多表查询

准备

建表与数据准备

#建表
create table department(
id int,
name varchar(20) 
);

create table employee(
id int primary key auto_increment,
name varchar(20),
sex enum(
'male','female') not null default 'male',
age int,
dep_id int
);

#插入数据
insert into department values
(
200,'技术'),
(
201,'人力资源'),
(
202,'销售'),
(
203,'运营');

insert into employee(name,sex,age,dep_id) values
('egon','male',18,200),
(
'alex','female',48,201),
(
'wupeiqi','male',38,201),
(
'yuanhao','female',28,202),
(
'liwenzhou','male',18,200),
(
'jingliyang','female',18,204)
;

#查看表结构和数据
mysql> desc department;
+-------+-------------+------+-----+---------+-------+
| Field | Type | Null | Key | Default | Extra |
+-------+-------------+------+-----+---------+-------+
| id | int(11) | YES | | NULL | |
| name | varchar(20) | YES | | NULL | |
+-------+-------------+------+-----+---------+-------+

mysql> desc employee;
+--------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | YES | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(11) | YES | | NULL | |
| dep_id | int(11) | YES | | NULL | |
+--------+-----------------------+------+-----+---------+----------------+

mysql> select * from department;
+------+--------------+
| id | name |
+------+--------------+
| 200 | 技术 |
| 201 | 人力资源 |
| 202 | 销售 |
| 203 | 运营 |
+------+--------------+

mysql> select * from employee;
+----+------------+--------+------+--------+
| 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 |
+----+------------+--------+------+--------+

表department与employee

sql示例

 

多表连接查询

#重点:外链接语法

SELECT 字段列表
FROM 表1 INNER|LEFT|RIGHT JOIN 表2
ON 表1.字段 = 表2.字段;

1 交叉连接:不适用任何匹配条件。生成笛卡尔积

复制代码
mysql> select * from employee,department;
+----+------------+--------+------+--------+------+--------------+
| id | name       | sex    | age  | dep_id | id   | name         |
+----+------------+--------+------+--------+------+--------------+
|  1 | egon       | male   |   18 |    200 |  200 | 技术         |
|  1 | egon       | male   |   18 |    200 |  201 | 人力资源     |
|  1 | egon       | male   |   18 |    200 |  202 | 销售         |
|  1 | egon       | male   |   18 |    200 |  203 | 运营         |
|  2 | alex       | female |   48 |    201 |  200 | 技术         |
|  2 | alex       | female |   48 |    201 |  201 | 人力资源     |
|  2 | alex       | female |   48 |    201 |  202 | 销售         |
|  2 | alex       | female |   48 |    201 |  203 | 运营         |
|  3 | wupeiqi    | male   |   38 |    201 |  200 | 技术         |
|  3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
|  3 | wupeiqi    | male   |   38 |    201 |  202 | 销售         |
|  3 | wupeiqi    | male   |   38 |    201 |  203 | 运营         |
|  4 | yuanhao    | female |   28 |    202 |  200 | 技术         |
|  4 | yuanhao    | female |   28 |    202 |  201 | 人力资源     |
|  4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
|  4 | yuanhao    | female |   28 |    202 |  203 | 运营         |
|  5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|  5 | liwenzhou  | male   |   18 |    200 |  201 | 人力资源     |
|  5 | liwenzhou  | male   |   18 |    200 |  202 | 销售         |
|  5 | liwenzhou  | male   |   18 |    200 |  203 | 运营         |
|  6 | jingliyang | female |   18 |    204 |  200 | 技术         |
|  6 | jingliyang | female |   18 |    204 |  201 | 人力资源     |
|  6 | jingliyang | female |   18 |    204 |  202 | 销售         |
|  6 | jingliyang | female |   18 |    204 |  203 | 运营         |
+----+------------+--------+------+--------+------+--------------+
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2 内连接:只连接匹配的行

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#找两张表共有的部分,相当于利用条件从笛卡尔积结果中筛选出了正确的结果
#department没有204这个部门,因而employee表中关于204这条员工信息没有匹配出来
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; 
+----+-----------+------+--------+--------------+
| id | name      | age  | sex    | name         |
+----+-----------+------+--------+--------------+
|  1 | egon      |   18 | male   | 技术         |
|  2 | alex      |   48 | female | 人力资源     |
|  3 | wupeiqi   |   38 | male   | 人力资源     |
|  4 | yuanhao   |   28 | female | 销售         |
|  5 | liwenzhou |   18 | male   | 技术         |
+----+-----------+------+--------+--------------+

