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sql常见50道查询练习题

2024/11/30 10:49:34 来源：https://blog.csdn.net/m0_46926492/article/details/140520386  浏览:  次   关键词：sql常见50道查询练习题

sql常见50道查询练习题

  • 1. 表创建
    • 1.1 表创建
    • 1.2 数据插入
  • 2. 简单查询例题(3题）
    • 2.1 查询"李"姓老师的数量
    • 2.2 查询男生、女生人数
    • 2.3 查询名字中含有"风"字的学生信息
  • 3. 日期相关例题(6题）
    • 3.1 查询各学生的年龄
    • 3.2 查询本周过生日的学生
    • 3.3 查询下周过生日的学生
    • 3.4 查询本月过生日的学生
    • 3.5 查询下月过生日的学生
    • 3.6 查询1990年出生的学生名单
  • 4. 开窗函数查询（7题）
    • 4.1 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
    • 4.2 按各科成绩进行排序，并显示排名(实现不完全)
    • 4.3 查询学生的总成绩并进行排名
    • 4.4 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
    • 4.5 查询学生平均成绩及其名次
    • 4.6 查询各科成绩前三名的记录
    • 4.7 查询每门功成绩最好的前两名
  • 5. 表连接+子查询+聚合函数查询(34题）
    • 5.1 查询"01"课程比"02"课程成绩高的学生的信息及课程分数
    • 5.2 查询"01"课程比"02"课程成绩低的学生的信息及课程分数
    • 5.3 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
    • 5.4 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)
    • 5.5 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
    • 5.6 查询学过"张三"老师授课的同学的信息
    • 5.7 查询没学过"张三"老师授课的同学的信息
    • 5.8 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
    • 5.9 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
    • 5.10 查询没有学全所有课程的同学的信息
    • 5.11 查询至少有一门课与学号为"01"的同学所学相同的同学的信息
    • 5.12 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
    • 5.13 查询没学过"张三"老师讲授的任一门课程的学生姓名
    • 5.14 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
    • 5.15 检索"01"课程分数小于60，按分数降序排列的学生信息
    • 5.16 查询各科成绩最高分、最低分和平均分
    • 5.17 统计各科成绩各分数段人数
    • 5.18 查询不同老师所教不同课程平均分从高到低显示
    • 5.19 查询每门课程被选修的学生数
    • 5.20 查询出只有两门课程的全部学生的学号和姓名
    • 5.21 查询同名同性学生名单，并统计同名人数
    • 5.22 查询每门课程的平均成绩
    • 5.23 查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩
    • 5.24 查询课程名称为"数学"，且分数低于60的学生姓名和分数
    • 5.25 查询所有学生的课程及分数情况
    • 5.26 查询任何一门课程成绩在70分以上的姓名、课程名称和分数(学生的每门课都大于70)
    • 5.27 查询不及格的课程
    • 5.28 查询课程编号为01且课程成绩在80分以上的学生的学号和姓名
    • 5.29 求每门课程的学生人数
    • 5.30 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及其成绩
    • 5.31 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
    • 5.32 统计每门课程的学生选修人数（超过5人的课程才统计）
    • 5.33 检索至少选修两门课程的学生学号
    • 5.34 查询选修了全部课程的学生信息

1. 表创建

在这里插入图片描述

1.1 表创建

#–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
CREATE TABLE `Student` (`s_id` VARCHAR(20),s_name VARCHAR(20) NOT NULL DEFAULT '',s_brith VARCHAR(20) NOT NULL DEFAULT '',s_sex VARCHAR(10) NOT NULL DEFAULT '',PRIMARY KEY(s_id)
);#–2.课程表 
#Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号 
create table Course(c_id varchar(20),c_name VARCHAR(20) not null DEFAULT '',t_id VARCHAR(20) NOT NULL,PRIMARY KEY(c_id)
);/*
–3.教师表 
Teacher(t_id,t_name) –教师编号,教师姓名 
*/
CREATE TABLE Teacher(t_id VARCHAR(20),t_name VARCHAR(20) NOT NULL DEFAULT '',PRIMARY KEY(t_id)
);/*
–4.成绩表 
Score(s_id,c_id,s_score) –学生编号,课程编号,分数
*/
Create table Score(s_id VARCHAR(20),c_id VARCHAR(20) not null default '',s_score INT(3),primary key(`s_id`,`c_id`)
);

