SQL练习
1、表构造
–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, 老师编号 –3.老师表 Teacher(t_id,t_name) –老师编号,老师姓名 –4.成绩表 Score(s_id,c_id,s_score) –学生编号,课程编号,分数2、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`)) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- 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`)) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- 老师表CREATE TABLE `Teacher`( `t_id` VARCHAR(20), `t_name` VARCHAR(20) NOT NULL DEFAULT '', PRIMARY KEY(`t_id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- 成绩表CREATE TABLE `Score`( `s_id` VARCHAR(20), `c_id` VARCHAR(20), `s_score` INT(3), PRIMARY KEY(`s_id`,`c_id`)) ENGINE=InnoDB DEFAULT CHARSET=utf8;-- 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');-- 老师表测试数据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);3、测试题
-- 1、查问"01"课程比"02"课程问题高的学生的信息及课程分数 select a.* ,b.s_score as 01_score,c.s_score as 02_score from Student a join Score b on a.s_id=b.s_id and b.c_id='01' left join Score c on a.s_id=c.s_id and c.c_id='02' or c.c_id = NULL where b.s_score>c.s_score -- 2、查问"01"课程比"02"课程问题低的学生的信息及课程分数 select a.* ,b.s_score as 01_score,c.s_score as 02_score from student a left join score b on a.s_id=b.s_id and b.c_id='01' or b.c_id=NULL join score c on a.s_id=c.s_id and c.c_id='02' where b.s_score<c.s_score-- 3、查问均匀问题大于等于60分的同学的学生编号和学生姓名和均匀问题select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from student b join score a on b.s_id = a.s_id GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)>=60; -- 4、查问均匀问题小于60分的同学的学生编号和学生姓名和均匀问题 -- (包含有问题的和无问题的) select b.s_id,b.s_name,ROUND(AVG(a.s_score),2) as avg_score from student b left join score a on b.s_id = a.s_id GROUP BY b.s_id,b.s_name HAVING ROUND(AVG(a.s_score),2)<60 unionselect a.s_id,a.s_name,0 as avg_score from student a where a.s_id not in ( select distinct s_id from score);-- 5、查问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 sum_course,sum(b.s_score) as sum_score from student a left join score b on a.s_id=b.s_id GROUP BY a.s_id,a.s_name;-- 6、查问"李"姓老师的数量 select count(t_id) from teacher where t_name like '李%';-- 7、查问学过"张三"老师授课的同学的信息 select a.* from student a join score b on a.s_id=b.s_id where b.c_id in( select c_id from course where t_id =( select t_id from teacher where t_name = '张三'));-- 8、查问没学过"张三"老师授课的同学的信息 select * from student c where c.s_id not in( select a.s_id from student a join score b on a.s_id=b.s_id where b.c_id in( select c_id from course where t_id =( select t_id from teacher where t_name = '张三')));-- 9、查问学过编号为"01"并且也学过编号为"02"的课程的同学的信息 select a.* from student a,score b,score c where a.s_id = b.s_id and a.s_id = c.s_id and b.c_id='01' and c.c_id='02'; -- 10、查问学过编号为"01"然而没有学过编号为"02"的课程的同学的信息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')-- 11、查问没有学全所有课程的同学的信息 select s.* from student s where s.s_id in( select s_id from score where s_id not in( select a.s_id from score a join score b on a.s_id = b.s_id and b.c_id='02' join score c on a.s_id = c.s_id and c.c_id='03' where a.c_id='01'))-- 12、查问至多有一门课与学号为"01"的同学所学雷同的同学的信息 select * from student where 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') );-- 13、查问和"01"号的同学学习的课程完全相同的其他同学的信息 select a.