MySQL之多表查询
多表查询思路
1、子查询
将SQL语句查询的结果括号括起来当做另外一条SQL语句的条件
根据部门编号查询部门名称 select name from dep where id=(select dep_id from emp where name='jason');
2、连表查询
先将需要使用到的表拼接成一张大表 之后基于单表查询完成
inner join 内连接
left join 左连接
right join 右连接
union 全连接
# 涉及到多表查询的时候 字段名称容易冲突 需要使用表名点字段的方式区分 # inner join:只拼接两张表中共有的部分
select * from emp inner join dep on emp.dep_id = dep.id; # left join:以左表为基准展示所有的内容 没有的NULL填充
select * from emp left join dep on emp.dep_id = dep.id; # right join:以右表为基准展示所有的内容 没有的NULL填充
select * from emp right join dep on emp.dep_id = dep.id; # union:左右表所有的数据都在 没有的NULL填充
select * from emp left join dep on emp.dep_id = dep.id union select * from emp right join dep on emp.dep_id = dep.id;
多表查询练习题
1、数据准备
#建表 create table dep( id int primary key auto_increment, name varchar(20) ); create table emp( id int primary key auto_increment, name varchar(20), sex enum('male','female') not null default 'male', age int, dep_id int ); #插入数据 insert into dep values (200,'技术'), (201,'人力资源'), (202,'销售'), (203,'运营'), (205,'保洁') ; insert into emp(name,sex,age,dep_id) values ('jason','male',18,200), ('egon','female',48,201), ('kevin','male',18,201), ('nick','male',28,202), ('owen','male',18,203), ('jerry','female',18,204);
2、练习题
-- 1、查询所有的课程的名称以及对应的任课老师姓名 -- SELECT -- teacher.tname, -- course.cname -- FROM -- teacher -- INNER JOIN course ON teacher.tid = course.teacher_id; -- 4、查询平均成绩大于八十分的同学的姓名和平均成绩 # 1.先确定需要使用到的表 # 2.在思考多表查询的方式 # 第一步先查询成绩表中 平均成绩大于80的学生编号 # 1.1 按照学生id分组并获取平均成绩 -- select student_id,avg(num) from score group by student_id; # 1.2 筛选出平均成绩大于80的数据 (针对聚合函数的字段结果 最好起别名防止冲突) -- select student_id,avg(num) as avg_num from score group by student_id having avg(num) > 80; # 1.3 将上述SQL的结果与student表拼接 -- SELECT -- student.sname, -- t1.avg_num -- FROM -- student -- INNER JOIN ( SELECT student_id, avg( num ) AS avg_num FROM score GROUP BY student_id HAVING avg( num ) > 80 ) AS t1 ON student.sid = t1.student_id; -- 7、查询没有报李平老师课的学生姓名 # 1.先查询李平老师教授的课程编号 -- select course.cid from course where teacher_id = -- (select tid from teacher where tname ='李平老师'); # 2.根据课程id号筛选出所有报了的学生id号 -- select distinct score.student_id from score where course_id in (select course.cid from course where teacher_id = -- (select tid from teacher where tname ='李平老师')); # 3.去学生表中根据id号取反筛选学生姓名 -- SELECT -- student.sname -- FROM -- student -- WHERE -- sid NOT IN ( -- SELECT DISTINCT -- score.student_id -- FROM -- score -- WHERE -- course_id IN ( SELECT course.cid FROM course WHERE teacher_id = ( SELECT tid FROM teacher WHERE tname = '李平老师' ) ) -- ); -- 8、查询没有同时选修物理课程和体育课程的学生姓名(只要了报了一门的 两门和一门没报的都不要) # 1.先获取两门课程的id号 -- select course.cid from course where cname in ('物理','体育'); # 2.再去分数表中先筛选出所有报了物理和体育的学生id(两门 一门) -- select * from score where course_id in (select course.cid from course where cname in ('物理','体育')); # 3.如何筛选出只报了一门的学生id 按照学生id分组 然后计数 并过滤出计数结果为1的数据 -- select score.student_id from score where course_id in (select course.cid from course where cname in ('物理','体育')) -- group by score.student_id -- having count(score.course_id) = 1 -- ; # 4.根据学生id号去student表中筛选学生姓名 -- SELECT -- student.sname -- FROM -- student -- WHERE -- sid IN ( -- SELECT -- score.student_id -- FROM -- score -- WHERE -- course_id IN ( SELECT course.cid FROM course WHERE cname IN ( '物理', '体育' ) ) -- GROUP BY -- score.student_id -- HAVING -- count( score.course_id ) = 1 -- ); -- 9、查询挂科超过两门(包括两门)的学生姓名和班级 # 1.先筛选出小于60分的数据 -- select * from score where num < 60; # 2.按照学生id分组 然后统计挂科数量 -- select student_id,count(course_id) from score where num < 60 group by student_id; # 3.筛选出挂科超过两门的学生id -- select student_id from score where num < 60 group by student_id -- having count(course_id) >=2; # 4.先将上述结果放在一边 去连接student和class表 SELECT student.sname, class.caption FROM class INNER JOIN student ON class.cid = student.class_id WHERE student.sid IN ( SELECT student_id FROM score WHERE num < 60 GROUP BY student_id HAVING count( course_id ) >= 2 );