1.2 数据插入

#--插入学生表测试数据
#('01' , '赵雷' , '1990-01-01' , '男')
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);

2. 简单查询例题(3题）

2.1 查询"李"姓老师的数量

SELECTcount(1) as cnt
FROMteacher
WHEREt_name like "李%"

2.2 查询男生、女生人数

SELECTs.s_sex,count(1) as 人数
FROMstudent s
group bys.s_sex

2.3 查询名字中含有"风"字的学生信息

SELECT*
FROMstudent
WHEREs_name like "%风%"

3. 日期相关例题(6题）

3.1 查询各学生的年龄

  • (按照出生日期来算，当前月日 < 出生年月的月日则，年龄减一)
    -- if函数selecta.*,year(NOW())-year(a.s_brith)-if(DATE_FORMAT(now(),"%m%d") >DATE_FORMAT(a.s_brith,"%m%d"),0,1) as age
FROMstudent a-- case函数select s_brith,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_brith,'%Y') - (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_brith,'%m%d') then 0 else 1 end)) as age
from student;

3.2 查询本周过生日的学生

SELECT*
FROMstudent
WHEREWEEKOFYEAR(STR_TO_DATE(concat(year(NOW()),DATE_FORMAT(s_brith,'%m%d')),"%Y%m%d"))=WEEKOFYEAR(NOW())
-- 	WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth)

3.3 查询下周过生日的学生

SELECT*
FROMstudent
WHEREWEEKOFYEAR(STR_TO_DATE(concat(year(NOW()),DATE_FORMAT(s_brith,'%m%d')),"%Y%m%d"))=WEEKOFYEAR(NOW()+interval "7" day)
-- 	WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1=WEEK(s_birth)

3.4 查询本月过生日的学生

SELECT*
FROMstudent
WHEREMONTH(now())=month(s_brith)

3.5 查询下月过生日的学生

SELECT*
FROMstudent
WHEREMONTH(now()+interval "1" month)=month(s_brith)

3.6 查询1990年出生的学生名单

SELECT*
FROMstudent
WHEREs_brith like "1990%"
-- 	left(s_brith,4)="1990"
-- 	year(s_brith)="1990"

4. 开窗函数查询（7题）

4.1 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

  • 方法一:开窗函数
    selecta.*,avg(a.s_score) over(PARTITION by a.s_id) as avg_score
FROMscore a
  • 方法二：临时表连接
    SELECTa.*,t.avg_score
FROMscore a,(SELECTa.s_id,round(avg(a.s_score),2) as avg_scoreFROMscore agroup bya.s_id) t
WHEREa.s_id=t.s_id
order byt.avg_score desc
  • 方法三:长型数据转为宽型数据
    SELECTa.s_id,ifnull((select s_score from score where s_id=a.s_id and c_id="01"),0) as "语文",ifnull((select s_score from score where s_id=a.s_id and c_id="02"),0) as "数学",ifnull((select s_score from score where s_id=a.s_id and c_id="03"),0) as "英语",ifnull(round(avg(a.s_score),2),0) as avg_score
FROMscore a
group bya.s_id
order byifnull(round(avg(a.s_score),2),0) desc

4.2 按各科成绩进行排序，并显示排名(实现不完全)

  • 方法一：开窗函数
    SELECTa.*,rank() over(PARTITION by c_id order by s_score desc) rank排名,row_number() over(PARTITION by c_id order by s_score desc) row_number排名,dense_rank() over(PARTITION by c_id order by s_score desc) dense_rank排名
FROMscore a
  • 方法二：子查询
    SELECT	a.*,(select count(s_score) from score b where a.c_id=b.c_id and  a.s_score<b.s_score)+1 rk,(select count(distinct s_score) from score b where a.c_id=b.c_id and  a.s_score<=b.s_score) den_rk
FROMscore a
order byc_id,s_score desc

