说明
工作多年，对mysql查询语句的使用一直停留在将就用，遇到难一点的、复杂一点的就束手无策的状态。这几天闲着没事做，找了一些练习题做下，感觉对select语句稍微熟悉一点了。原文转载自其他地方，但原文没有答案，所以写此文记录一下。
文章转载自: mysql查询练习
### 数据准备
/*
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', '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;
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
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
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
11、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
参考答案
SELECT c.cname, t.tname
FROM course c
LEFT JOIN teacher t
ON c.teacher_id = t.tid
SELECT gender, COUNT(sid)
FROM student
GROUP BY gender
SELECT s.sname
FROM student s
LEFT JOIN score sc
ON sc.student_id = s.sid
LEFT JOIN course c
ON c.cid = sc.course_id
WHERE c.cname = "物理"
AND sc.num = 100
SELECT s.sname 姓名, AVG(sc.num) 平均成绩
FROM student s
LEFT JOIN score sc
ON sc.student_id = s.sid
GROUP BY sc.student_id HAVING AVG(sc.num) > 80;
SELECT s.sid 学号, s.sname 姓名, COUNT(sc.course_id) 选课总数, SUM(sc.num) 总成绩
FROM student s
LEFT JOIN score sc
ON sc.student_id = s.sid
GROUP BY s.sid;
SELECT COUNT(1)
FROM teacher
WHERE tname LIKE '李%';
SELECT s.sname
FROM student s
WHERE NOT EXISTS (
SELECT *
FROM score sc
INNER JOIN course c
ON sc.course_id = c.cid
INNER JOIN teacher t
ON t.tid = c.teacher_id
WHERE t.tname = "李平老师"
AND sc.student_id = s.sid
)
SELECT s.sname, wl.num 物理成绩, sw.num 生物成绩
FROM student s
LEFT JOIN (SELECT sc.student_id, sc.num
FROM score sc
INNER JOIN course c
ON c.cid = sc.course_id
AND c.cname = "物理") wl
ON wl.student_id = s.sid
LEFT JOIN (SELECT sc.student_id, sc.num
FROM score sc
INNER JOIN course c
ON c.cid = sc.course_id
AND c.cname = "生物" ) sw
ON sw.student_id = wl.student_id
WHERE wl.num > sw.num;
SELECT s.sname
FROM student s
LEFT JOIN score sc
ON s.sid = sc.student_id
INNER JOIN (SELECT cid
FROM course
WHERE cname IN ("物理", "体育")) cc
ON cc.cid = sc.course_id
GROUP BY s.sid HAVING COUNT(sc.course_id) != 2;
SELECT s.sname 姓名, cl.caption 班级
FROM class cl
LEFT JOIN student s
ON cl.cid = s.class_id
LEFT JOIN score sc
ON s.sid = sc.student_id
WHERE sc.num < 60
GROUP BY sc.student_id
HAVING SUM(sc.course_id) >=2;
SELECT s.sname 姓名
FROM course c
INNER JOIN score sc
ON sc.course_id = c.cid
INNER JOIN student s
ON s.sid = sc.student_id
GROUP BY sc.student_id
HAVING COUNT(sc.course_id) = (SELECT COUNT(1) FROM course)
SELECT sc.*
FROM score sc
WHERE sc.course_id IN
(
SELECT c.cid
FROM teacher t
LEFT JOIN course c
ON c.teacher_id = t.tid
WHERE t.tname = "李平老师"
)
SELECT c.cid, c.cname
FROM course c
LEFT JOIN score sc
ON sc.course_id = c.cid
GROUP BY c.cid
HAVING COUNT(sc.student_id) = (SELECT COUNT(1) FROM student);
# 查询结果为空
SELECT c.cname 课程, COUNT(sc.course_id) 被选修次数
FROM course c
LEFT JOIN score sc
