Nombre de tabla y nombre de campo:
–1.学生表
Student(s_id,s_name,s_birth,s_sex) –学生编号,学生姓名, 出生年月,学生性别
–2.课程表
Course(c_id,c_name,t_id) – –课程编号, 课程名称, 教师编号
–3.教师表
Teacher(t_id,t_name) –教师编号,教师姓名
–4.成绩表
Score(s_id,c_id,s_score) –学生编号,课程编号,分数
script sql:
/*
Navicat MySQL Data Transfer
Source Server : 192.168.160.132
Source Server Version : 50725
Source Host : 192.168.160.132:3306
Source Database : test
Target Server Type : MYSQL
Target Server Version : 50725
File Encoding : 65001
Date: 2019-03-26 17:28:51
*/
SET FOREIGN_KEY_CHECKS=0;
-- ----------------------------
-- Table structure for Course
-- ----------------------------
DROP TABLE IF EXISTS `Course`;
CREATE TABLE `Course` (
`c_id` varchar(20) NOT NULL,
`c_name` varchar(20) NOT NULL DEFAULT '',
`t_id` varchar(20) NOT NULL,
PRIMARY KEY (`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of Course
-- ----------------------------
INSERT INTO `Course` VALUES ('01', '语文', '02');
INSERT INTO `Course` VALUES ('02', '数学', '01');
INSERT INTO `Course` VALUES ('03', '英语', '03');
-- ----------------------------
-- Table structure for Score
-- ----------------------------
DROP TABLE IF EXISTS `Score`;
CREATE TABLE `Score` (
`s_id` varchar(20) NOT NULL,
`c_id` varchar(20) NOT NULL,
`s_score` int(3) DEFAULT NULL,
PRIMARY KEY (`s_id`,`c_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of Score
-- ----------------------------
INSERT INTO `Score` VALUES ('01', '01', '80');
INSERT INTO `Score` VALUES ('01', '02', '90');
INSERT INTO `Score` VALUES ('01', '03', '99');
INSERT INTO `Score` VALUES ('02', '01', '70');
INSERT INTO `Score` VALUES ('02', '02', '60');
INSERT INTO `Score` VALUES ('02', '03', '80');
INSERT INTO `Score` VALUES ('03', '01', '80');
INSERT INTO `Score` VALUES ('03', '02', '80');
INSERT INTO `Score` VALUES ('03', '03', '80');
INSERT INTO `Score` VALUES ('04', '01', '50');
INSERT INTO `Score` VALUES ('04', '02', '30');
INSERT INTO `Score` VALUES ('04', '03', '20');
INSERT INTO `Score` VALUES ('05', '01', '76');
INSERT INTO `Score` VALUES ('05', '02', '87');
INSERT INTO `Score` VALUES ('06', '01', '31');
INSERT INTO `Score` VALUES ('06', '03', '34');
INSERT INTO `Score` VALUES ('07', '02', '89');
INSERT INTO `Score` VALUES ('08', '03', '90');
-- ----------------------------
-- Table structure for Student
-- ----------------------------
DROP TABLE IF EXISTS `Student`;
CREATE TABLE `Student` (
`s_id` varchar(20) NOT NULL,
`s_name` varchar(20) NOT NULL DEFAULT '',
`s_birth` varchar(20) NOT NULL DEFAULT '',
`s_sex` varchar(10) NOT NULL DEFAULT '',
PRIMARY KEY (`s_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of Student
-- ----------------------------
INSERT INTO `Student` VALUES ('01', '赵雷', '1990-01-01', '男');
INSERT INTO `Student` VALUES ('02', '钱电', '1990-12-21', '男');
INSERT INTO `Student` VALUES ('03', '孙风', '1990-05-20', '男');
INSERT INTO `Student` VALUES ('04', '李云', '1990-08-06', '男');
INSERT INTO `Student` VALUES ('05', '周梅', '1991-12-01', '女');
INSERT INTO `Student` VALUES ('06', '吴兰', '1992-03-01', '女');
INSERT INTO `Student` VALUES ('07', '郑竹', '1989-07-01', '女');
INSERT INTO `Student` VALUES ('08', '王菊', '1990-01-20', '女');
-- ----------------------------
-- Table structure for Teacher
-- ----------------------------
DROP TABLE IF EXISTS `Teacher`;
CREATE TABLE `Teacher` (
`t_id` varchar(20) NOT NULL,
`t_name` varchar(20) NOT NULL DEFAULT '',
PRIMARY KEY (`t_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of Teacher
-- ----------------------------
INSERT INTO `Teacher` VALUES ('01', '张三');
INSERT INTO `Teacher` VALUES ('02', '李四');
INSERT INTO `Teacher` VALUES ('03', '王五');
Practique preguntas y respuestas sql: (Escribí el sql yo mismo, solo como referencia, tal vez tenga una mejor manera de escribir)
-- 1、查询"01"课程比"02"课程成绩高的学生的信息及课程分数
SELECT s.*,sc.s_score,sc1.s_score from