4.3 查询学生的总成绩并进行排名

  • 方法一：开窗函数
    SELECTt.*,rank() over(order by sum_score desc) rank排名
FROM(SELECTs_id,sum(s_score) as sum_scoreFROMscoregroup bys_id) t

4.4 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

  • 方法一：子查询+开窗函数
    SELECTa.*,t.c_id,t.rk,t.s_score
FROMstudent a,(SELECTa.s_id,a.c_id,a.s_score,dense_rank() over(PARTITION by a.c_id order by a.s_score desc) as rkFROMscore a) t
WHEREt.rk in (2,3)
ANDa.s_id=t.s_id

4.5 查询学生平均成绩及其名次

  • 方法一： 开窗函数
    SELECTt.*,rank() over(order by t.avg_score desc) 排名
FROM(SELECTa.s_id,round(avg(a.s_score),2) as avg_scoreFROMscore agroup bya.s_id) t

4.6 查询各科成绩前三名的记录

  • 方法一：开窗函数
    SELECTt.*
from(SELECTa.c_id,a.c_name,b.s_score,rank() over(PARTITION by a.c_id order by b.s_score desc) rkFROMcourse aLEFT JOINscore bONa.c_id=b.c_id) t
WHEREt.rk<=3;
  • 方法二：子查询
    SELECT*
from (SELECTa.c_id,a.c_name,b.s_score,(select count(c.s_score) from score c where a.c_id=c.c_id and b.s_score<c.s_score)+1 as rkFROMcourse aLEFT JOINscore bONa.c_id=b.c_id) t
WHEREt.rk<=3
order byt.c_name,t.rk asc;

4.7 查询每门功成绩最好的前两名

  • 方法一：开窗函数
    SELECTt.s_id,t.c_id,t.s_score
FROM(SELECT*,rank() over(PARTITION by b.c_id order by b.s_score desc) rkFROMscore b) t
WHEREt.rk<=2;
  • 方法二：自连接
    SELECTt.s_id,t.c_id,t.s_score
FROM(SELECTa.*,(select count(1) from score b where b.c_id=a.c_id and a.s_score<b.s_score)+1 as rkFROMscore aorder bya.c_id,rk) t
WHERE	t.rk<=2
  • 方法三：条件查询+子查询
    SELECTa.*
FROMscore a
WHERE(select count(1) from score b where b.c_id=a.c_id and a.s_score<b.s_score)+1<=2
order bya.c_id

5. 表连接+子查询+聚合函数查询(34题）

5.1 查询"01"课程比"02"课程成绩高的学生的信息及课程分数

  • 方法一：自连接，同列比较，使用自查询
    • 思路：先找出查询条件的学生信息及分数，根据子查询得到最终结果
    SELECT
st.*,t1.sc1,t1.sc2
FROMstudent st,(SELECTs1.s_id,s1.s_score as sc1,s2.s_score as sc2FROMscore s1,score s2WHERE	s1.c_id="01"AND	s2.c_id="02"ANDs1.s_id=s2.s_idANDs1.s_score>s2.s_score) t1
WHEREst.s_id=t1.s_id;
  • 方法二：表连接
    SELECT
st.*,s1.s_score as sc1,s2.s_score as sc2
FROMstudent st
left JOINscore s1
ONs1.s_id=st.s_id
left JOINscore s2
ONs2.s_id=st.s_id
WHERE	s1.c_id="01"
AND	s2.c_id="02"
ANDs1.s_id=s2.s_id
ANDs1.s_score>s2.s_score
  • 数据长型数据变为宽型数据
    -- IF函数或case函数
SELECTa.*,t.s01,t.s02
fromstudent a,(SELECTa.s_id,max(case when a.c_id="01" then a.s_score end) as s01,max(case when a.c_id="02" then a.s_score end) as s02
--  max(if(a.c_id="01",a.s_score,null)) as s01,
--  max(if(a.c_id="02",a.s_score,null)) as s02fromscore agroup bya.s_id) t
WHEREa.s_id=t.s_id
ANDt.s01>t.s02

