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MySQL五-2:多表查询

时间:2021-07-01 10:21:17 帮助过:16人阅读

department表中存在dept_id=203,Ture mysql> select * from employee -> where exists -> (select id from department where id=200); +----+------------+--------+------+--------+ | id | name | sex | age | dep_id | +----+------------+--------+------+--------+ | 1 | egon | male | 18 | 200 | | 2 | alex | female | 48 | 201 | | 3 | wupeiqi | male | 38 | 201 | | 4 | yuanhao | female | 28 | 202 | | 5 | liwenzhou | male | 18 | 200 | | 6 | jingliyang | female | 18 | 204 | +----+------------+--------+------+--------+ #department表中存在dept_id=205,False mysql> select * from employee -> where exists -> (select id from department where id=204); Empty set (0.00 sec)

五 综合练习

init.sql文件内容

技术分享
/*
 数据导入:
 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, , 1, 理解), (2, , 1, 钢蛋), (3, , 1, 张三), (4, , 1, 张一), (5, , 1, 张二), (6, , 1, 张四), (7, , 2, 铁锤), (8, , 2, 李三), (9, , 2, 李一), (10, , 2, 李二), (11, , 2, 李四), (12, , 3, 如花), (13, , 3, 刘三), (14, , 3, 刘一), (15, , 3, 刘二), (16, , 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;
View Code

从init.sql文件中导入数据

#准备表、记录
mysql> create database db1;
mysql> use db1;
mysql> source /root/init.sql

!!!重中之重:练习之前务必搞清楚sql逻辑查询语句的执行顺序

链接:http://www.cnblogs.com/bingpan/p/7491338.html

技术分享
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

题目
习题 技术分享
参考答案:

1、SELECT cname,tname FROM course LEFT JOIN teacher on course.teacher_id=teacher.tid

2、SELECT gender 性别 ,COUNT(gender) 人数 from student GROUP BY gender;

3、SELECT sname from student where sid in (SELECT student_id from score LEFT JOIN course ON  

                   score.course_id=course.cid WHERE num=100 AND course.cname="物理");

 4、SELECT sname 姓名,平均成绩 from student RIGHT JOIN (SELECT student_id,avg(num) 平均成绩 from score                                            GROUP BY student_id HAVING avg(num)>80) as A
                                on student.sid=A.student_id;

5、SELECT  student_id 学号,sname 姓名,COUNT(course_id) 课程数,SUM(num) 总分 FROM score LEFT JOIN student 
                                ON score.student_id=student.sid GROUP BY student_id;
6、略

7、SELECT * FROM student WHERE sid not in (SELECT student_id FROM score WHERE course_id in (SELECT cid                   FROM course LEFT JOIN teacher ON teacher.tid=course.teacher_id WHERE tname like "李平%"))


8、     SELECT * FROM

        (SELECT * FROM score LEFT JOIN course ON score.course_id=course.cid WHERE cname="生物") as A

    INNER JOIN

        (SELECT * FROM score LEFT JOIN course ON score.course_id=course.cid WHERE cname="物理") as B


    ON A.student_id=B.student_id

    WHERE a.num<b.num

9、SELECT sname from student WHERE sid NOT IN (SELECT student_id FROM score LEFT JOIN course ON 
        score.course_id=course.cid WHERE cname=体育  OR cname=物理 GROUP BY student_id HAVING COUNT(course_id)>1) ;

10、SELECT sname,gender from student WHERE sid in (SELECT student_id from score WHERE num<60 GROUP BY 
        student_id HAVING COUNT(student_id)>1);

11、SELECT sid 学号,sname 姓名 from student WHERE sid in (SELECT student_id FROM score GROUP BY 
                student_id HAVING COUNT(student_id)!= (SELECT COUNT(cid) FROM course))

12、SELECT * FROM score WHERE course_id NOT IN (SELECT cid from teacher INNER JOIN course on
                      teacher.tid=course.teacher_id WHERE teacher.tname LIKE "李平%");
            
13、SELECT course_id from score GROUP BY course_id HAVING COUNT(course_id)=(SELECT COUNT(sid) FROM 
             student)

14、--  SELECT course_id,COUNT(course_id) from score GROUP BY course_id


    SELECT cname,COUNT(course_id) FROM score LEFT JOIN course ON score.course_id=course.cid GROUP BY course_id

15、SELECT sid,sname FROM student  WHERE sid in (SELECT student_id from score GROUP BY student_id 
        HAVING COUNT(sid)=1);


16、SELECT DISTINCT num FROM score ORDER BY num DESC

17、SELECT sname from student WHERE sid in (SELECT student_id FROM score GROUP BY student_id  HAVING
         AVG(num)>85)
       注意:用这种方式我们无法查看对应的平均成绩,所以不得不使用连表:
       SELECT sname,AVG(num) FROM student LEFT JOIN score ON student.sid=score.student_id GROUP BY student_id HAVING AVG(num)>85;

18、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 score.num<60 AND cname="生物"19、
      方法1:
      SELECT student_id,avg(num),sname FROM score 
     LEFT JOIN student ON score.student_id=student.sid
             WHERE course_id in (SELECT cid FROM course LEFT JOIN teacher ON teacher.tid=course.teacher_id WHERE tname LIKE "李平%")
             GROUP BY student_id
     ORDER BY AVG(num) DESC
     LIMIT 1

     方法2:

     SELECT sname,MAX(B) FROM 

    (SELECT student_id,sname,AVG(num) as B FROM score 
                             LEFT JOIN student ON score.student_id=student.sid

                 WHERE course_id in (SELECT cid FROM course LEFT JOIN teacher ON teacher.tid=course.teacher_id WHERE tname LIKE "李平%")
                 
                            
                             GROUP BY student_id

    ) as A

 WHERE B=(

    SELECT MAX(B) FROM
     (SELECT student_id,sname,AVG(num) as B FROM score 
                             LEFT JOIN student ON score.student_id=student.sid

                 WHERE course_id in (SELECT cid FROM course LEFT JOIN teacher ON teacher.tid=course.teacher_id WHERE tname LIKE "李平%")
                 
                            
                             GROUP BY student_id

    ) as A
                     )
答案

参考答案:http://www.cnblogs.com/wupeiqi/articles/5748496.html

MySQL五-2:多表查询

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