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排序 order by
desc降序排序
asc 升序排序 默认不写就是升序排序
1.查询所有学员信息 并按照年龄进行降序排序
select * from students order by age desc;
2.查询班级为python的男生信息 并按照年龄进行降序排序
select * from students where class='python' and gender='1' order by age desc;
分页获取数据 limit
limit 1 ; 从前面获取前一条数据
limit 1,1; 跳过第一条数据取一条数据
1.获取前三个学员的数据
select * from students limit 3;
2.获取第二个学员的信息
select * from students limit 1,1;
使用limit实现分页效果
要求 没两条数据为一页
取第一页数据
select * from students limit 0,2; （1-1）*每一页数据的条数
取第二页数据
select * from students limit 2,2; (2-1)*2=2
取第三页
select * from students limit 4,2; （3-1）*2=4
取第五页数据 （5-1）*2=8
select * from students limit (5-1)*2,2;
练习：
1.获取Python班级年龄最大的2个学生
select * from students where class='python' order by age desc limit 2;
2.获取每个班级的平均年龄，并按照平均年龄进行降序排序
select class,avg(age) from students group by class order by avg(age) desc;
3.统计每个班级的人数 获取人数最多的两个班级
select class,count(*) from students group by class order by count(*) desc limit 2;
顺序问题：
weher group by order by limit;
嵌套子查询
将一条查询一句 嵌入到另一条查询语句当中去
1.可做为数据源
2.可以作为查询条件
1.查询年龄大于18岁的男生
select * from students where age>18 and gender='1';
select * from students where age>18;
select * from (select * from students where age>18) as newtable where newtable.gender='1';
查询每个班级的男生和女生的人数 获取女生人数为0的班级
select class,gender,count(*) from students group by class,gender; 就把没有女生的班级排出掉了
查询的有女生的班级
1.select class from students group by class,gender having gender!='1'
2.拿1和原表数据做取反 最终得到是没有女生的班级
select * from students where class not in (select class from students group by class,gender having gender!='1');
第二种方式
select class,count(*) from students group by class having count(*)=sum(gender);
数据库关系搭建
一对一
一张表中唯一一条数据对应另一张表中唯一一条数据
users
id uname phone password age email pic address
把一张表的数据库查分成两张表，并通过外键字段将两张表中的数据进行关联
user
id uname password
1 张三 123
2 李四 123
uinfo
id phone age emial pic address u_id
1 110 18 [email protected] 1.jpg beijing 1
2 120 25 123 123 231 2
1.创建两个表
create table user(
id int unsigned not null auto_increment primary key,
uname char(32)
)
create table uinfo(
id int unsigned not null auto_increment primary key,
phone varchar(11),
age int,
email varchar(32),
gender varchar(10),
address varchar(255) default '北京',
u_id int
)
2.插入数据
insert into user values(null,'老张'),(null,'老牛'),(null,'老刘'),(null,'老钟');
insert into uinfo values(null,'110',18,'[email protected]','1','北京昌平',1),
(null,'120',28,'[email protected]','1','北京朝阳',2),
(null,'119',38,'[email protected]','1','黑龙江',3);
通过用户去查询用户的详情
根据用户名老张 查询老张的详细信息
1.select id from user where uname='老张';
2.select * from uinfo where u_id=1;
select * from uinfo where u_id=(select id from user where uname='老刘');
通过用户详情去查询用户
已知手机号为120 查询手机号对应的用户
select * from user where id=(select u_id from uinfo where phone='120');
查询年龄为28和38的用户信息
1.select u_id from uinfo where age=28 or age=38;
2.select * from user where id in(select u_id from uinfo where age=28 or age=38);
一对多
一张表中的一条数据 对应另一张表中的多条数据
parent
id name
1 老张
2 老李
child
id name p_id
1 张1 1
2 张2 1
3 李1 2
4 李2 2
根据父亲找儿子
查找老张所有的孩子
select * from child where p_id=(select id from parent where name='老张');
根据儿子找父亲
查询张1的父亲
select * from parent where id=(select p_id from child where name='张1' );