5.2 查询"01"课程比"02"课程成绩低的学生的信息及课程分数

  • 与上一题思路一致，条件大于变小于
  • 方法一：自连接
    SELECT
st.*,t1.sc1,t1.sc2
FROMstudent st,(SELECTs1.s_id,s1.s_score as sc1,s2.s_score as sc2FROMscore s1,score s2WHERE	s1.c_id="01"AND	s2.c_id="02"ANDs1.s_id=s2.s_idANDs1.s_score<s2.s_score) t1
WHEREst.s_id=t1.s_id;
  • 方法二：表连接
    SELECTst.*,s1.s_score as sc1,s2.s_score as sc2
FROMstudent st
left JOINscore s1
ONs1.s_id=st.s_id
left JOINscore s2
ONs2.s_id=st.s_id
WHERE	s1.c_id="01"
AND	s2.c_id="02"
ANDs1.s_id=s2.s_id
ANDs1.s_score<s2.s_score-- 方法二
SELECTst.*,s1.s_score as sc1,s2.s_score as sc2
FROMstudent st
left JOINscore s1
ONs1.s_id=st.s_id
ANDs1.c_id="01"
left JOINscore s2
ONs2.s_id=st.s_id
ANDs2.c_id="02"
ANDs1.s_id=s2.s_id
WHEREs1.s_score<s2.s_score
  • 方法三:数据长型数据变为宽型数据
    -- IF函数或case函数
SELECTa.*,t.s01,t.s02
fromstudent a,(SELECTa.s_id,max(case when a.c_id="01" then a.s_score end) as s01,max(case when a.c_id="02" then a.s_score end) as s02
--  max(if(a.c_id="01",a.s_score,null)) as s01,
--  max(if(a.c_id="02",a.s_score,null)) as s02fromscore agroup bya.s_id) t
WHEREt.s01<t.s02
ANDa.s_id=t.s_id

5.3 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

  • 方法一：子查询
    -- 子查询一
SELECT
st.s_id,st.s_name,t.avg_s
FROMstudent ST,(SELECTs.s_id,round(avg(s.s_score),2) as avg_sFROM	score sGROUP BYs.s_idHAVINGround(avg(s.s_score),2)>=60) t
WHEREst.s_id=t.s_id-- 方法二：子查询二SELECTs.s_id,(select s_name from student where s_id=s.s_id) as s_name,round(avg(s.s_score),2) as avg_s
FROM	score s
GROUP BYs.s_id
HAVINGavg_s>=60
  • 方法二：表连接
    SELECTa.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
GROUP BYa.s_id
HAVINGround(avg(b.s_score),2)>=60;

5.4 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)

  • 方法一：子查询
    -- 有成绩的SELECTa.s_id,a.s_name,t.avg_acore
FROMstudent a,(SELECTa.s_id,round(avg(a.s_score),2) as avg_acoreFROMscore aGROUP BYa.s_idHAVINGround(avg(a.s_score),2)<60) t
WHEREa.s_id=t.s_idUNION
-- 没有成绩的：没有成绩的s_id不存在
SELECTa.s_id,a.s_name,0 as avg_acore
FROMstudent a
WHEREa.s_id not in (SELECT DISTINCT s_id FROM score);
  • 方法二：表连接
    SELECTa.s_id,a.s_name,ifnull(round(avg(b.s_score),2),0) as avg_score
FROMstudent a
LEFT JOINscore b
on a.s_id=b.s_id
GROUP BYa.s_id
HAVINGavg_score<60

5.5 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

SELECTa.s_id,a.s_name,count(b.c_id) as cnt_course,ifnull(sum(b.s_score),0) as sum_score
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
group bya.s_id

5.6 查询学过"张三"老师授课的同学的信息

  • 方法一：表连接+子查询单层嵌套
    SELECTa.*
FROMstudent a
LEFT JOINscore b
on a.s_id=b.s_id
LEFT JOINcourse c
ONb.c_id=c.c_id
where c.t_id in(SELECT t_id FROM teacher WHERE t_name = "张三")
  • 方法二：表连接+子查询多层嵌套
    SELECTa.*
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
WHEREb.c_id in (
SELECTc_id
FROMcourse where t_id in(SELECT t_id from teacher where t_name="张三")
);
  • 方法三：多表连接
    selecta.*
fromstudent a,score b,course c,teacher d
WHEREa.s_id=b.s_id
ANDb.c_id=c.c_id
ANDc.t_id=d.t_id
ANDd.t_name="张三"