更多练习
题目
1、查询所有的课程的名称以及对应的任课老师姓名 2、查询学生表中男女生各有多少人 3、查询物理成绩等于100的学生的姓名 4、查询平均成绩大于八十分的同学的姓名和平均成绩 5、查询所有学生的学号,姓名,选课数,总成绩 6、 查询姓李老师的个数 7、 查询没有报李平老师课的学生姓名 8、 查询物理课程比生物课程高的学生的学号 9、 查询没有同时选修物理课程和体育课程的学生姓名 10、查询挂科超过两门(包括两门)的学生姓名和班级 、查询选修了所有课程的学生姓名 12、查询李平老师教的课程的所有成绩记录 13、查询全部学生都选修了的课程号和课程名 14、查询每门课程被选修的次数 15、查询之选修了一门课程的学生姓名和学号 16、查询所有学生考出的成绩并按从高到低排序(成绩去重) 17、查询平均成绩大于85的学生姓名和平均成绩 18、查询生物成绩不及格的学生姓名和对应生物分数 19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名 20、查询每门课程成绩最好的前两名学生姓名 21、查询不同课程但成绩相同的学号,课程号,成绩 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称; 23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名; 24、任课最多的老师中学生单科成绩最高的学生姓名
答案
#1、查询所有的课程的名称以及对应的任课老师姓名 SELECT course.cname, teacher.tname FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid; #2、查询学生表中男女生各有多少人 SELECT gender 性别, count(1) 人数 FROM student GROUP BY gender; #3、查询物理成绩等于100的学生的姓名 SELECT student.sname FROM student WHERE sid IN ( SELECT student_id FROM score INNER JOIN course ON score.course_id = course.cid WHERE course.cname = '物理' AND score.num = 100 ); #4、查询平均成绩大于八十分的同学的姓名和平均成绩 SELECT student.sname, t1.avg_num FROM student INNER JOIN ( SELECT student_id, avg(num) AS avg_num FROM score GROUP BY student_id HAVING avg(num) > 80 ) AS t1 ON student.sid = t1.student_id; #5、查询所有学生的学号,姓名,选课数,总成绩(注意:对于那些没有选修任何课程的学生也算在内) SELECT student.sid, student.sname, t1.course_num, t1.total_num FROM student LEFT JOIN ( SELECT student_id, COUNT(course_id) course_num, sum(num) total_num FROM score GROUP BY student_id ) AS t1 ON student.sid = t1.student_id; #6、 查询姓李老师的个数 SELECT count(tid) FROM teacher WHERE tname LIKE '李%'; #7、 查询没有报李平老师课的学生姓名(找出报名李平老师课程的学生,然后取反就可以) SELECT student.sname FROM student WHERE sid NOT IN ( SELECT DISTINCT student_id FROM score WHERE course_id IN ( SELECT course.cid FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid WHERE teacher.tname = '李平老师' ) ); #8、 查询物理课程比生物课程高的学生的学号(分别得到物理成绩表与生物成绩表,然后连表即可) SELECT t1.student_id FROM ( SELECT student_id, num FROM score WHERE course_id = ( SELECT cid FROM course WHERE cname = '物理' ) ) AS t1 INNER JOIN ( SELECT student_id, num FROM score WHERE course_id = ( SELECT cid FROM course WHERE cname = '生物' ) ) AS t2 ON t1.student_id = t2.student_id WHERE t1.num > t2.num; #9、 查询没有同时选修物理课程和体育课程的学生姓名(没有同时选修指的是选修了一门的,思路是得到物理+体育课程的学生信息表,然后基于学生分组,统计count(课程)=1) SELECT student.sname FROM student WHERE sid IN ( SELECT student_id FROM score WHERE course_id IN ( SELECT cid FROM course WHERE cname = '物理' OR cname = '体育' ) GROUP BY student_id HAVING COUNT(course_id) = 1 ); #10、查询挂科超过两门(包括两门)的学生姓名和班级(求出<60的表,然后对学生进行分组,统计课程数目>=2) SELECT student.sname, class.caption FROM student INNER JOIN ( SELECT student_id FROM score WHERE num < 60 GROUP BY student_id HAVING count(course_id) >= 2 ) AS t1 INNER JOIN class ON student.sid = t1.student_id AND student.class_id = class.cid; #11、查询选修了所有课程的学生姓名(先从course表统计课程的总数,然后基于score表按照student_id分组,统计课程数据等于课程总数即可) SELECT student.sname FROM student