SQL笔试50题

Posted 2020-12-17 unuliha

工具：Navicat Premium 封装的mysql。

1.表的创建

-- 创建数据库
create database school;
use school;

-- 建表
-- 学生表：学生编号,学生姓名, 出生年月,学生性别
create table Student(s_id varchar(10),s_name nvarchar(10),s_birth datetime,s_sex nvarchar(10));
insert into Student values(\'01\' , N\'赵雷\' , \'1990-01-01\' , N\'男\');
insert into Student values(\'02\' , N\'钱电\' , \'1990-12-21\' , N\'男\');
insert into Student values(\'03\' , N\'孙风\' , \'1990-05-20\' , N\'男\');
insert into Student values(\'04\' , N\'李云\' , \'1990-08-06\' , N\'男\');
insert into Student values(\'05\' , N\'周梅\' , \'1991-12-01\' , N\'女\');
insert into Student values(\'06\' , N\'吴兰\' , \'1992-03-01\' , N\'女\');
insert into Student values(\'07\' , N\'郑竹\' , \'1989-07-01\' , N\'女\');
insert into Student values(\'08\' , N\'王菊\' , \'1990-01-20\' , N\'女\');
-- 课程表：课程编号, 课程名称, 教师编号
create table Course(c_id varchar(10),c_name nvarchar(10),t_id varchar(10));
insert into Course values(\'01\' , N\'语文\' , \'02\');
insert into Course values(\'02\' , N\'数学\' , \'01\');
insert into Course values(\'03\' , N\'英语\' , \'03\');
-- 教师表：教师编号,教师姓名
create table Teacher(t_id varchar(10),t_name nvarchar(10));
insert into Teacher values(\'01\' , N\'张三\');
insert into Teacher values(\'02\' , N\'李四\');
insert into Teacher values(\'03\' , N\'王五\');
-- 成绩表：学生编号,课程编号,分数
create table Score(s_id varchar(10),c_id varchar(10),s_score decimal(18,1));
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);

2.表的结构

 

 

3.笔试50题

-- 1.查询“01”课程比“02”课程成绩高的所有学生的学号
SELECT st.*, sc1.s_score as "课程1", sc2.s_score as "课程2", sc3.s_score as "课程3"
From student st 
    JOIN score sc1 on st.s_id=sc1.S_id AND sc1.c_id = "01"
    JOIN score sc2 on st.s_id=sc2.S_id AND  sc2.c_id = "02"
    JOIN score sc3 on st.s_id=sc3.S_id AND sc3.c_id ="03"
WHERE sc1.s_score>sc2.s_score

-- 2.查询平均成绩大于60分的同学的学号和平均成绩
SELECT st.s_id,s_name,ROUND(AVG(s_score),2) as avg_score
FROM student st JOIN score on st.s_id = score.s_id
GROUP BY s_id HAVING avg(s_score)>=60
#HAVING子句给出了选择组的条件;where作用于基本表或视图，having作用于组;WHERE子句中不能用聚集函数做条件表达式

-- 3.查询所有同学的学号、姓名、选课数、总成绩
SELECT st.s_id,st.s_name,count(c_id) as "选课总数",sum(s_score) as total_score
from student st left join score on st.s_id=score.s_id
GROUP BY st.s_id
#有学生未出现在成绩表上，用左连接，保证出现在学生表的学生都被输出

-- 4.查询姓“李”的老师的个数
SELECT count(t_id) from teacher where t_name like \'李%\'

-- 5.查询没学过“张三”老师课的同学的学号、姓名
SELECT st1.* 
from student st1 where s_id not in 
    (SELECT st.s_id
    from student st,score sc,course c,teacher te
    WHERE st.s_id = sc.s_id and sc.c_id = c.c_id and c.t_id = te.t_id and     te.t_name =     "张三")

-- 6.查询学过“张三”老师所教的课的同学的学号、姓名；
SELECT st.* 
from student st 
    join score on st.s_id = score.s_id
    join course on score.c_id = course.c_id
    join teacher on course.t_id = teacher.t_id AND teacher.t_name = "张三"        

SELECT  st.* 
from student st,score sc,course c,teacher te
WHERE st.s_id = sc.s_id and sc.c_id = c.c_id and c.t_id = te.t_id and te.t_name = "张三"

-- 7.查询学过编号“01”并且也学过编号“02”课程的同学的学号、姓名；
SELECT st.* 
from student st 
    join score sc1 on st.s_id = sc1.s_id and sc1.c_id = 01
    join score sc2 on st.s_id = sc2.s_id and sc2.c_id = 02 

-- 8.查询学过01但是没有学过02的同学的信息
SELECT st.* 
from student st join score s1 on st.s_id = s1.s_id and s1.c_id = 01 
where st.s_id not in (SELECT  s_id from  score WHERE c_id = 02)
#字段前最好都跟上表名

-- 9.查询所有课程成绩小于60分的同学的学号、姓名；
SELECT st.s_id,s_name
from student st 
WHERE s_id in(
    SELECT s_id from score 
    GROUP BY s_id HAVING max(s_score)<60
    ) 

-- 10.查询没有学全所有课的同学的学号、姓名
select * from student 
where s_id not in(
    SELECT s_id from score 
    GROUP BY s_id HAVING count(c_id) = (
        select count(c_id) from course
        )
    )