5.7 查询没学过"张三"老师授课的同学的信息

  • 注意：一个学生有几门课程包含张三课程，不是张三课程的，根据没学过的查询不出来，因为一个人有多个老师的课程
  • 方法一：多层嵌套子查询
    SELECTs.*
FROMstudent s
WHEREs.s_id NOT IN (-- 查找学的学生SELECT DISTINCTa.s_idFROMstudent aLEFT JOIN score b ON a.s_id = b.s_idWHEREb.c_id IN (-- 查找学过的课程SELECT c_idFROM courseWHERE t_id IN ( SELECT t_id FROM teacher WHERE t_name = "张三")))
  • 方法二:条件查询+子表连接
    SELECT*
FROMstudent s
WHEREs.s_id not in (selecta.s_idfromscore a,course b,teacher cWHEREa.c_id=b.c_idANDb.t_id=c.t_idANDc.t_name="张三")

5.8 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

  • 方法一：子查询+自连接,同列对比可以用自连接
    SELECT*
FROMstudent s
WHEREs.s_id in(SELECTa.s_idFROMscore a,score bWHEREa.c_id="01" ANDb.c_id="02"ANDa.s_id=b.s_id)
  • 方法二：连表+自连接,同列对比可以用自连接
    SELECTs.*
FROMstudent s
LEFT JOIN score a
ONs.s_id=a.s_id
LEFT JOINscore b
ONa.s_id=b.s_id
WHEREa.c_id="01" 
ANDb.c_id="02"
  • 方法三：条件查询+子查询
    SELECT*
FROMstudent
WHEREs_id in (SELECTs_idFROMscorewherec_id="01" or c_id="02"GROUP BYs_idHAVINGcount(1)=2)
  • 方法四：自连接，条件连接
    SELECTs.*
FROMstudent s,score a,score b
WHEREs.s_id=a.s_id
ANDa.s_id=b.s_id
ANDa.c_id="01" 
ANDb.c_id="02"
  • 方法五：子查询+数据长型数据变为宽型数据
    SELECTa.*
FROMstudent a,(selecta.s_id,max(if(a.c_id="01",a.s_score,0)) as s01,max(if(a.c_id="02",a.s_score,0)) as s02fromscore agroup bya.s_id) t
WHEREa.s_id=t.s_id
ANDt.s01>0
ANDt.s02>0

5.9 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

  • 方法一：条件查询+子查询
    selecta.*
fromstudent a
WHERE	a.s_id in(select s_id from score where c_id="01") 
ANDa.s_id not in (select s_id from score where c_id="02")
  • 方法二： 子查询+分组聚合
    SELECTs.*
FROMstudent s,(SELECTa.s_id,max(case when a.c_id="01" then a.s_score end) s01,max(case when a.c_id="02" then a.s_score end) s02FROMscore agroup bya.s_id) t
WHEREs.s_id=t.s_id
ANDt.s01 is not NULL
ANDt.s02 is null
  • 方法三：数据长型数据变为宽型数据
    SELECTa.*
FROMstudent a,(selecta.s_id,max(if(a.c_id="01",a.s_score,null)) as s01,max(if(a.c_id="02",a.s_score,null)) as s02fromscore agroup bya.s_id) t
WHEREa.s_id=t.s_id
ANDt.s01 is not null
ANDt.s02 is null

5.10 查询没有学全所有课程的同学的信息

  • 方法一:条件查询+子查询
    SELECTs.*
FROMstudent s
WHEREs.s_id in(SELECTa.s_idFROMscore agroup bya.s_idhavingcount(1)<(select count(1) from course))
  • 方法二:表连接
    SELECTs.*,count(a.c_id) cnt
FROMstudent s
LEFT JOINscore a
ONa.s_id=s.s_id
group bys.s_id
HAVINGcount(a.c_id)<(select count(1) from course)