多对多
一张表中的多条数据 对应另一张表中的多条数据
teachers
id name
1 老张
2 老李
3 老边
students
id name
1 小黄
2 小兰
3 小绿
4 小花
5 小红
t_s
t_id s_id
1 1
1 2
1 3
3 5
1 5
2 5
1.创建表
create table teacher(
id int unsigned not null auto_increment primary key,
name varchar(10)
);
create table teacher_student(
id int unsigned not null auto_increment primary key,
t_id int,
s_id int
);
2.添加数据
insert into teacher values(null,'老刘'),(null,'老边'),(null,'老温');
表数据：
老师：
+----+------+
| id | name |
+----+------+
| 1 | 老刘 |
| 2 | 老边 |
| 3 | 老温 |
+----+------+
学生：
+----+--------+------+--------+-----------------+--------+
| id | name | age | gender | email | class |
+----+--------+------+--------+-----------------+--------+
| 2 | 小张 | 19 | 1 | [email protected] | python |
| 5 | 赵六 | 30 | 1 | [email protected] | web |
| 6 | 田七 | NULL | NULL | NULL | NULL |
| 7 | 王八 | 28 | 0 | [email protected] | ui |
| 8 | 九九 | 19 | 1 | [email protected] | python |
| 9 | 后羿 | 80 | 1 | [email protected] | python |
| 10 | 庄周 | NULL | NULL | NULL | NULL |
| 11 | 阿珂 | 18 | 0 | NULL | python |
| 12 | 贾克斯 | 24 | 1 | NULL | 上单 |
+----+--------+------+--------+-----------------+--------+
关系：
添加关系
1 2
1 5
1 12
2 2
2 5
2 12
2 7
insert into teacher_student values(null,1,2),(null,1,5),(null,1,12),(null,2,2),(null,2,5),(null,2,12),(null,2,7);
+----+------+------+
| id | t_id | s_id |
+----+------+------+
| 1 | 1 | 2 |
| 2 | 1 | 5 |
| 3 | 1 | 12 |
| 4 | 2 | 2 |
| 5 | 2 | 5 |
| 6 | 2 | 12 |
| 7 | 2 | 7 |
+----+------+------+
查询：
查老刘老师带过的所有学生的信息
1.查老师的id
select id from teacher where name='老刘';
2.根据老师的id去关系表查对应学生id;
select s_id from teacher_student where t_id=(select id from teacher where name='老刘');
3.根据学生id去查学生表
select * from students where id in(select s_id from teacher_student where t_id=(select id from teacher where name='老刘'));
查老刘老师带过的年龄大于19岁的学生信息
select * from students where id in (select s_id from teacher_student where t_id in(select id from teacher where name='老刘')) and age>19;
1.根据名字查询学生id
select id from students where name='贾克斯'
2.根据学生的id去关系表查询对应t_id
select t_id from teacher_student where s_id=(select id from students where name='贾克斯');
3.根据t_id 去老师表查
select * from teacher where id in(select t_id from teacher_student where s_id=(select id from students where name='贾克斯'));
外键
逻辑外键
物理外检 会增加表和表之间的复杂度，会降低查询的速度
外键约束：foreign key(当前表的外键字段) references 关联的表名(关联表的字段)
外键主要使用约束一对多关系
一般情况下 多表中的外键字段 关联一表中的主键
学员和宿舍的关系
student dormitory
id num
1.创建表
create table dormitory(
id int unsigned not null auto_increment primary key,
num int
);
2.学生
create table student(
id int unsigned not null auto_increment primary key,
name char(32),
age int,
gender int,
d_id int unsigned,
foreign key(d_id) references dormitory(id)
);
专业mojar 课程course mojar_course
create table mojar(
id int unsigned not null auto_increment primary key,
name char(32)
);
create table course(
id int unsigned not null auto_increment primary key,
name char(32)
);
create table mojar_course(
id int unsigned not null auto_increment primary key,
m_id int unsigned,
c_id int unsigned,
foreign key(m_id) references mojar(id),
foreign key(c_id) references course(id)
);
注意：添加外键约束时
1.要先有外键字段