-- 11.查询至少有一门课与学号为“01”的同学所学相同的同学的学号和姓名
SELECT st.* 
from student st 
where s_id  in (
    SELECT s_id from score where c_id in(
        SELECT c_id from score WHERE s_id = 01
        ) and s_id not in ("01")
    )

#思路类似，但将所有表放在一起可以精简过程
SELECT DISTINCT st.* 
FROM student st JOIN score sc 
    ON st.s_id=sc.s_id AND sc.c_id IN (
        SELECT sc.c_id FROM score sc WHERE sc.s_id="01" 
    ) AND sc.s_id NOT IN ("01");

-- 12.查询和"01"号的同学学习的课程完全相同的其他同学的学号和姓名
SELECT st.* 
from student st join score sc on st.s_id = sc.s_id
WHERE st.s_id not in (
    SELECT s_id from student WHERE c_id not in(
        SELECT c_id from score where s_id = "01"
    )
)
GROUP BY st.s_id
HAVING COUNT(c_id)=(SELECT count(c_id) from score WHERE s_id = "01")
#虚拟表的名称为局部变量

-- 14.查询没学过"张三"老师讲授的任一门课程的学生姓名
SELECT s_name from student  WHERE s_id not in(
    SELECT st.s_id  FROM student st 
            join score sc on st.s_id = sc.s_id
        join course c on sc.c_id = c.c_id
            join teacher te on te.t_id = c.t_id and 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,s_name,ROUND(avg(sc.s_score),2) as avg_score
from student st join score sc on st.s_id = sc.s_id
WHERE st.s_id in (
    SELECT s_id from score WHERE s_score < "60"
    GROUP BY s_id 
    HAVING COUNT(*)> 1
)
GROUP BY st.s_id

#有分组汇总函数就要有group by，除非只有一个组
SELECT st.s_id,st.s_name,ROUND(avg(sc.s_score),2) as avg_score
from student st 
    join score sc on st.s_id = sc.s_id and sc.s_score<\'60\'
GROUP BY s_id HAVING count(s_score) > 1

-- 16.检索"01"课程分数小于60，按分数降序排列的学生信息
SELECT st.* from student st JOIN score sc on st.s_id = sc.s_id
WHERE sc.s_score < 60 and sc.c_id = "01"
ORDER BY sc.s_score DESC

-- 17.按平均成绩从高到低显示所有学生的平均成绩
SELECT st.s_id, st.s_name,sc1.s_score as "01",sc2.s_score as "02",sc3.s_score as "03",ROUND(avg(sc.s_score),2) as "avg_score"
from student st 
    left join score sc on st.s_id = sc.s_id 
    left join score sc1 on st.s_id = sc1.s_id and sc1.c_id = "01"
    left join score sc2 on st.s_id = sc2.s_id and sc2.c_id = "02"
    left join score sc3 on st.s_id = sc3.s_id and sc3.c_id = "03" 
GROUP BY st.s_id
ORDER BY avg(sc.s_score) DESC;

-- 18.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
SELECT a.c_id "课程ID",a.c_name "课程名",
    max(b.s_score) "最高分",
    min(b.s_score) "最低分",
    ROUND(avg(b.s_score),2)  "平均分",
    sum(case when b.s_score>60 then 1 else 0 end)/count(1) "及格率",
    sum(case when b.s_score>=70 and b.s_score < 80 then 1 else 0 end)/count(1) "中等率",
    sum(case when b.s_score>=80 and b.s_score < 90 then 1 else 0 end)/count(1) "优良率",
    sum(case when b.s_score>=90 then 1 else 0 end)/count(1) "优秀率"
from course a join score b on a.c_id = b.c_id
GROUP BY 1

-- 19.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 sc.c_id,c_name,round(avg(s_score),2) \'avg_score\',
    concat(round(sum(case when s_score >= 60 then 1 else 0 end)/count(1)*100,2),\'%\') "及格率"
from score sc join course c on sc.c_id = c.c_id
GROUP BY sc.c_id
ORDER BY avg_score,"及格率" desc

-- 20.查询学生的总成绩并进行排名
--有rank函数时
SELECT sc.s_id,st.s_name,sum(s_score) as sum_score,
    rank() over(ORDER BY sum(sc.s_score) desc) as score_rank
FROM score sc join student st on sc.s_id = st.s_id 
GROUP BY sc.s_id;

--无rank函数时
SELECT a.s_id,a.s_name,
    @i := @i +1 as 序号,
    @k := (case when  @score = a.total_score then  @k else @i end) as 排名,
    @score := a.total_score as total_score
from(
    SELECT st.s_id ,s_name ,sum(sc.s_score) as total_score
    from student st join score sc on st.s_id = sc.s_id 
    GROUP BY st.s_id
    ORDER BY total_score desc
    )a,
    (SELECT @i:=0,@k:=0,@score:=0)b 
    
-- 21.查询不同老师所教不同课程平均分从高到低显示
SELECT t.t_id,t_name,c.c_id,c.c_name,
    round(avg(s_score),2) avg_score
from teacher t 
    join course c on t.t_id =某公司笔试题解读

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