ON sc.course_id = c.cid
GROUP BY c.cid;
SELECT s.sname 姓名, s.sid 学号
FROM student s
INNER JOIN score sc
ON s.sid = sc.student_id
GROUP BY s.sid
HAVING COUNT(sc.course_id) = 1;
SELECT DISTINCT num
FROM score
ORDER BY num DESC;
SELECT s.sname 姓名, AVG(sc.num) 平均成绩
FROM student s
LEFT JOIN score sc
ON sc.student_id = s.sid
GROUP BY s.sid
HAVING AVG(sc.num) > 85;
SELECT s.sname 姓名, sc.num 生物成绩
FROM course c
LEFT JOIN score sc
ON sc.course_id = c.cid
LEFT JOIN student s
ON s.sid = sc.student_id
WHERE c.cname = "生物"
AND sc.num < 60;
SELECT s.sname 姓名, AVG(sc.num) 李平老师课程平均分
FROM teacher t
LEFT JOIN course c
ON c.teacher_id = t.tid
INNER JOIN score sc
ON sc.course_id = c.cid
LEFT JOIN student s
ON s.sid = sc.student_id
WHERE t.tname = "李平老师"
GROUP BY sc.student_id
HAVING AVG(sc.num) = (SELECT AVG(sc.num)
FROM teacher t
LEFT JOIN course c
ON c.teacher_id = t.tid
INNER JOIN score sc
ON sc.course_id = c.cid
WHERE t.tname = "李平老师"
GROUP BY sc.student_id
ORDER BY AVG(sc.num) DESC
LIMIT 1);
SELECT c.cname 课程, top.t1 第一名成绩, top.t2 第二名成绩, GROUP_CONCAT(s.sname) 前两名
FROM (SELECT f1.course_id, f1.mnum t1, MAX(sc.num) t2
FROM (SELECT course_id, MAX(num) mnum
FROM score
GROUP BY course_id) f1
LEFT JOIN score sc
ON sc.course_id = f1.course_id
WHERE sc.num != f1.mnum
GROUP BY sc.course_id) top
INNER JOIN score sc2
ON sc2.course_id = top.course_id
LEFT JOIN student s
ON sc2.student_id = s.sid
LEFT JOIN course c
ON c.cid = sc2.course_id
WHERE sc2.num = top.t1 OR sc2.num = top.t2
GROUP BY top.course_id;
SELECT s.sname, sc1.course_id, sc2.course_id, sc1.num
FROM score sc1
LEFT JOIN student s
ON s.sid = sc1.student_id
LEFT JOIN score sc2
ON sc1.num = sc2.num
AND s.sid = sc2.student_id
WHERE sc1.course_id != sc2.course_id
PS: 有点问题，两条记录重复了
SELECT s.sname, GROUP_CONCAT(c2.cname)
FROM student s
LEFT JOIN score sc2
ON s.sid = sc2.student_id
LEFT JOIN course c2
ON c2.cid = sc2.course_id
WHERE NOT EXISTS (SELECT c.cid
FROM teacher t
LEFT JOIN course c
ON t.tid = c.teacher_id
LEFT JOIN score sc
ON sc.course_id = c.cid
WHERE t.tname = "李平老师"
AND s.sid = sc.student_id)
GROUP BY s.sid;
SELECT s.sid 学号, s.sname 姓名
FROM score sc
LEFT JOIN student s
ON s.sid = sc.student_id
WHERE EXISTS (SELECT course_id
FROM score
WHERE student_id = 1
AND course_id = sc.course_id
AND student_id != sc.student_id)
GROUP BY sc.student_id;
SELECT sc.course_id, MAX(sc.num), GROUP_CONCAT(s.sname)
FROM score sc
INNER JOIN course c2
ON c2.cid = sc.course_id
INNER JOIN
(SELECT t.tid
FROM course c
LEFT JOIN teacher t
ON t.tid = c.teacher_id
GROUP BY t.tid
HAVING COUNT(c.cid) = (SELECT COUNT(cid)
FROM course
GROUP BY teacher_id
ORDER BY COUNT(cid) DESC
LIMIT 1)) t2
ON c2.teacher_id = t2.tid
INNER JOIN student s
ON s.sid = sc.student_id
GROUP BY sc.course_id;