SQL逻辑查询语句执行顺序—练习题

在做练习题之前要重点熟悉select 的执行顺序

1.SELECT语句关键字的定义顺序

SELECT DISTINCT <select_list>
FROM <left_table>
<join_type> JOIN <right_table>
ON <join_condition>
WHERE <where_condition>
GROUP BY <group_by_list>
HAVING <having_condition>
ORDER BY <order_by_condition>
LIMIT <limit_number>

2.SELECT语句关键字的执行顺序

(7)     SELECT
(8)     DISTINCT <select_list>
(1)     FROM <left_table>
(3)     <join_type> JOIN <right_table>
(2)     ON <join_condition>
(4)     WHERE <where_condition>
(5)     GROUP BY <group_by_list>
(6)     HAVING <having_condition>
(9)     ORDER BY <order_by_condition>
(10)    LIMIT <limit_number>

1.先从库,表里找 
2.on 后面加两表连接的限制条件
3.将两表连接起来（内左右全等）
4.从约束条件where里过滤出数据
5.然后交给group by 进行分组，
6.分完组后 用having 过滤
7.之后才是运行select 后面的语句，
8.distinct进行去重
9.接着轮到order by 排序
10.limit 最后运行

练习题

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

题目

建立表

#建立表
/*
 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;

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,count(sid)
from student
group by gender

3、查询物理成绩等于100的学生的姓名

select sname
from student
where sid in
    (   select student_id
        from score
        where num=100 and course_id=2);

select student.sname
from student
where sid in
(select score.student_id
from score
inner join course
on course.cid =score.course_id
where score.num =100 and course.cname="物理");

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 t1.student_id=student.sid;

5、查询所有学生的学号，姓名，选课数，总成绩

select t1.sid,t1.sname,t2.count_id,t2.sum_num from student as t1 inner join
(select student_id,count(course_id)as count_id,sum(num)as sum_num from score
group by student_id)as t2 on t1.sid= t2.student_id;

6、查询姓李老师的个数

select count(tid) from teacher
where tname regexp "^李.*老师$";

7、 查询没有报李平老师课的学生姓名

select sname from student where sid not in
(select student_id from score
where course_id=2 or course_id=4);

8、 查询物理课程比生物课程高的学生的学号

select t1.student_id from
(select student_id,num from score
where course_id=1)as t1 inner join
(select student_id,num from score
where course_id=2)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

select sname from student where sid not in
(select student_id from  score
where course_id =2 or course_id=3
group by student_id
having count(student_id)=2);

10、查询挂科超过两门(包括两门)的学生姓名和班级

select sname,caption from
(select * from student inner join class on student.class_id= class.cid)as t1
where t1.sid in
(select student_id from score
where num<60
group by student_id
having count(student_id)>=2);

11 、查询选修了所有课程的学生姓名

select sname from student where sid in
(select student_id from score
group by student_id
having count(student_id)>=4);

12、查询李平老师教的课程的所有成绩记录

select * from score
where course_id=2 or course_id=4;

#13、查询全部学生都选修了的课程号和课程名

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 count(course_id) from score
group by course_id

15、查询只选修了一门课程的学生姓名和学号

select sname,sid from student where sid in
(select student_id from score
group by student_id
having count(student_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),student_id from score
group by student_id
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 student.sname,t2.avg_num from student inner join
(select student_id,avg(num)as avg_num from score
group by student_id
having avg(num)>85)as t2 on student.sid= t2.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 student.sname,t2.num from student inner join
(select student_id,num from score
where course_id =1 and num<60)as t2 on student.sid=t2.student_id;

#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 student.sname,max(avg_num) from student inner join
(select student_id,avg(num)as avg_num from score
where course_id=2 or course_id =4
group by student_id
order by avg(num) DESC)as t2 on student.sid=t2.student_id;

20、查询每门课程成绩最好的前两名学生姓名

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;

中间过程

最后查询的结果。说实话这个是看了别人的答案的，抄的，最后的查询的结果有点无语，

总感觉并不是想要的查询，只是在凑表合表

如果真的要查询

select * from score
order by course_id,num DESC;

上面也能看的出来的。绕的头疼

贴上egon老师的链接

http://www.cnblogs.com/linhaifeng/articles/7895711.html

可能我自己写的答案，格式不漂亮，也不完美，可以看看上面的，作业这个还是要自己操手，实在想不出来再看

感谢观看，如有不对，一定要指出。小木在这里谢谢了