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sql常见50道查询练习题
- 1. 表创建
- 2. 简单查询例题(3题）
- 3. 日期相关例题(6题）
- 4. 开窗函数查询（7题）
- 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
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
SELECT count(1) as cnt FROM teacher WHERE t_name like "李%" 2.2 查询男生、女生人数
SELECT s.s_sex, count(1) as 人数 FROM student s group by s.s_sex 2.3 查询名字中含有"风"字的学生信息
SELECT * FROM student WHERE s_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 select a.*, year(NOW())-year(a.s_brith)-if(DATE_FORMAT(now(),"%m%d") >DATE_FORMAT(a.s_brith,"%m%d"),0,1) as age FROM student 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 * FROM student WHERE WEEKOFYEAR(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 * FROM student WHERE WEEKOFYEAR(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 * FROM student WHERE MONTH(now())=month(s_brith) 3.5 查询下月过生日的学生
SELECT * FROM student WHERE MONTH(now()+interval "1" month)=month(s_brith) 3.6 查询1990年出生的学生名单
SELECT * FROM student WHERE s_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
select a.*, avg(a.s_score) over(PARTITION by a.s_id) as avg_score FROM score a - 方法二：临时表连接
SELECT a.*, t.avg_score FROM score a, (SELECT a.s_id, round(avg(a.s_score),2) as avg_score FROM score a group by a.s_id) t WHERE a.s_id=t.s_id order by t.avg_score desc - 方法三:长型数据转为宽型数据
SELECT a.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 FROM score a group by a.s_id order by ifnull(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
SELECT a.*, 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 FROM score 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 FROM score a order by c_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
SELECT t.*, rank() over(order by sum_score desc) rank排名 FROM (SELECT s_id, sum(s_score) as sum_score FROM score group by s_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
SELECT a.*, t.c_id, t.rk, t.s_score FROM student a, (SELECT a.s_id, a.c_id, a.s_score, dense_rank() over(PARTITION by a.c_id order by a.s_score desc) as rk FROM score a) t WHERE t.rk in (2,3) AND a.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
SELECT t.*, rank() over(order by t.avg_score desc) 排名 FROM (SELECT a.s_id, round(avg(a.s_score),2) as avg_score FROM score a group by a.s_id) t
4.6 查询各科成绩前三名的记录
- 方法一：开窗函数
SELECT t.* from (SELECT a.c_id, a.c_name, b.s_score, rank() over(PARTITION by a.c_id order by b.s_score desc) rk FROM course a LEFT JOIN score b ON a.c_id=b.c_id) t WHERE t.rk<=3; - 方法二：子查询
SELECT * from ( SELECT a.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 rk FROM course a LEFT JOIN score b ON a.c_id=b.c_id) t WHERE t.rk<=3 order by t.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
SELECT t.s_id, t.c_id, t.s_score FROM (SELECT *, rank() over(PARTITION by b.c_id order by b.s_score desc) rk FROM score b) t WHERE t.rk<=2; - 方法二：自连接
SELECT t.s_id, t.c_id, t.s_score FROM (SELECT a.*, (select count(1) from score b where b.c_id=a.c_id and a.s_score<b.s_score)+1 as rk FROM score a order by a.c_id,rk) t WHERE t.rk<=2 - 方法三：条件查询+子查询
SELECT a.* FROM score 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 by a.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 FROM student st, (SELECT s1.s_id,s1.s_score as sc1,s2.s_score as sc2 FROM score s1,score s2 WHERE s1.c_id="01" AND s2.c_id="02" AND s1.s_id=s2.s_id AND s1.s_score>s2.s_score) t1 WHERE st.s_id=t1.s_id; - 方法二：表连接
SELECT st.*,s1.s_score as sc1,s2.s_score as sc2 FROM student st left JOIN score s1 ON s1.s_id=st.s_id left JOIN score s2 ON s2.s_id=st.s_id WHERE s1.c_id="01" AND s2.c_id="02" AND s1.s_id=s2.s_id AND s1.s_score>s2.s_score - 数据长型数据变为宽型数据
