only_full_group_by
它要求：某字段使用聚合函数后，必须同时使用group by
select student.*,
course.c_name,
max(score.s_score) as maxscore
from student
join score on student.s_id = score.s_id
join course on score.c_id = course.c_id
join teacher on teacher.t_id = course.t_id and t_name = '张三';
| s_id | s_name | s_birth | s_sex | c_name | maxscore |
|---|---|---|---|---|---|
| 01 | 赵雷 | 1990-01-01 00:00:00 | 男 | 数学 | 90.0 |
—> 查询选修"张三"老师所授课程的学生中，成绩最高的学生姓名，成绩，以及其它课程的成绩
mysql8.x系列语法更为严格，以下语句会报错： sql_mode=only_full_group_by
select student.s_name,
t1.mathmaxscore,
sum(case when score.c_id = '01' then score.s_score else 0 end) as chinese,
sum(case when score.c_id = '03' then score.s_score else 0 end) as english
from score
join (select s_id,
max(s_score) as mathmaxscore
from score
join course on score.c_id = course.c_id
join teacher on teacher.t_id = course.t_id and teacher.t_name = '张三') t1
on score.s_id = t1.s_id
join student on student.s_id = score.s_id;
之所以不用in，是因为用了以后无法确定那门课是最高成绩，故：用join比较合适
| s_name | mathmaxscore | chinese | english |
|---|---|---|---|
| 赵雷 | 90.0 | 80.0 | 99.0 |
—> 查询1990年出生的学生名单
select *
from student
where s_birth like '1990%';
| s_id | s_name | s_birth | s_sex |
|---|---|---|---|
| 01 | 赵雷 | 1990-01-01 | 男 |
| 02 | 钱电 | 1990-12-21 | 男 |
| 03 | 孙风 | 1990-05-20 | 男 |
| 04 | 李云 | 1990-08-06 | 男 |
| 08 | 王菊 | 1990-01-20 | 女 |
—> 查询各学生的年龄
-- 张三生日：2020-05-04
-- 李四生日：2020-03-04
-- 当前日期：2022-04-04
-- 通常计算年龄的方式（张三，李四） 2022-2020 = 2岁
select s_name, s_birth, year(current_date) - year(s_birth) as age
from student;
| s_name | s_birth | age |
|---|---|---|
| 赵雷 | 1990-01-01 00:00:00 | 32 |
| 钱电 | 1990-12-21 00:00:00 | 32 |
| 孙风 | 1990-05-20 00:00:00 | 32 |
| 李云 | 1990-08-06 00:00:00 | 32 |
| 周梅 | 1991-12-01 00:00:00 | 31 |
| 吴兰 | 1992-03-01 00:00:00 | 30 |
| 郑竹 | 1989-07-01 00:00:00 | 33 |
| 王菊 | 1990-01-20 00:00:00 | 32 |
-- 有时候较真计算年龄：张三实际年龄2岁少1个月：2022-2020-1=1岁
-- 李四实际年龄2岁多1个月：2022-2020-0=2岁
select s_name,
s_birth,
(year(current_date) - year(s_birth) - (case
when month(CURRENT_DATE) < month(s_birth) then 1
when month(CURRENT_DATE) = month(s_birth) and
day(CURRENT_DATE) < day(s_birth) then 1
else 0 end)
) as age
from student;
| s_name | s_birth | age |
|---|---|---|
| 赵雷 | 1990-01-01 00:00:00 | 32 |
| 钱电 | 1990-12-21 00:00:00 | 31 |
| 孙风 | 1990-05-20 00:00:00 | 32 |
| 李云 | 1990-08-06 00:00:00 | 32 |
| 周梅 | 1991-12-01 00:00:00 | 30 |
| 吴兰 | 1992-03-01 00:00:00 | 30 |
| 郑竹 | 1989-07-01 00:00:00 | 33 |
| 王菊 | 1990-01-20 00:00:00 | 32 |
—> 查询本周过生日的学生
select *
from student
where weekofyear(current_date) = weekofyear(s_birth);
—> 查询下周过生日的学生
select *
from student
where weekofyear(CURRENT_DATE) + 1 = weekofyear(s_birth);
—>查询12月份过生日的学生
--方法1
select s_name, s_sex, s_birth
from student
where substring(s_birth, 6, 2) = '12';
--方法2
select s_name, s_sex, s_birth
from student
where month(s_birth) = '12';
| s_name | s_sex | s_birth |
|---|---|---|