Student s
left join Score sc on s.s_id=sc.s_id and sc.c_id='01'
left join Score sc1 on s.s_id=sc1.s_id and sc1.c_id='02'
where sc.s_score>sc1.s_score
-- 2、查询学生男性和女性分别多少人？（这个笔试题遇到过）
写法一：
SELECT s_sex,count(*) from Student GROUP BY s_sex
写法二:
select count(s1.s_id) as nan,count(s2.s_id) as nv
from Student s
left join Student s1 on s.s_id=s1.s_id and s1.s_sex='男'
left join Student s2 on s.s_id=s2.s_id and s2.s_sex='女'
-- 3、查询平均成绩大于等于60分的同学的学生编号和学生姓名和平均成绩（这个笔试题遇到过）
select s.s_id,s.s_name,ROUND(avg(sc.s_score),1)
from Student s left join Score sc
on s.s_id=sc.s_id
group by s.s_id,s.s_name
having avg(sc.s_score)>60
注：因为分数一般保留1位小数，所以这里用ROUND()函数将其平均成绩保留一位小数；
-- 4、查询"李"姓老师的数量 （这个笔试题遇到过）
select count(*)
from Teacher t
where t.t_name like '李%'
-- 5、查询每门课程被选修的学生数
select c_id,count(s_id)
from Score
group by c_id
-- 6、查询选修了全部课程的学生信息
写法一：
select *
from Student where s_id in (
select sc.s_id
from Score sc
group by sc.s_id
having count(*)=(select count(c_id) from Course))
写法二：显示的内连接，一般称为内连接，有INNER JOIN
SELECT *
from Student inner join (SELECT Score.s_id from Score
group by Score.s_id HAVING COUNT(Score.c_id)=(SELECT COUNT(1) FROM Course)) t
on Student.s_id=t.s_id
写法三：其实与写法二意思一样，只是隐式的内连接，不需要INNER JOIN
SELECT * from Student,(SELECT Score.s_id from Score group by Score.s_id HAVING COUNT(Score.c_id)=(SELECT COUNT(1) FROM Course)) t
where Student.s_id=t.s_id
-- 7、查询老师"张三"教的学生信息
select s.*
from Student s left join Score sc
on s.s_id=sc.s_id
left join Course c
on sc.c_id =c.c_id
left join Teacher t
on c.t_id=t.t_id
where t.t_name='张三'
-- 8、查询每门选修课最高的分数、平均分、最低分；
写法一：
SELECT c_id,max(s_score),avg(s_score),min(s_score)
FROM Score
group by c_id
写法二：
select c.c_name,max(s_score),AVG(s_score),MIN(s_score)
from Score sc left join Course c
on sc.c_id=c.c_id
GROUP BY c.c_name
-- 9、查询老师"张三"教的学生中该选修课成绩最高的分数和最低的分数
写法一：
select t.t_name,c.c_name,max(sc.s_score),min(sc.s_score)
from Score sc left join Course c
on sc.c_id=c.c_id
left join Teacher t
ON c.t_id=t.t_id
group by t.t_name,c.c_name
having t.t_name='张三'
写法二：
SELECT Course.c_name,Course.t_id,MAX(Score.s_score),MIN(Score.s_score)
FROM Score,Course
where Score.c_id=Course.c_id
GROUP BY Course.c_name,Course.t_id
HAVING Course.t_id=(SELECT t_id FROM Teacher where t_name="张三")
-- 10、查询和"01"号的同学学习的课程完全相同的其他同学的信息
写法一：
select *
from Student where s_id in(
select sc.s_id
from Score sc
where sc.s_id !='01' and sc.c_id in (select c_id from Score where s_id='01')
group by sc.s_id
having count(c_id)=(select count(c_id) from Score where s_id='01')
注：课程完全相同，那么必须最少满足两个条件：1、课程的数量相同 2、课程必须在01号同学学习的课程当中
写法二：
SELECT Student.*
FROM Student LEFT JOIN Score
on Student.s_id=Score.s_id
where Student.s_id !=01
GROUP BY Student.s_id
HAVING count(Score.c_id)= (SELECT count(c_id) FROM Score WHERE s_id=01)
-- 11、查询没有学过'张三'老师任何一门课的学生
SELECT * from Student where s_id not in(
SELECT s.s_id
from Teacher t INNER JOIN Course c
on t.t_id=c.c_id
INNER JOIN Score s
on c.c_id=s.c_id
where t.t_name='张三'
)
思路：这个思路很简单，先用Teacher 、Course 、Score 三表进行内连接关联，条件查询出'张三'老师教过的学生，然后用子查询，查询学生表中id不在子查询查询结果集中的；
-- 12、查询大于28岁的学生信息
写法一：
select s.*
from Student s
where DATE_FORMAT(NOW(),'%Y-%m-%d')-s.s_birth >28
写法二：
SELECT *,DATE_ADD(s_birth,INTERVAL 28 YEAR) as newDate
FROM Student
GROUP BY s_id
HAVING newDate>NOW()
-- 13、查询每门课程都在70分以上的学生信息
SELECT Student.* FROM Student where s_id in (
SELECT s_id
FROM Score
where s_score>70
GROUP BY s_id
HAVING COUNT(s_score)=(SELECT count(*) FROM Course)
)
-- 14、查询既是'张三'又是'王五'的学生信息
SELECT * from Student where s_id in(
SELECT s.s_id
from Teacher t INNER JOIN Course c
on t.t_id=c.c_id
INNER JOIN Score s
on c.c_id=s.c_id
where t.t_name in ('张三','王五')
GROUP BY s.s_id
HAVING count(c.c_id)=2
)
思路：先用Teacher 、Course 、Score 三表进行内连接关联，然后条件查询教师名称在('张三','王五')
之中，因为既是'张三'又是'王五'的学生，那么根据学生进行分组，再对结果用HAVING 条件查询出课程编
号有2个的学生id，那么这个学生肯定既是'张三'又是'王五'。