5.11 查询至少有一门课与学号为"01"的同学所学相同的同学的信息

  • 方法一:子查询
    SELECTs.*
FROMstudent s
WHEREs.s_id in(SELECTdistinct a.s_idFROMscore aWHEREa.c_id in(SELECTb.c_idFROMscore bWHEREb.s_id="01"))
ANDs.s_id!='01'
  • 方法二:表连接+子查询
    SELECTa.*
FROMstudent a
LEFT JOINscore b
on a.s_id=b.s_id
WHEREb.c_id in (SELECTb.c_idFROMscore bWHEREb.s_id="01")
group by 1,2,3,4

5.12 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

  • 筛选课程与01号一样的数据，计算课程数与01一致的
SELECTs.*
FROMstudent s
WHEREs.s_id in(SELECT distinct a.s_idFROMscore aWHEREa.c_id in(SELECTa.c_idFROMscore aWHEREa.s_id="01")ANDa.s_id!="01"group by a.s_idHAVINGcount(distinct a.c_id)=(select count(1) from score a where a.s_id="01"))

5.13 查询没学过"张三"老师讲授的任一门课程的学生姓名

  • 查询学过张三老师的学生，在学生表中反向查询
SELECTs.s_name
FROMstudent s
WHEREs.s_id not in(SELECTa.s_idFROMscore aWHEREa.c_id in (SELECTa.c_idFROMcourse aWHEREa.t_id in (SELECT t.t_id FROM teacher t WHERE t.t_name="张三")))

5.14 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

  • 方法一：表连接+分组+having条件
    SELECTa.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
group bya.s_id
havingsum(if(b.s_score>=60,0,1))>=2
  • 方法二：自连接+子查询
    selecta.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
FROMstudent a,score b
WHEREa.s_id=b.s_id
ANDa.s_id in(SELECTa.s_idFROMscore aWHEREa.s_score<60group bya.s_idHAVINGcount(1)>=2)
group bya.s_id
  • 方法三：表连接+子查询
    selecta.s_id,a.s_name,round(avg(b.s_score),2) as avg_score
FROMstudent a
LEFT JOINscore b
ona.s_id=b.s_id
wherea.s_id in(SELECTa.s_idFROMscore aWHEREa.s_score<60group bya.s_idHAVINGcount(1)>=2)
group bya.s_id

5.15 检索"01"课程分数小于60，按分数降序排列的学生信息

  • 方法一：表连接
    SELECTa.*,b.c_id,b.s_score
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
WHEREb.c_id="01" and b.s_score<60
order byb.s_score desc

5.16 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
  • 方法一：if语句
    SELECTa.c_id,a.c_name,max(b.s_score) as max_score,min(b.s_score) as min_score,round(avg(b.s_score),2) as avg_score,round(100*sum(if(b.s_score>=60,1,0))/count(1),2) as "及格率",round(100*sum(if(b.s_score>=70 and b.s_score<80,1,0))/count(1),2) as "中等率",round(100*sum(if(b.s_score>=80 and b.s_score<90,1,0))/count(1),2) as "优良率",round(100*sum(if(b.s_score>=90,1,0))/count(1),2) as "优秀率"
FROMcourse a,score b
WHEREa.c_id=b.c_id
group bya.c_id
  • 方法二：case when
    SELECTa.c_id,a.c_name,max(b.s_score) as max_score,min(b.s_score) as min_score,round(avg(b.s_score),2) as avg_score,round(100*sum(case when b.s_score>=60 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "及格率",round(100*sum(case when b.s_score>=70 and b.s_score<80 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "中等率",round(100*sum(case when b.s_score>=80 and b.s_score<90 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "优良率",round(100*sum(case when b.s_score>=90 then 1 else 0 end)/sum(case when b.s_score then 1 else 0 end),2) as "优秀率"
FROMcourse a,score b
WHEREa.c_id=b.c_id
group bya.c_id