* from student a where a.s_id in( select distinct s_id from score where s_id!='01' and c_id in(select c_id from score where s_id='01') group by s_id having count(1)=(select count(1) from score where s_id='01'));-- 14、查问没学过"张三"老师讲授的任一门课程的学生姓名 select a.s_name from student a where a.s_id not in ( select s_id from score where c_id = (select c_id from course where t_id =( select t_id from teacher where t_name = '张三')) group by s_id);-- 15、查问两门及其以上不及格课程的同学的学号，姓名及其均匀问题 select a.s_id,a.s_name,ROUND(AVG(b.s_score)) from student a left join score b on a.s_id = b.s_id where a.s_id in( select s_id from score where s_score<60 GROUP BY s_id having count(1)>=2) GROUP BY a.s_id,a.s_name -- 16、检索"01"课程分数小于60，按分数降序排列的学生信息select a.*,b.c_id,b.s_score from student a,score b where a.s_id = b.s_id and b.c_id='01' and b.s_score<60 ORDER BY b.s_score DESC;-- 17、按均匀问题从高到低显示所有学生的所有课程的问题以及均匀问题select a.s_id,(select s_score from score where s_id=a.s_id and c_id='01') as 语文, (select s_score from score where s_id=a.s_id and c_id='02') as 数学, (select s_score from score where s_id=a.s_id and c_id='03') as 英语, round(avg(s_score),2) as 平均分 from score a GROUP BY a.s_id ORDER BY 平均分 DESC;-- 18.查问各科问题最高分、最低分和平均分：以如下模式显示：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，低劣为：80-90，优良为：>=90select a.c_id,b.c_name,MAX(s_score),MIN(s_score),ROUND(AVG(s_score),2), ROUND(100*(SUM(case when a.s_score>=60 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 及格率, ROUND(100*(SUM(case when a.s_score>=70 and a.s_score<=80 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 中等率, ROUND(100*(SUM(case when a.s_score>=80 and a.s_score<=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优良率, ROUND(100*(SUM(case when a.s_score>=90 then 1 else 0 end)/SUM(case when a.s_score then 1 else 0 end)),2) as 优秀率 from score a left join course b on a.c_id = b.c_id GROUP BY a.c_id,b.c_name-- 19、按各科问题进行排序，并显示排名(实现不齐全)-- mysql没有rank函数 select a.s_id,a.c_id, @i:=@i +1 as i保留排名, @k:=(case when @score=a.s_score then @k else @i end) as rank不保留排名, @score:=a.s_score as score from ( select s_id,c_id,s_score from score WHERE c_id='01' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC)a,(select @k:=0,@i:=0,@score:=0)s union select a.s_id,a.c_id, @i:=@i +1 as i, @k:=(case when @score=a.s_score then @k else @i end) as rank, @score:=a.s_score as score from ( select s_id,c_id,s_score from score WHERE c_id='02' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC)a,(select @k:=0,@i:=0,@score:=0)s union select a.s_id,a.c_id, @i:=@i +1 as i, @k:=(case when @score=a.s_score then @k else @i end) as rank, @score:=a.s_score as score from ( select s_id,c_id,s_score from score WHERE c_id='03' GROUP BY s_id,c_id,s_score ORDER BY s_score DESC)a,(select @k:=0,@i:=0,@score:=0)s-- 20、查问学生的总成绩并进行排名select a.s_id, @i:=@i+1 as i, @k:=(case when @score=a.sum_score then @k else @i end) as rank, @score:=a.sum_score as scorefrom (select s_id,SUM(s_score) as sum_score from score GROUP BY s_id ORDER BY sum_score DESC)a, (select @k:=0,@i:=0,@score:=0)s-- 21、查问不同老师所教不同课程平均分从高到低显示 select a.t_id,c.t_name,a.c_id,ROUND(avg(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 a.c_id,a.t_id,c.t_name ORDER BY avg_score DESC;-- 22、查问所有课程的问题第2名到第3名的学生信息及该课程问题 select d.