#上述sql等同于
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;
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3 外链接之左连接:优先显示左表全部记录

复制代码
#以左表为准,即找出所有员工信息,当然包括没有部门的员工
#本质就是:在内连接的基础上增加左边有右边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
+----+------------+--------------+
| id | name       | depart_name  |
+----+------------+--------------+
|  1 | egon       | 技术         |
|  5 | liwenzhou  | 技术         |
|  2 | alex       | 人力资源     |
|  3 | wupeiqi    | 人力资源     |
|  4 | yuanhao    | 销售         |
|  6 | jingliyang | NULL         |
+----+------------+--------------+
复制代码

4 外链接之右连接:优先显示右表全部记录

复制代码
#以右表为准,即找出所有部门信息,包括没有员工的部门
#本质就是:在内连接的基础上增加右边有左边没有的结果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
+------+-----------+--------------+
| id   | name      | depart_name  |
+------+-----------+--------------+
|    1 | egon      | 技术         |
|    2 | alex      | 人力资源     |
|    3 | wupeiqi   | 人力资源     |
|    4 | yuanhao   | 销售         |
|    5 | liwenzhou | 技术         |
| NULL | NULL      | 运营         |
+------+-----------+--------------+
复制代码

5 全外连接:显示左右两个表全部记录

复制代码
全外连接:在内连接的基础上增加左边有右边没有的和右边有左边没有的结果
#注意:mysql不支持全外连接 full JOIN
#强调:mysql可以使用此种方式间接实现全外连接
select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id
;
#查看结果
+------+------------+--------+------+--------+------+--------------+
| id   | name       | sex    | age  | dep_id | id   | name         |
+------+------------+--------+------+--------+------+--------------+
|    1 | egon       | male   |   18 |    200 |  200 | 技术         |
|    5 | liwenzhou  | male   |   18 |    200 |  200 | 技术         |
|    2 | alex       | female |   48 |    201 |  201 | 人力资源     |
|    3 | wupeiqi    | male   |   38 |    201 |  201 | 人力资源     |
|    4 | yuanhao    | female |   28 |    202 |  202 | 销售         |
|    6 | jingliyang | female |   18 |    204 | NULL | NULL        |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 运营         |
+------+------------+--------+------+--------+------+--------------+

#注意 union与union all的区别:union会去掉相同的纪录
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符合条件连接查询

复制代码
#示例1:以内连接的方式查询employee和department表,并且employee表中的age字段值必须大于25,即找出年龄大于25岁的员工以及员工所在的部门
select employee.name,department.name from employee inner join department
    on employee.dep_id = department.id
    where age > 25;

#示例2:以内连接的方式查询employee和department表,并且以age字段的升序方式显示
select employee.id,employee.name,employee.age,department.name from employee,department
where employee.dep_id
= department.id
and age > 25
order by age asc;

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子查询

#1:子查询是将一个查询语句嵌套在另一个查询语句中。
#2:内层查询语句的查询结果,可以为外层查询语句提供查询条件。
#3:子查询中可以包含:IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等关键字
#4:还可以包含比较运算符:= 、 !=、> 、<等

1 带IN关键字的子查询

复制代码
#查询平均年龄在25岁以上的部门名
select id,name from department
    where id in 
        (select dep_id from employee group by dep_id having avg(age) > 25);

查看技术部员工姓名

select name from employee
where dep_id in
(
select id from department where name='技术');

查看不足1人的部门名(子查询得到的是有人的部门id)

select name from department where id not in (select distinct dep_id from employee);

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2 带比较运算符的子查询

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#比较运算符:=!=>>=<<=<>
#查询大于所有人平均年龄的员工名与年龄
mysql> select name,age from emp where age > (select avg(age) from emp);
+---------+------+
| name    | age  |
+---------+------+
| alex    | 48   |
| wupeiqi | 38   |
+---------+------+
2 rows in set (0.00 sec)

查询大于部门内平均年龄的员工名、年龄

select t1.name,t1.age from emp t1
inner join
(
select dep_id,avg(age) avg_age from emp group by dep_id) t2
on t1.dep_id = t2.dep_id
where t1.age > t2.avg_age;

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3 带EXISTS关键字的子查询

EXISTS关字键字表示存在。在使用EXISTS关键字时,内层查询语句不返回查询的记录。
而是返回一个真假值。True或False
当返回True时,外层查询语句将进行查询;当返回值为False时,外层查询语句不进行查询