5.17 统计各科成绩各分数段人数

  • 课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]个数及所占百分比
  • 方法一：if函数
    SELECTb.c_id,a.c_name,round(100*sum(if(b.s_score>85 and b.s_score<=100,1,0))/count(1),2) as "[100-85]百分比",sum(if(b.s_score>85 and b.s_score<=100,1,0)) as "[100-85]",round(100*sum(if(b.s_score>70 and b.s_score<=85,1,0))/count(1),2) as "[85-70]百分比",sum(if(b.s_score>70 and b.s_score<=85,1,0)) as "[85-70]",round(100*sum(if(b.s_score>60 and b.s_score<=70,1,0))/count(1),2) as "[70-60]百分比",sum(if(b.s_score>60 and b.s_score<=70,1,0)) as "[70-60]",round(100*sum(if(b.s_score>0 and b.s_score<=60,1,0))/count(1),2) as "[0-60]百分比",sum(if(b.s_score>=0 and b.s_score<=60,1,0)) as "[0-60]"
FROMcourse a,score b
WHEREa.c_id=b.c_id
group byb.c_id

5.18 查询不同老师所教不同课程平均分从高到低显示

  • 方法一：表连接
    SELECTc.t_name,a.c_name,round(avg(b.s_score),2) as avg_score
FROMcourse a
left JOINscore b
ONa.c_id=b.c_id
LEFT JOINteacher c
ONa.t_id=c.t_id
group byc.t_name,a.c_name
order byavg_score DESC

5.19 查询每门课程被选修的学生数

SELECTa.c_id,a.c_name,count(1) as cnt
FROMcourse a
LEFT JOINscore b
ON	a.c_id=b.c_id
group bya.c_id

5.20 查询出只有两门课程的全部学生的学号和姓名

  • 方法一:连表
    SELECTdistinct a.s_id,a.s_name
FROMstudent a,score b
WHEREa.s_id=b.s_id
group bya.s_id
HAVINGcount(b.c_id)=2
  • 方法二：条件查询
    select s_id,s_name 
from student 
where s_id in (select s_id from score GROUP BY s_id HAVING COUNT(c_id)=2);

5.21 查询同名同性学生名单，并统计同名人数

  • 方法一:分组条件查询
    SELECTs_name,count(1) as "人数"
FROMstudent
group bys_name,s_sex
HAVINGcount(1)>1
  • 方法二:自连接(同列比较可以用自连接）
    select a.s_name,a.s_sex,count(*) 
from student a  
JOIN student b 
on a.s_id !=b.s_id 
and a.s_name = b.s_name 
and a.s_sex = b.s_sex
GROUP BY a.s_name,a.s_sex

5.22 查询每门课程的平均成绩

  • 结果按平均成绩降序排列，平均成绩相同时，按课程编号升序排列
    SELECTa.c_id,round(avg(a.s_score),2) as avg_score
FROMscore a
group bya.c_id
order byavg_score desc,a.c_id asc

5.23 查询平均成绩大于等于85的所有学生的学号、姓名和平均成绩

  • 方法一：子查询
    SELECTa.s_id,a.s_name,t.avg_score
FROMstudent a,(SELECTa.s_id,round(avg(a.s_score),2) as avg_scoreFROMscore agroup bya.s_idHAVINGavg_score>=85) t
WHERE	a.s_id=t.s_id
ANDt.avg_score is not null
  • 方法二：表连接
    selecta.s_id,b.s_name,ifnull(round(avg(a.s_score),2),0) as avg_score
FROMscore a
LEFT JOINstudent b
ONa.s_id=b.s_id
GROUP BYa.s_id
HAVINGavg_score>=85

5.24 查询课程名称为"数学"，且分数低于60的学生姓名和分数

  • 方法一：条件查询+子查询
    SELECTb.s_name,a.s_score
FROMscore a
LEFT JOINstudent b
ONa.s_id=b.s_id
WHEREc_id in (SELECT c_id FROM course where c_name="数学")
ANDa.s_score<60
  • 方法二：多表连接
    SELECTb.s_name,a.s_score
FROMscore a
LEFT JOINstudent b
ONa.s_id=b.s_id
LEFT JOINcourse c
ONa.c_id=c.c_id
WHEREc.c_name="数学"
ANDa.s_score<60