*,c.排名,c.s_score,c.c_id from ( select a.s_id,a.s_score,a.c_id,@i:=@i+1 as 排名 from score a,(select @i:=0)s where a.c_id='01' )c left join student d on c.s_id=d.s_id where 排名 BETWEEN 2 AND 3 UNION select d.*,c.排名,c.s_score,c.c_id from ( select a.s_id,a.s_score,a.c_id,@j:=@j+1 as 排名 from score a,(select @j:=0)s where a.c_id='02' )c left join student d on c.s_id=d.s_id where 排名 BETWEEN 2 AND 3 UNION select d.*,c.排名,c.s_score,c.c_id from ( select a.s_id,a.s_score,a.c_id,@k:=@k+1 as 排名 from score a,(select @k:=0)s where a.c_id='03' )c left join student d on c.s_id=d.s_id where 排名 BETWEEN 2 AND 3;-- 23、统计各科问题各分数段人数：课程编号,课程名称,[100-85],[85-70],[70-60],[0-60]及所占百分比 select distinct f.c_name,a.c_id,b.`85-100`,b.百分比,c.`70-85`,c.百分比,d.`60-70`,d.百分比,e.`0-60`,e.百分比 from score a left join (select c_id,SUM(case when s_score >85 and s_score <=100 then 1 else 0 end) as `85-100`, ROUND(100*(SUM(case when s_score >85 and s_score <=100 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)b on a.c_id=b.c_id left join (select c_id,SUM(case when s_score >70 and s_score <=85 then 1 else 0 end) as `70-85`, ROUND(100*(SUM(case when s_score >70 and s_score <=85 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)c on a.c_id=c.c_id left join (select c_id,SUM(case when s_score >60 and s_score <=70 then 1 else 0 end) as `60-70`, ROUND(100*(SUM(case when s_score >60 and s_score <=70 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)d on a.c_id=d.c_id left join (select c_id,SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end) as `0-60`, ROUND(100*(SUM(case when s_score >=0 and s_score <=60 then 1 else 0 end)/count(*)),2) as 百分比 from score GROUP BY c_id)e on a.c_id=e.c_id left join course f on a.c_id = f.c_id-- 24、查问学生均匀问题及其名次 select a.s_id, @i:=@i+1 as '不保留空缺排名', @k:=(case when @avg_score=a.avg_s then @k else @i end) as '保留空缺排名', @avg_score:=avg_s as '平均分' from (select s_id,ROUND(AVG(s_score),2) as avg_s from score GROUP BY s_id)a,(select @avg_score:=0,@i:=0,@k:=0)b;-- 25、查问各科问题前三名的记录 -- 1.选出b表比a表问题大的所有组 -- 2.选出比以后id问题大的 小于三个的 select a.s_id,a.c_id,a.s_score from score a left join score b on a.c_id = b.c_id and a.s_score<b.s_score group by a.s_id,a.c_id,a.s_score HAVING COUNT(b.s_id)<3 ORDER BY a.c_id,a.s_score DESC -- 26、查问每门课程被选修的学生数 select c_id,count(s_id) from score a GROUP BY c_id -- 27、查问出只有两门课程的全副学生的学号和姓名 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); -- 28、查问男生、女生人数 select s_sex,COUNT(s_sex) as 人数 from student GROUP BY s_sex-- 29、查问名字中含有"风"字的学生信息 select * from student where s_name like '%风%';-- 30、查问同名同性学生名单，并统计同名人数 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-- 31、查问1990年出世的学生名单 select s_name from student where s_birth like '1990%' -- 32、查问每门课程的均匀问题，后果按均匀问题降序排列，均匀问题雷同时，按课程编号升序排列 select c_id,ROUND(AVG(s_score),2) as avg_score from score GROUP BY c_id ORDER BY avg_score DESC,c_id ASC-- 33、查问均匀问题大于等于85的所有学生的学号、姓名和均匀问题 select a.s_id,b.s_name,ROUND(avg(a.s_score),2) as avg_score from score a left join student b on a.s_id=b.s_id GROUP BY s_id