-- IF函数或case函数 SELECT a.*, t.s01, t.s02 from student a, (SELECT a.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 s02 from score a group by a.s_id) t WHERE a.s_id=t.s_id AND t.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 FROM student st, (SELECT s1.s_id,s1.s_score as sc1,s2.s_score as sc2 FROM score s1,score s2 WHERE s1.c_id="01" AND s2.c_id="02" AND s1.s_id=s2.s_id AND s1.s_score<s2.s_score) t1 WHERE st.s_id=t1.s_id; - 方法二：表连接
SELECT st.*,s1.s_score as sc1,s2.s_score as sc2 FROM student st left JOIN score s1 ON s1.s_id=st.s_id left JOIN score s2 ON s2.s_id=st.s_id WHERE s1.c_id="01" AND s2.c_id="02" AND s1.s_id=s2.s_id AND s1.s_score<s2.s_score -- 方法二 SELECT st.*,s1.s_score as sc1,s2.s_score as sc2 FROM student st left JOIN score s1 ON s1.s_id=st.s_id AND s1.c_id="01" left JOIN score s2 ON s2.s_id=st.s_id AND s2.c_id="02" AND s1.s_id=s2.s_id WHERE s1.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 SELECT a.*, t.s01, t.s02 from student a, (SELECT a.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 s02 from score a group by a.s_id) t WHERE t.s01<t.s02 AND a.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 FROM student ST, (SELECT s.s_id,round(avg(s.s_score),2) as avg_s FROM score s GROUP BY s.s_id HAVING round(avg(s.s_score),2)>=60) t WHERE st.s_id=t.s_id -- 方法二：子查询二 SELECT s.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 BY s.s_id HAVING avg_s>=60 - 方法二：表连接
SELECT a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score FROM student a LEFT JOIN score b ON a.s_id=b.s_id GROUP BY a.s_id HAVING round(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 SELECT a.s_id,a.s_name,t.avg_acore FROM student a, (SELECT a.s_id,round(avg(a.s_score),2) as avg_acore FROM score a GROUP BY a.s_id HAVING round(avg(a.s_score),2)<60) t WHERE a.s_id=t.s_id UNION -- 没有成绩的：没有成绩的s_id不存在 SELECT a.s_id,a.s_name,0 as avg_acore FROM student a WHERE a.s_id not in (SELECT DISTINCT s_id FROM score); - 方法二：表连接
SELECT a.s_id,a.s_name,ifnull(round(avg(b.s_score),2),0) as avg_score FROM student a LEFT JOIN score b on a.s_id=b.s_id GROUP BY a.s_id HAVING avg_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
SELECT a.s_id, a.s_name, count(b.c_id) as cnt_course, ifnull(sum(b.s_score),0) as sum_score FROM student a LEFT JOIN score b ON a.s_id=b.s_id group by a.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
SELECT a.* FROM student a LEFT JOIN score b on a.s_id=b.s_id LEFT JOIN course c ON b.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
SELECT a.* FROM student a LEFT JOIN score b ON a.s_id=b.s_id WHERE b.c_id in ( SELECT c_id FROM course where t_id in(SELECT t_id from teacher where t_name="张三") ); - 方法三：多表连接
select a.* from student a,score b,course c,teacher d WHERE a.s_id=b.s_id AND b.c_id=c.c_id AND c.t_id=d.t_id AND d.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
SELECT s.* FROM student s WHERE s.s_id NOT IN ( -- 查找学的学生 SELECT DISTINCT a.s_id FROM student a LEFT JOIN score b ON a.s_id = b.s_id WHERE b.c_id IN ( -- 查找学过的课程 SELECT c_id FROM course WHERE 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 * FROM student s WHERE s.s_id not in ( select a.s_id from score a, course b, teacher c WHERE a.c_id=b.c_id AND b.t_id=c.t_id AND c.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 * FROM student s WHERE s.s_id in( SELECT a.s_id FROM score a,score b WHERE a.c_id="01" AND b.c_id="02" AND a.s_id=b.s_id) - 方法二：连表+自连接,同列对比可以用自连接
SELECT s.* FROM student s LEFT JOIN score a ON s.s_id=a.s_id LEFT JOIN score b ON a.s_id=b.s_id WHERE a.c_id="01" AND b.c_id="02" - 方法三：条件查询+子查询
SELECT * FROM student WHERE s_id in (SELECT s_id FROM score where c_id="01" or c_id="02" GROUP BY s_id HAVING count(1)=2) - 方法四：自连接，条件连接
SELECT s.* FROM student s,score a,score b WHERE s.s_id=a.s_id AND a.s_id=b.s_id AND a.c_id="01" AND b.c_id="02" - 方法五：子查询+数据长型数据变为宽型数据
