【10.1】多表查询练习题
【一】SQL文件
/*
Navicat Premium Data Transfer
Source Server : localhost
Source Server Type : MySQL
Source Server Version : 50624
Source Host : localhost
Source Database : sqlexam
Target Server Type : MySQL
Target Server Version : 50624
File Encoding : utf-8
Date: 10/21/2016 06:46:46 AM
*/
SET NAMES utf8;
SET FOREIGN_KEY_CHECKS = 0;
-- ----------------------------
-- Table structure for `class`
-- ----------------------------
DROP TABLE IF EXISTS `class`;
CREATE TABLE `class` (
`cid` int(11) NOT NULL AUTO_INCREMENT,
`caption` varchar(32) NOT NULL,
PRIMARY KEY (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of `class`
-- ----------------------------
BEGIN;
INSERT INTO `class` VALUES ('1', '三年二班'), ('2', '三年三班'), ('3', '一年二班'), ('4', '二年九班');
COMMIT;
-- ----------------------------
-- Table structure for `course`
-- ----------------------------
DROP TABLE IF EXISTS `course`;
CREATE TABLE `course` (
`cid` int(11) NOT NULL AUTO_INCREMENT,
`cname` varchar(32) NOT NULL,
`teacher_id` int(11) NOT NULL,
PRIMARY KEY (`cid`),
KEY `fk_course_teacher` (`teacher_id`),
CONSTRAINT `fk_course_teacher` FOREIGN KEY (`teacher_id`) REFERENCES `teacher` (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=5 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of `course`
-- ----------------------------
BEGIN;
INSERT INTO `course` VALUES ('1', '生物', '1'), ('2', '物理', '2'), ('3', '体育', '3'), ('4', '美术', '2');
COMMIT;
-- ----------------------------
-- Table structure for `score`
-- ----------------------------
DROP TABLE IF EXISTS `score`;
CREATE TABLE `score` (
`sid` int(11) NOT NULL AUTO_INCREMENT,
`student_id` int(11) NOT NULL,
`course_id` int(11) NOT NULL,
`num` int(11) NOT NULL,
PRIMARY KEY (`sid`),
KEY `fk_score_student` (`student_id`),
KEY `fk_score_course` (`course_id`),
CONSTRAINT `fk_score_course` FOREIGN KEY (`course_id`) REFERENCES `course` (`cid`),
CONSTRAINT `fk_score_student` FOREIGN KEY (`student_id`) REFERENCES `student` (`sid`)
) ENGINE=InnoDB AUTO_INCREMENT=53 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of `score`
-- ----------------------------
BEGIN;
INSERT INTO `score` VALUES ('1', '1', '1', '10'), ('2', '1', '2', '9'), ('5', '1', '4', '66'), ('6', '2', '1', '8'), ('8', '2', '3', '68'), ('9', '2', '4', '99'), ('10', '3', '1', '77'), ('11', '3', '2', '66'), ('12', '3', '3', '87'), ('13', '3', '4', '99'), ('14', '4', '1', '79'), ('15', '4', '2', '11'), ('16', '4', '3', '67'), ('17', '4', '4', '100'), ('18', '5', '1', '79'), ('19', '5', '2', '11'), ('20', '5', '3', '67'), ('21', '5', '4', '100'), ('22', '6', '1', '9'), ('23', '6', '2', '100'), ('24', '6', '3', '67'), ('25', '6', '4', '100'), ('26', '7', '1', '9'), ('27', '7', '2', '100'), ('28', '7', '3', '67'), ('29', '7', '4', '88'), ('30', '8', '1', '9'), ('31', '8', '2', '100'), ('32', '8', '3', '67'), ('33', '8', '4', '88'), ('34', '9', '1', '91'), ('35', '9', '2', '88'), ('36', '9', '3', '67'), ('37', '9', '4', '22'), ('38', '10', '1', '90'), ('39', '10', '2', '77'), ('40', '10', '3', '43'), ('41', '10', '4', '87'), ('42', '11', '1', '90'), ('43', '11', '2', '77'), ('44', '11', '3', '43'), ('45', '11', '4', '87'), ('46', '12', '1', '90'), ('47', '12', '2', '77'), ('48', '12', '3', '43'), ('49', '12', '4', '87'), ('52', '13', '3', '87');
COMMIT;
-- ----------------------------
-- Table structure for `student`
-- ----------------------------
DROP TABLE IF EXISTS `student`;
CREATE TABLE `student` (
`sid` int(11) NOT NULL AUTO_INCREMENT,
`gender` char(1) NOT NULL,
`class_id` int(11) NOT NULL,
`sname` varchar(32) NOT NULL,
PRIMARY KEY (`sid`),
KEY `fk_class` (`class_id`),
CONSTRAINT `fk_class` FOREIGN KEY (`class_id`) REFERENCES `class` (`cid`)
) ENGINE=InnoDB AUTO_INCREMENT=17 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of `student`
-- ----------------------------
BEGIN;
INSERT INTO `student` VALUES ('1', '男', '1', '理解'), ('2', '女', '1', '钢蛋'), ('3', '男', '1', '张三'), ('4', '男', '1', '张一'), ('5', '女', '1', '张二'), ('6', '男', '1', '张四'), ('7', '女', '2', '铁锤'), ('8', '男', '2', '李三'), ('9', '男', '2', '李一'), ('10', '女', '2', '李二'), ('11', '男', '2', '李四'), ('12', '女', '3', '如花'), ('13', '男', '3', '刘三'), ('14', '男', '3', '刘一'), ('15', '女', '3', '刘二'), ('16', '男', '3', '刘四');
COMMIT;
-- ----------------------------
-- Table structure for `teacher`
-- ----------------------------
DROP TABLE IF EXISTS `teacher`;
CREATE TABLE `teacher` (
`tid` int(11) NOT NULL AUTO_INCREMENT,
`tname` varchar(32) NOT NULL,
PRIMARY KEY (`tid`)
) ENGINE=InnoDB AUTO_INCREMENT=6 DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of `teacher`
-- ----------------------------
BEGIN;
INSERT INTO `teacher` VALUES ('1', '张磊老师'), ('2', '李平老师'), ('3', '刘海燕老师'), ('4', '朱云海老师'), ('5', '李杰老师');
COMMIT;
SET FOREIGN_KEY_CHECKS = 1;
【二】习题
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
【三】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
本文来自博客园，作者：Chimengmeng，转载请注明原文链接：https://www.cnblogs.com/dream-ze/p/17995419