HAVING avg_score>=85-- 34、查问课程名称为"数学"，且分数低于60的学生姓名和分数 select a.s_name,b.s_score from score b LEFT JOIN student a on a.s_id=b.s_id where b.c_id=( select c_id from course where c_name ='数学') and b.s_score<60 -- 35、查问所有学生的课程及分数状况； select a.s_id,a.s_name, SUM(case c.c_name when '语文' then b.s_score else 0 end) as '语文', SUM(case c.c_name when '数学' then b.s_score else 0 end) as '数学', SUM(case c.c_name when '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 b.c_id = c.c_id GROUP BY a.s_id,a.s_name -- 36、查问任何一门课程问题在70分以上的姓名、课程名称和分数； select a.s_name,b.c_name,c.s_score from course b left join score c on b.c_id = c.c_id left join student a on a.s_id=c.s_id where c.s_score>=70 -- 37、查问不及格的课程 select a.s_id,a.c_id,b.c_name,a.s_score from score a left join course b on a.c_id = b.c_id where a.s_score<60 -- 38、查问课程编号为01且课程问题在80分以上的学生的学号和姓名； select a.s_id,b.s_name from score a LEFT JOIN student b on a.s_id = b.s_id where a.c_id = '01' and a.s_score>80-- 39、求每门课程的学生人数 select count(*) from score GROUP BY c_id; -- 40、查问选修"张三"老师所授课程的学生中，问题最高的学生信息及其问题 -- 查问老师id select c_id from course c,teacher d where c.t_id=d.t_id and d.t_name='张三' -- 查问最高分（可能有雷同分数） select MAX(s_score) from score where c_id='02' -- 查问信息 select a.*,b.s_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 b.c_id=c.c_id where b.c_id =(select c_id from Course c,Teacher d where c.t_id=d.t_id and d.t_name='张三') and b.s_score in (select MAX(s_score) from Score where c_id=(select c_id from Course c,Teacher d where c.t_id=d.t_id and d.t_name='张三'))-- 41、查问不同课程问题雷同的学生的学生编号、课程编号、学生问题 select DISTINCT b.s_id,b.c_id,b.s_score from score a,score b where a.c_id != b.c_id and a.s_score = b.s_score-- 42、查问每门功问题最好的前两名 -- 牛逼的写法 select a.s_id,a.c_id,a.s_score from Score a where (select COUNT(1) from Score b where b.c_id=a.c_id and b.s_score>=a.s_score)<=2 ORDER BY a.c_id-- 43、统计每门课程的学生选修人数（超过5人的课程才统计）。要求输入课程号和选修人数，查问后果按人数降序排列，若人数雷同，按课程号升序排列 select c_id,count(*) as total from score GROUP BY c_id HAVING total>5 ORDER BY total,c_id ASC-- 44、检索至多选修两门课程的学生学号 select s_id,count(*) as sel from score GROUP BY s_id HAVING sel>=2-- 45、查问选修了全副课程的学生信息 select * from student where s_id in( select s_id from score GROUP BY s_id HAVING count(*)=(select count(*) from course))-- 46、查问各学生的年龄 -- 依照出生日期来算，以后月日 < 出生年月的月日则，年龄减一 select s_birth,(DATE_FORMAT(NOW(),'%Y')-DATE_FORMAT(s_birth,'%Y') - (case when DATE_FORMAT(NOW(),'%m%d')>DATE_FORMAT(s_birth,'%m%d') then 0 else 1 end)) as age from student;-- 47、查问本周过生日的学生 select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))=WEEK(s_birth) select * from student where YEARWEEK(s_birth)=YEARWEEK(DATE_FORMAT(NOW(),'%Y%m%d')) select WEEK(DATE_FORMAT(NOW(),'%Y%m%d')) -- 48、查问下周过生日的学生 select * from student where WEEK(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =WEEK(s_birth) -- 49、查问本月过生日的学生 select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d')) =MONTH(s_birth)-- 50、查问下月过生日的学生 select * from student where MONTH(DATE_FORMAT(NOW(),'%Y%m%d'))+1 =MONTH(s_birth)练习题问题
-- 为什么用left join而且加判断 OR c.c_id = NULL? 右表数据为空为什么思考？
总结
思路
- 表连贯就是先join失去一个表，而后再join条件失去另一个表
- 个别应用left join，过滤用join
- 用left join时，关联条件须要判断右表数据有没有为NULL的