SELECT a.* FROM student a, (select a.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 s02 from score a group by a.s_id) t WHERE a.s_id=t.s_id AND t.s01>0 AND t.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
select a.* from student a WHERE a.s_id in(select s_id from score where c_id="01") AND a.s_id not in (select s_id from score where c_id="02") - 方法二： 子查询+分组聚合
SELECT s.* FROM student s, ( SELECT a.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) s02 FROM score a group by a.s_id) t WHERE s.s_id=t.s_id AND t.s01 is not NULL AND t.s02 is null - 方法三：数据长型数据变为宽型数据
SELECT a.* FROM student a, (select a.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 s02 from score a group by a.s_id) t WHERE a.s_id=t.s_id AND t.s01 is not null AND t.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
SELECT s.* FROM student s WHERE s.s_id in( SELECT a.s_id FROM score a group by a.s_id having count(1)<(select count(1) from course)) - 方法二:表连接
SELECT s.*, count(a.c_id) cnt FROM student s LEFT JOIN score a ON a.s_id=s.s_id group by s.s_id HAVING count(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
SELECT s.* FROM student s WHERE s.s_id in( SELECT distinct a.s_id FROM score a WHERE a.c_id in( SELECT b.c_id FROM score b WHERE b.s_id="01")) AND s.s_id!='01' - 方法二:表连接+子查询
SELECT a.* FROM student a LEFT JOIN score b on a.s_id=b.s_id WHERE b.c_id in ( SELECT b.c_id FROM score b WHERE b.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
SELECT s.* FROM student s WHERE s.s_id in( SELECT distinct a.s_id FROM score a WHERE a.c_id in( SELECT a.c_id FROM score a WHERE a.s_id="01") AND a.s_id!="01" group by a.s_id HAVING count(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
SELECT s.s_name FROM student s WHERE s.s_id not in( SELECT a.s_id FROM score a WHERE a.c_id in ( SELECT a.c_id FROM course a WHERE a.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
SELECT a.s_id, a.s_name, round(avg(b.s_score),2) as avg_score FROM student a LEFT JOIN score b ON a.s_id=b.s_id group by a.s_id having sum(if(b.s_score>=60,0,1))>=2 - 方法二：自连接+子查询
select a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score FROM student a,score b WHERE a.s_id=b.s_id AND a.s_id in( SELECT a.s_id FROM score a WHERE a.s_score<60 group by a.s_id HAVING count(1)>=2) group by a.s_id - 方法三：表连接+子查询
select a.s_id,a.s_name,round(avg(b.s_score),2) as avg_score FROM student a LEFT JOIN score b on a.s_id=b.s_id where a.s_id in( SELECT a.s_id FROM score a WHERE a.s_score<60 group by a.s_id HAVING count(1)>=2) group by a.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
SELECT a.*,b.c_id,b.s_score FROM student a LEFT JOIN score b ON a.s_id=b.s_id WHERE b.c_id="01" and b.s_score<60 order by b.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
SELECT a.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" FROM course a, score b WHERE a.c_id=b.c_id group by a.c_id - 方法二：case when
SELECT a.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" FROM course a, score b WHERE a.c_id=b.c_id group by a.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
SELECT b.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]" FROM course a, score b WHERE a.c_id=b.c_id group by b.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
SELECT c.t_name, a.c_name, round(avg(b.s_score),2) as avg_score FROM course a left JOIN score b ON a.c_id=b.c_id LEFT JOIN teacher c ON a.t_id=c.t_id group by c.t_name,a.c_name order by avg_score DESC
5.19 查询每门课程被选修的学生数
SELECT a.c_id, a.c_name, count(1) as cnt FROM course a LEFT JOIN score b ON a.c_id=b.c_id group by a.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