WHERE sid IN ( SELECT student_id FROM score GROUP BY student_id HAVING COUNT(course_id) = (SELECT count(cid) FROM course) ); #12、查询李平老师教的课程的所有成绩记录 SELECT * FROM score WHERE course_id IN ( SELECT cid FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid WHERE teacher.tname = '李平老师' ); #13、查询全部学生都选修了的课程号和课程名(取所有学生数,然后基于score表的课程分组,找出count(student_id)等于学生数即可) SELECT cid, cname FROM course WHERE cid IN ( SELECT course_id FROM score GROUP BY course_id HAVING COUNT(student_id) = ( SELECT COUNT(sid) FROM student ) ); #14、查询每门课程被选修的次数 SELECT course_id, COUNT(student_id) FROM score GROUP BY course_id; #15、查询之选修了一门课程的学生姓名和学号 SELECT sid, sname FROM student WHERE sid IN ( SELECT student_id FROM score GROUP BY student_id HAVING COUNT(course_id) = 1 ); #16、查询所有学生考出的成绩并按从高到低排序(成绩去重) SELECT DISTINCT num FROM score ORDER BY num DESC; #17、查询平均成绩大于85的学生姓名和平均成绩 SELECT sname, t1.avg_num FROM student INNER JOIN ( SELECT student_id, avg(num) avg_num FROM score GROUP BY student_id HAVING AVG(num) > 85 ) t1 ON student.sid = t1.student_id; #18、查询生物成绩不及格的学生姓名和对应生物分数 SELECT sname 姓名, num 生物成绩 FROM score LEFT JOIN course ON score.course_id = course.cid LEFT JOIN student ON score.student_id = student.sid WHERE course.cname = '生物' AND score.num < 60; #19、查询在所有选修了李平老师课程的学生中,这些课程(李平老师的课程,不是所有课程)平均成绩最高的学生姓名 SELECT sname FROM student WHERE sid = ( SELECT student_id FROM score WHERE course_id IN ( SELECT course.cid FROM course INNER JOIN teacher ON course.teacher_id = teacher.tid WHERE teacher.tname = '李平老师' ) GROUP BY student_id ORDER BY AVG(num) DESC LIMIT 1 ); #20、查询每门课程成绩最好的前两名学生姓名 #查看每门课程按照分数排序的信息,为下列查找正确与否提供依据 SELECT * FROM score ORDER BY course_id, num DESC; #表1:求出每门课程的课程course_id,与最高分数first_num SELECT course_id, max(num) first_num FROM score GROUP BY course_id; #表2:去掉最高分,再按照课程分组,取得的最高分,就是第二高的分数second_num SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id; #将表1和表2联合到一起,得到一张表t3,包含课程course_id与该们课程的first_num与second_num SELECT t1.course_id, t1.first_num, t2.second_num FROM ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t1 INNER JOIN ( SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id ) AS t2 ON t1.course_id = t2.course_id; #查询前两名的学生(有可能出现并列第一或者并列第二的情况) SELECT score.student_id, t3.course_id, t3.first_num, t3.second_num FROM score INNER JOIN ( SELECT t1.course_id, t1.first_num, t2.second_num FROM ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t1 INNER JOIN ( SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id ) AS t2 ON t1.course_id = t2.course_id ) AS t3 ON score.course_id = t3.course_id WHERE score.num >= t3.second_num AND score.num <= t3.first_num; #排序后可以看的明显点 SELECT score.student_id, t3.course_id, t3.first_num, t3.second_num FROM score INNER JOIN ( SELECT t1.course_id, t1.first_num, t2.second_num FROM ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t1 INNER JOIN ( SELECT score.course_id, max(num) second_num FROM score INNER JOIN ( SELECT course_id, max(num) first_num FROM score GROUP BY course_id ) AS t ON score.course_id = t.course_id WHERE score.num < t.first_num GROUP BY course_id ) AS t2 ON t1.course_id = t2.course_id ) AS t3 ON score.course_id = t3.course_id WHERE score.num >= t3.second_num AND score.num <= t3.first_num ORDER BY course_id; #可以用以下命令验证上述查询的正确性 SELECT * FROM score ORDER BY course_id, num DESC; -- 21、查询不同课程但成绩相同的学号,课程号,成绩 -- 22、查询没学过“叶平”老师课程的学生姓名以及选修的课程名称; -- 23、查询所有选修了学号为1的同学选修过的一门或者多门课程的同学学号和姓名; -- 24、任课最多的老师中学生单科成绩最高的学生姓名