| 钱电 | 男 | 1990-12-21 |
| 周梅 | 女 | 1991-12-01 |
总结：mysql8.x语法更为严格体现在哪些方面
order by 中被排序的字段，must be either aggregated, or mentioned in GROUP BY clause(例如查询不同老师所教课程的平均成绩)- having 同 order by 一样
- 某字段使用聚合函数后，必须同时使用 group by（在student中，筛选出id最大的同学的名字）
—> 数据&建表语句(hive)
create table student
(
s_id string,
s_name string,
s_birth string,
s_sex string
);
create table course
(
c_id string,
c_name string,
t_id string
);
create table teacher
(
t_id string,
t_name string
);
create table score
(
s_id string,
c_id string,
s_score int
);
insert into student (s_id, s_name, s_birth, s_sex) values ('01', '赵雷', '1990-01-01 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('02', '钱电', '1990-12-21 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('03', '孙风', '1990-05-20 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('04', '李云', '1990-08-06 00:00:00', '男');
insert into student (s_id, s_name, s_birth, s_sex) values ('05', '周梅', '1991-12-01 00:00:00', '女');
insert into student (s_id, s_name, s_birth, s_sex) values ('06', '吴兰', '1992-03-01 00:00:00', '女');
insert into student (s_id, s_name, s_birth, s_sex) values ('07', '郑竹', '1989-07-01 00:00:00', '女');
insert into student (s_id, s_name, s_birth, s_sex) values ('08', '王菊', '1990-01-20 00:00:00', '女');
insert into score (s_id, c_id, s_score) values ('01', '01', 80.0);
insert into score (s_id, c_id, s_score) values ('01', '02', 90.0);
insert into score (s_id, c_id, s_score) values ('01', '03', 99.0);
insert into score (s_id, c_id, s_score) values ('02', '01', 70.0);
insert into score (s_id, c_id, s_score) values ('02', '02', 60.0);
insert into score (s_id, c_id, s_score) values ('02', '03', 80.0);
insert into score (s_id, c_id, s_score) values ('03', '01', 80.0);
insert into score (s_id, c_id, s_score) values ('03', '02', 80.0);
insert into score (s_id, c_id, s_score) values ('03', '03', 80.0);
insert into score (s_id, c_id, s_score) values ('04', '01', 50.0);
insert into score (s_id, c_id, s_score) values ('04', '02', 30.0);
insert into score (s_id, c_id, s_score) values ('04', '03', 20.0);
insert into score (s_id, c_id, s_score) values ('05', '01', 76.0);
insert into score (s_id, c_id, s_score) values ('05', '02', 87.0);
insert into score (s_id, c_id, s_score) values ('06', '01', 31.0);
insert into score (s_id, c_id, s_score) values ('06', '03', 34.0);
insert into score (s_id, c_id, s_score) values ('07', '02', 89.0);
insert into score (s_id, c_id, s_score) values ('07', '03', 98.0);
insert into course (c_id, c_name, t_id) values ('01', '语文', '02');
insert into course (c_id, c_name, t_id) values ('02', '数学', '01');
insert into course (c_id, c_name, t_id) values ('03', '英语', '03');
insert into teacher (t_id, t_name) values ('01', '张三');
insert into teacher (t_id, t_name) values ('02', '李四');
insert into teacher (t_id, t_name) values ('03', '王五');
sql 面试题-求职
sql 面试题-求职