SELECT distinct a.s_id,a.s_name FROM student a, score b WHERE a.s_id=b.s_id group by a.s_id HAVING count(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
SELECT s_name, count(1) as "人数" FROM student group by s_name,s_sex HAVING count(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
SELECT a.c_id, round(avg(a.s_score),2) as avg_score FROM score a group by a.c_id order by avg_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
SELECT a.s_id, a.s_name, t.avg_score FROM student a, (SELECT a.s_id, round(avg(a.s_score),2) as avg_score FROM score a group by a.s_id HAVING avg_score>=85) t WHERE a.s_id=t.s_id AND t.avg_score is not null - 方法二：表连接
select a.s_id, b.s_name, ifnull(round(avg(a.s_score),2),0) as avg_score FROM score a LEFT JOIN student b ON a.s_id=b.s_id GROUP BY a.s_id HAVING avg_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
SELECT b.s_name, a.s_score FROM score a LEFT JOIN student b ON a.s_id=b.s_id WHERE c_id in (SELECT c_id FROM course where c_name="数学") AND a.s_score<60 - 方法二：多表连接
SELECT b.s_name, a.s_score FROM score a LEFT JOIN student b ON a.s_id=b.s_id LEFT JOIN course c ON a.c_id=c.c_id WHERE c.c_name="数学" AND a.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
SELECT a.s_name,c.c_name,b.s_score FROM student a LEFT JOIN score b ON a.s_id=b.s_id LEFT JOIN course c ON c.c_id=b.c_id - 方法二：if函数
SELECT a.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" FROM student a LEFT JOIN score b ON a.s_id=b.s_id LEFT JOIN course c ON c.c_id=b.c_id group by a.s_id,a.s_name - 方法三：case函数
select a.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" FROM student a LEFT JOIN score b ON a.s_id=b.s_id LEFT JOIN course c ON c.c_id=b.c_id group by a.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
SELECT a.s_name, c.c_name, b.s_score FROM student a LEFT JOIN score b ON a.s_id=b.s_id LEFT JOIN course c ON c.c_id=b.c_id WHERE a.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 distinct b.s_id, b.c_id, a.c_name, b.s_score from course a LEFT JOIN score b ON a.c_id=b.c_id WHERE b.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
SELECT t.s_id, t.s_name FROM student t WHERE t.s_id in( SELECT a.s_id FROM score a WHERE a.c_id="01" AND a.s_score>80) - 方法二：表连接
select a.s_id, a.s_name from student a LEFT JOIN score b ON a.s_id=b.s_id WHERE b.c_id="01" AND b.s_score>80
5.29 求每门课程的学生人数
SELECT a.c_name, count(1) as "人数" FROM course a LEFT JOIN score b ON a.c_id=b.c_id group by a.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
SELECT a.*, b.s_score as max_score, b.c_id, c.c_name FROM student a LEFT JOIN score b ON a.s_id=b.s_id LEFT JOIN course c ON c.c_id=b.c_id WHERE -- 查询id b.c_id in ( SELECT c_id FROM course WHERE t_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
SELECT a.*, b.s_score as max_score, b.c_id, c.c_name FROM student a LEFT JOIN score b ON a.s_id=b.s_id LEFT JOIN course c ON c.c_id=b.c_id LEFT JOIN teacher d ON d.t_id=c.t_id WHERE d.t_name="张三" order by max_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 distinct a.* FROM score a, score b WHERE a.c_id!=b.c_id AND a.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
SELECT c_id, count(1) as "选修人数" FROM score group by c_id HAVING count(1) >5 order by "选修人数" desc,c_id asc - 方法二：连表+分组聚合
SELECT a.c_id, count(b.s_id) cnt FROM course a LEFT JOIN score b ON a.c_id=b.c_id group by a.c_id HAVING count(b.s_id)>5 order by cnt desc,a.c_id asc
5.33 检索至少选修两门课程的学生学号
SELECT s_id FROM score group by s_id HAVING count(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
SELECT a.* FROM student a, score b WHERE a.s_id=b.s_id group by s_id HAVING count(1)=(select count(1) from course) - 方法二：子查询
SELECT * FROM student a WHERE a.s_id in( select s_id FROM score group by s_id HAVING count(1)=(select count(1) from course))