5.25 查询所有学生的课程及分数情况

  • 方法一：表连接
    SELECTa.s_name,c.c_name,b.s_score
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
LEFT JOINcourse c
ONc.c_id=b.c_id
  • 方法二：if函数
    SELECTa.s_id,a.s_name,sum(if(c.c_name="语文",b.s_score,0)) as "语文",sum(if(c.c_name="数学",b.s_score,0)) as "数学",sum(if(c.c_name="英语",b.s_score,0)) as "英语",sum(b.s_score) as "总分"
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
LEFT JOINcourse c
ONc.c_id=b.c_id
group bya.s_id,a.s_name
  • 方法三：case函数
    selecta.s_id,a.s_name,sum(case when c.c_name="语文" then b.s_score else 0 end) as "语文",sum(case when c.c_name="数学" then b.s_score else 0 end) as "数学",sum(case when c.c_name="英语" then b.s_score else 0 end) as "英语",sum(b.s_score) as "总分"
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
LEFT JOINcourse c
ONc.c_id=b.c_id
group bya.s_id,a.s_name

5.26 查询任何一门课程成绩在70分以上的姓名、课程名称和分数(学生的每门课都大于70)

  • 方法一：表连接+子查询
    SELECTa.s_name,c.c_name,b.s_scoreFROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
LEFT JOINcourse c
ONc.c_id=b.c_id
WHEREa.s_id in (select s_id from score group by s_id having min(s_score)>70);

5.27 查询不及格的课程

  • 方法一：表连接
SELECT
distinctb.s_id,b.c_id,a.c_name,b.s_score
fromcourse a
LEFT JOINscore b
ONa.c_id=b.c_id
WHEREb.s_score<60

5.28 查询课程编号为01且课程成绩在80分以上的学生的学号和姓名

  • 方法一：子查询
    SELECTt.s_id,t.s_name
FROMstudent t
WHEREt.s_id in(SELECTa.s_idFROMscore aWHEREa.c_id="01" ANDa.s_score>80)
  • 方法二：表连接
    selecta.s_id,a.s_name
fromstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
WHEREb.c_id="01"
ANDb.s_score>80

5.29 求每门课程的学生人数

SELECTa.c_name,count(1) as "人数"
FROMcourse a
LEFT JOINscore b
ONa.c_id=b.c_id
group bya.c_id

5.30 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及其成绩

  • 方法一：表连接+子查询
    SELECTa.*,b.s_score as max_score,b.c_id,c.c_name
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
LEFT JOINcourse c
ONc.c_id=b.c_id
WHERE
-- 查询idb.c_id in (SELECTc_idFROMcourse WHEREt_id in (select t_id from teacher where t_name="张三"))
AND
-- 查询最大分数b.s_score=(select distinct max(s_score) from score where c_id="02")
  • 方法二：表连接
     SELECTa.*,b.s_score as max_score,b.c_id,c.c_name
FROMstudent a
LEFT JOINscore b
ONa.s_id=b.s_id
LEFT JOINcourse c
ONc.c_id=b.c_id
LEFT JOINteacher d
ONd.t_id=c.t_id
WHEREd.t_name="张三"
order bymax_score desc
limit 1;

5.31 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

SELECT
distincta.*
FROMscore a,score b
WHEREa.c_id!=b.c_id
ANDa.s_score=b.s_score

5.32 统计每门课程的学生选修人数（超过5人的课程才统计）

  • 要求输出课程号和选修人数，查询结果按人数降序排列，若人数相同，按课程号升序排列
  • 方法一： 分组聚合
    SELECTc_id,count(1) as "选修人数"
FROMscore
group byc_id
HAVINGcount(1) >5
order by"选修人数" desc,c_id asc
  • 方法二：连表+分组聚合
    SELECTa.c_id,count(b.s_id) cnt
FROMcourse a
LEFT JOINscore b
ONa.c_id=b.c_id
group bya.c_id
HAVINGcount(b.s_id)>5
order bycnt desc,a.c_id asc

5.33 检索至少选修两门课程的学生学号

SELECTs_id
FROMscore
group bys_id
HAVINGcount(c_id)>=2;

5.34 查询选修了全部课程的学生信息

  • 方法一：连表查询
    SELECTa.*
FROMstudent a,score b
WHEREa.s_id=b.s_id
group bys_id
HAVINGcount(1)=(select count(1) from course)
  • 方法二：子查询
    SELECT*
FROMstudent a
WHEREa.s_id in(select s_idFROMscoregroup bys_idHAVING	count(1)=(select count(1) from course))

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