2.外键字段的类型 必须和要关联的字段类型保持一至
多表查询
1.嵌套查询
2.where 关联查询
只能查询两张表都有对应数据的信息
1.查看每个学生的宿舍信息
select * from student s,dormitory d where s.d_id=d.id;
2.查询小红住在哪个宿舍
select s.id,s.name,d.num from student s,dormitory d where s.d_id=d.id and name='小红';
3.所有专业对应的课程
select * from mojar m,mojar_course m_c,course c where m.id=m_c.m_id and m_c.c_id=c.id;
4.查询python都要学习那些课程
select m.id,m.name,c.name from mojar m,mojar_course m_c,course c where m.id=m_c.m_id and m_c.c_id=c.id and m.name='python';
5.查询毛概都有哪些专业在学
连接查询
inner join 内链接 结果和where关联查询结果一样
格式： select * from 表1 inner join 表2 on 关联条件
1.查看每个学生的宿舍信息
select * from student s inner join dormitory d on s.d_id=d.id;
2.查询小红住在哪个宿舍
select s.id,s.name,d.num from student s inner join dormitory d on s.d_id=d.id and s.name='小红';
3.所有专业对应的课程
select * from
mojar m
inner join
mojar_course m_c
on
m.id=m_c.m_id
inner join
course c
on
m_c.c_id=c.id;
left join 外左连接
以左表为准，将坐表中所有的数据查询出来，去右表匹配信息，
如果没有对应的信息以null占位
1.查看每个学生的宿舍信息
select * from student s left join dormitory d on s.d_id=d.id;
2.哪些学员没有宿舍
select * from student s left join dormitory d on s.d_id=d.id where d.num is null;
3.哪些宿舍是空宿舍
select * from dormitory d left join student s on s.d_id=d.id where s.id is null;
4.统计每个宿舍住了多少人(有问题)
方法一：直接去统计字段 s.id就ok了
select d.num,count(s.id) from dormitory d left join student s on s.d_id=d.id group by d.num;
select d.num,count(s.id) from dormitory d left join student s on s.d_id=d.id group by d.num;
方法2：
1.先查询学员按照 外键宿舍id号进行分组 得到空宿舍
mysql> select d_id,count(*) from student group by d_id;
+------+----------+
| d_id | count(*) |
+------+----------+
| NULL | 1 |
| 1 | 3 |
| 2 | 2 |
| 3 | 2 |
+------+----------+
4 rows in set (0.00 sec)
2.把返回的结果 和宿舍进行关联 返回时判断d_id如果为空就输出 0，不为空就输出统计的值
select d.id,d.num,if(news.d_id is null,0,news.num) from
dormitory d
left join
(select d_id,count(*) num from student group by d_id) as news
on
news.d_id=d.id;
+----+------+----------------------------------+
| id | num | if(news.d_id is null,0,news.num) |
+----+------+----------------------------------+
| 1 | 208 | 3 |
| 2 | 209 | 2 |
| 3 | 520 | 2 |
| 4 | 911 | 0 |
+----+------+----------------------------------+
right join 外右链接
以右表为准，将右表中所有的数据查询出来，去左表匹配信息
1.查看每个学生的宿舍信息
select * from dormitory d right join student s on d.id=s.d_id;
练习：
1.有表 演员表 和电影表
一部电影有多个演员
一个演员可以演多部电影
actors:
id name
films:
id name
actors_films:
actors_id films_id
查询两张表所有的信息
select * from actors a inner join actors_films af on a.id=af.actors_id inner join films f on af.films_id=f.id;
查询演员 甄子丹 演出的电影
select * from actors a inner join actors_films af on a.id=af.actors_id inner join films f on af.films_id=f.id where a.name='甄子丹';
统计 甄子丹 演出了多少部电影
select a.name,count(*) from actors a inner join actors_films af on a.id=af.actors_id inner join films f on af.films_id=f.id group by a.name having a.name='甄子丹';
查询 扫毒电影 都有哪些演员
select f.name,a.name from actors a inner join actors_films af on a.id=af.actors_id inner join films f on af.films_id=f.id where f.name='扫毒电影';
查统计 扫毒 一共有多少个演员
select f.name,count(a.name) from actors a inner join actors_films af on a.id=af.actors_id inner join films f on af.films_id=f.id group by f.name having f.name='扫毒';