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Mysql Sql 语句练习题(50道)_Mysql_
2023-05-27
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简介 Mysql Sql 语句练习题(50道)_Mysql_
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
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);
表数据如下
student 学生表:
s_id | s_name | s_birth | s_sex |
---|---|---|---|
01 | 赵雷 | 1990-01-01 | 男 |
02 | 钱电 | 1990-12-21 | 男 |
03 | 孙凤 | 1990-05-20 | 男 |
04 | 李云 | 1990-08-06 | 男 |
05 | 周梅 | 1991-12-12 | 女 |
06 | 吴兰 | 2017-12-13 | 女 |
07 | 郑竹 | 1989-07-01 | 女 |
08 | 王菊 | 1990-01-20 | 女 |
09 | 赵雷 | 1990-01-21 | 女 |
10 | 赵雷 | 1990-01-22 | 男 |
score 分数表:
s_id | c_id | s_score |
---|---|---|
01 | 01 | 80 |
01 | 02 | 90 |
01 | 03 | 99 |
02 | 01 | 70 |
02 | 02 | 60 |
02 | 03 | 80 |
03 | 01 | 80 |
03 | 02 | 80 |
03 | 03 | 80 |
04 | 01 | 50 |
04 | 02 | 30 |
04 | 03 | 20 |
05 | 01 | 76 |
05 | 03 | 87 |
06 | 01 | 31 |
06 | 03 | 34 |
07 | 03 | 89 |
07 | 01 | 98 |
course 课程表
c_id | c_name | t_id |
---|---|---|
01 | 语文 | 02 |
02 | 数学 | 01 |
03 | 英语 | 03 |
teacher 老师表:
t_id | t_name |
---|---|
01 | 张三 |
02 | 李四 |
03 | 王五 |
-- 准备条件,去掉 sql_mode 的 ONLY_FULL_GROUP_BY 否则此种情况下会报错: -- Expression #1 of select list is not in group by clause and contains nonaggregated column 'userinfo. -- 原因: -- MySQL 5.7.5和up实现了对功能依赖的检测。如果启用了only_full_group_by SQL模式(在默认情况下是这样), -- 那么MySQL就会拒绝选择列表、条件或顺序列表引用的查询,这些查询将引用组中未命名的非聚合列,而不是在功能上依赖于它们。 -- (在5.7.5之前,MySQL没有检测到功能依赖项,only_full_group_by在默认情况下是不启用的。关于前5.7.5行为的描述,请参阅MySQL 5.6参考手册。) -- 执行以下个命令,可以查看 sql_mode 的内容。 SHOW SESSION VARIABLES; SHOW GLOBAL VARIABLES; select @@sql_mode; -- 更改 set global sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION'; set session sql_mode='STRICT_TRANS_TABLES,NO_ZERO_IN_DATE,NO_ZERO_DATE,ERROR_FOR_DIVISION_BY_ZERO,NO_AUTO_CREATE_USER,NO_ENGINE_SUBSTITUTION';
练习题和sql
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数 select st.*,sc.s_score as '语文' ,sc2.s_score '数学' from student st left join score sc on sc.s_id=st.s_id and sc.c_id='01' left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02' where sc.s_score>sc2.s_score -- 2、查询"01"课程比"02"课程成绩低的学生的信息及课程分数 select st.*,sc.s_score '语文',sc2.s_score '数学' from student st left join score sc on sc.s_id=st.s_id and sc.c_id='01' left join score sc2 on sc2.s_id=st.s_id and sc2.c_id='02' where sc.s_score=60 -- 4、查询平均成绩小于60分的同学的学生编号和学生姓名和平均成绩 -- (包括有成绩的和无成绩的) select st.s_id,st.s_name,(case when ROUND(AVG(sc.s_score),2) is null then 0 else ROUND(AVG(sc.s_score)) end ) cjScore from student st left join score sc on sc.s_id=st.s_id group by st.s_id having AVG(sc.s_score)<60 or AVG(sc.s_score) is NULL -- 5、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 st.s_id,st.s_name,count(c.c_id),( case when SUM(sc.s_score) is null or sum(sc.s_score)="" then 0 else SUM(sc.s_score) end) 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 group by st.s_id -- 6、查询"李"姓老师的数量 select t.t_name,count(t.t_id) from teacher t group by t.t_id having t.t_name like "李%"; -- 7、查询学过"张三"老师授课的同学的信息 select st.* 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 left join teacher t on t.t_id=c.t_id where t.t_name="张三" -- 8、查询没学过"张三"老师授课的同学的信息 -- 张三老师教的课 select c.* from course c left join teacher t on t.t_id=c.t_id where t.t_name="张三" -- 有张三老师课成绩的st.s_id 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") -- 不在上面查到的st.s_id的学生信息,即没学过张三老师授课的同学信息 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") ) -- 9、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 student st inner 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" where st.s_id in ( select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02" ) 网友提供的思路(厉害呦~): SELECT st.* FROM student st INNER JOIN score sc ON sc.`s_id`=st.`s_id` GROUP BY st.`s_id` HAVING SUM(IF(sc.`c_id`="01" OR sc.`c_id`="02" ,1,0))>1 -- 10、查询学过编号为"01"但是没有学过编号为"02"的课程的同学的信息 select st.* from student st inner 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" where st.s_id not in ( select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id inner join course c2 on c2.c_id=sc2.c_id and c2.c_id="02" ) -- 11、查询没有学全所有课程的同学的信息 -- 太复杂,下次换一种思路,看有没有简单点方法 -- 此处思路为查学全所有课程的学生id,再内联取反面 select * from student where s_id not in ( select st.s_id from student st inner join score sc on sc.s_id = st.s_id and sc.c_id="01" where st.s_id in ( select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id="02" ) and st.s_id in ( select st2.s_id from student st2 inner join score sc2 on sc2.s_id = st2.s_id and sc2.c_id="03" )) -- 来自一楼网友的思路,左连接,根据学生id分组过滤掉 数量小于 课程表中总课程数量的结果(show me his code),简洁不少。 select st.* from Student st left join Score S on st.s_id = S.s_id group by st.s_id having count(c_id)<(select count(c_id) from Course) -- 12、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 st.* from student st left join score sc on sc.s_id=st.s_id where sc.c_id in ( select sc2.c_id from student st2 left join score sc2 on sc2.s_id=st2.s_id where st2.s_id ='01' ) -- 13、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 st.* from student st left join score sc on sc.s_id=st.s_id group by st.s_id having group_concat(sc.c_id) = ( select group_concat(sc2.c_id) from student st2 left join score sc2 on sc2.s_id=st2.s_id where st2.s_id ='01' ) -- 14、查询没学过"张三"老师讲授的任一门课程的学生姓名 select st.s_name from student st where st.s_id not in ( select sc.s_id from score sc inner join course c on c.c_id=sc.c_id inner join teacher t on t.t_id=c.t_id and t.t_name="张三" ) -- 15、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 st.s_id,st.s_name,avg(sc.s_score) from student st left join score sc on sc.s_id=st.s_id where sc.s_id in ( select sc.s_id from score sc where sc.s_score<60 or sc.s_score is NULL group by sc.s_id having COUNT(sc.s_id)>=2 ) group by st.
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