复制代码
#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)
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练习:查询每个部门最新入职的那位员工

company.employee
    员工id      id                  int             
    姓名        emp_name            varchar
    性别        sex                 enum
    年龄        age                 int
    入职日期     hire_date           date
    岗位        post                varchar
    职位描述     post_comment        varchar
    薪水        salary              double
    办公室       office              int
    部门编号     depart_id           int

#创建表
create table employee(
id int
not null unique auto_increment,
name varchar(
20) not null,
sex enum(
'male','female') not null default 'male', #大部分是男的
age int(3) unsigned not null default 28,
hire_date date
not null,
post varchar(
50),
post_comment varchar(
100),
salary double(
15,2),
office int,
#一个部门一个屋子
depart_id int
);

#查看表结构
mysql> desc employee;
+--------------+-----------------------+------+-----+---------+----------------+
| Field | Type | Null | Key | Default | Extra |
+--------------+-----------------------+------+-----+---------+----------------+
| id | int(11) | NO | PRI | NULL | auto_increment |
| name | varchar(20) | NO | | NULL | |
| sex | enum('male','female') | NO | | male | |
| age | int(3) unsigned | NO | | 28 | |
| hire_date | date | NO | | NULL | |
| post | varchar(50) | YES | | NULL | |
| post_comment | varchar(100) | YES | | NULL | |
| salary | double(15,2) | YES | | NULL | |
| office | int(11) | YES | | NULL | |
| depart_id | int(11) | YES | | NULL | |
+--------------+-----------------------+------+-----+---------+----------------+

#插入记录

三个部门:教学,销售,运营

insert into employee(name,sex,age,hire_date,post,salary,office,depart_id) values
(
'egon','male',18,'20170301','老男孩驻沙河办事处外交大使',7300.33,401,1), #以下是教学部
('alex','male',78,'20150302','teacher',1000000.31,401,1),
(
'wupeiqi','male',81,'20130305','teacher',8300,401,1),
(
'yuanhao','male',73,'20140701','teacher',3500,401,1),
(
'liwenzhou','male',28,'20121101','teacher',2100,401,1),
(
'jingliyang','female',18,'20110211','teacher',9000,401,1),
(
'jinxin','male',18,'19000301','teacher',30000,401,1),
(
'成龙','male',48,'20101111','teacher',10000,401,1),

('歪歪','female',48,'20150311','sale',3000.13,402,2),#以下是销售部门
('丫丫','female',38,'20101101','sale',2000.35,402,2),
(
'丁丁','female',18,'20110312','sale',1000.37,402,2),
(
'星星','female',18,'20160513','sale',3000.29,402,2),
(
'格格','female',28,'20170127','sale',4000.33,402,2),

('张野','male',28,'20160311','operation',10000.13,403,3), #以下是运营部门
('程咬金','male',18,'19970312','operation',20000,403,3),
(
'程咬银','female',18,'20130311','operation',19000,403,3),
(
'程咬铜','male',18,'20150411','operation',18000,403,3),
(
'程咬铁','female',18,'20140512','operation',17000,403,3)
;

#ps:如果在windows系统中,插入中文字符,select的结果为空白,可以将所有字符编码统一设置成gbk

准备表和记录

表与数据准备
SELECT
    *
FROM
    emp AS t1
INNER JOIN (
    SELECT
        post,
        max(hire_date) max_date
    FROM
        emp
    GROUP BY
        post
) AS t2 ON t1.post = t2.post
WHERE
    t1.hire_date = t2.max_date;
答案一(连表查询)
mysql> select (select t2.name from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post;
+---------------------------------------------------------------------------------------+
| (select t2.name from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) |
+---------------------------------------------------------------------------------------+
| 张野                                                                                  |
| 格格                                                                                  |
| alex                                                                                  |
| egon                                                                                  |
+---------------------------------------------------------------------------------------+
rows in set (0.00 sec)

mysql> select (select t2.id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post;
+-------------------------------------------------------------------------------------+
| (select t2.id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) |
+-------------------------------------------------------------------------------------+
| 14 |
| 13 |
| 2 |
| 1 |
+-------------------------------------------------------------------------------------+
rows
in set (0.00 sec)

#正确答案
mysql> select t3.name,t3.post,t3.hire_date from emp as t3 where id in (select (select id from emp as t2 where t2.post=t1.post order by hire_date desc limit 1) from emp as t1 group by post);
+--------+-----------------------------------------+------------+
| name | post | hire_date |
+--------+-----------------------------------------+------------+
| egon | 老男孩驻沙河办事处外交大使 | 2017-03-01 |
| alex | teacher | 2015-03-02 |
| 格格 | sale | 2017-01-27 |
| 张野 | operation | 2016-03-11 |
+--------+-----------------------------------------+------------+
rows
in set (0.00 sec)

答案二(子查询)

答案一为正确答案,答案二中的limit 1有问题(每个部门可能有>1个为同一时间入职的新员工),我只是想用该例子来说明可以在select后使用子查询

可以基于上述方法解决:比如某网站在全国各个市都有站点,每个站点一条数据,想取每个省下最新的那一条市的网站质量信息

综合练习

init.sql文件内容

/*
 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 <span style="color: #0000ff;">class</span><span style="color: #000000;">
-- ----------------------------
DROP TABLE IF EXISTS </span><span style="color: #0000ff;">class</span><span style="color: #000000;">;
CREATE TABLE </span><span style="color: #0000ff;">class</span><span style="color: #000000;"> (
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 <span style="color: #0000ff;">class</span><span style="color: #000000;">
-- ----------------------------
BEGIN;
INSERT INTO </span><span style="color: #0000ff;">class</span> 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 </span><span style="color: #0000ff;">class</span><span style="color: #000000;"> (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;

init.sql

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

表结构为

 

复制代码
1、查询男生、女生的人数;

2、查询姓“张”的学生名单;

3、课程平均分从高到低显示

4、查询有课程成绩小于60分的同学的学号、姓名;

5、查询至少有一门课与学号为1的同学所学课程相同的同学的学号和姓名;

6、查询出只选修了一门课程的全部学生的学号和姓名;

7、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分;

8、查询课程编号“2”的成绩比课程编号“1”课程低的所有同学的学号、姓名;

9、查询“生物”课程比“物理”课程成绩高的所有学生的学号;

10、查询平均成绩大于60分的同学的学号和平均成绩;

11、查询所有同学的学号、姓名、选课数、总成绩;

12、查询姓“李”的老师的个数;

13、查询没学过“张磊老师”课的同学的学号、姓名;

14、查询学过“1”并且也学过编号“2”课程的同学的学号、姓名;

15、查询学过“李平老师”所教的所有课的同学的学号、姓名;

复制代码
1、查询没有学全所有课的同学的学号、姓名;
2、查询和“002”号的同学学习的课程完全相同的其他同学学号和姓名;
3、删除学习“叶平”老师课的SC表记录;
4、向SC表中插入一些记录,这些记录要求符合以下条件:①没有上过编号“002”课程的同学学号;②插入“002”号课程的平均成绩; 
5、按平均成绩从低到高显示所有学生的“语文”、“数学”、“英语”三门的课程成绩,按如下形式显示: 学生ID,语文,数学,英语,有效课程数,有效平均分;
6、查询各科成绩最高和最低的分:以如下形式显示:课程ID,最高分,最低分;
7、按各科平均成绩从低到高和及格率的百分数从高到低顺序;
8、查询各科成绩前三名的记录:(不考虑成绩并列情况) 
9、查询每门课程被选修的学生数;
10、查询同名同姓学生名单,并统计同名人数;
11、查询每门课程的平均成绩,结果按平均成绩升序排列,平均成绩相同时,按课程号降序排列;
12、查询平均成绩大于85的所有学生的学号、姓名和平均成绩;
13、查询课程名称为“数学”,且分数低于60的学生姓名和分数;
14、查询课程编号为003且课程成绩在80分以上的学生的学号和姓名; 
15、求选了课程的学生人数
16、查询选修“杨艳”老师所授课程的学生中,成绩最高的学生姓名及其成绩;
17、查询各个课程及相应的选修人数;
18、查询不同课程但成绩相同的学生的学号、课程号、学生成绩;
19、查询每门课程成绩最好的前两名;
20、检索至少选修两门课程的学生学号;
21、查询全部学生都选修的课程的课程号和课程名;
22、查询没学过“叶平”老师讲授的任一门课程的学生姓名;
23、查询两门以上不及格课程的同学的学号及其平均成绩;
24、检索“004”课程分数小于60,按分数降序排列的同学学号;
25、删除“002”同学的“001”课程的成绩;
更多练习

 

posted @ 2019-09-17 11:52  若如初见_you  阅读(100)  评论(0编辑  收藏  举报