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  • 多表查询

    一:介绍

    多表连接查询

    复合条件查询

    子查询

    准备表

    #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
    

    二:多表联合查询

    一:多表联合语法

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

    二:交叉连接:不适用于任何匹配条件。生成笛卡尔积

    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         |
    +----+------------+--------+------+--------+------+--------------+
    

    三:内连接,只连接匹配的行

    #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;
    

    四:外连接之左连接:优先显示左表全部记录

    #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         |
    +----+------------+--------------+
    

    五:外连接之右连接,优先显示右表全部记录

    #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         |
    +------+-----------+--------------+
    

    六:全外连接:显示左右两个表全部记录

    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
    

    三:符合条件连接查询

    #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;
    

    四: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
    

    一:带IN关键字的子查询

    #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);
    

    注意问题

    not in 无法处理null的值,即子查询中如果存在null的值,not in将无法处理,如下
    
    mysql> select * from emp;
    +----+------------+--------+------+--------+
    | 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 |
    | 7 | xxx | male | 19 | NULL |
    +----+------------+--------+------+--------+
    7 rows in set (0.00 sec)
    
    mysql> select * from dep;
    +------+--------------+
    | id | name |
    +------+--------------+
    | 200 | 技术 |
    | 201 | 人力资源 |
    | 202 | 销售 |
    | 203 | 运营 |
    +------+--------------+
    4 rows in set (0.00 sec)
    
    # 子查询中存在null
    mysql> select * from dep where id not in (select distinct dep_id from emp);
    Empty set (0.00 sec)
    
    # 解决方案如下
    mysql> select * from dep where id not in (select distinct dep_id from emp where dep_id is not null);
    +------+--------+
    | id | name |
    +------+--------+
    | 203 | 运营 |
    +------+--------+
    1 row in set (0.00 sec)
    
    mysql>
    
    !!!注意not in
    

    二:带ANY关键字的子查询

    #在 SQL 中 ANY 和 SOME 是同义词,SOME 的用法和功能和 ANY 一模一样。
    
    # ANY 和 IN 运算符不同之处1
    ANY 必须和其他的比较运算符共同使用,而且ANY必须将比较运算符放在 ANY 关键字之前,所比较的值需要匹配子查询中的任意一个值,这也就是 ANY 在英文中所表示的意义
    
    例如:使用 IN 和使用 ANY运算符得到的结果是一致的
    select * from employee where salary = any (
    select max(salary) from employee group by depart_id);
    
    select * from employee where salary in (
    select max(salary) from employee group by depart_id);
    
    结论:也就是说“=ANY”等价于 IN 运算符,而“<>ANY”则等价于 NOT IN 运算符
    
    # ANY和 IN 运算符不同之处2
    ANY 运算符不能与固定的集合相匹配,比如下面的 SQL 语句是错误的
    
    SELECT
    *
    FROM
    T_Book
    WHERE
    FYearPublished < ANY (2001, 2003, 2005)
    

    三: 带ALL关键字的子查询

    # all同any类似,只不过all表示的是所有,any表示任一
    查询出那些薪资比所有部门的平均薪资都高的员工=》薪资在所有部门平均线以上的狗币资本家
    select * from employee where salary > all (
    select avg(salary) from employee group by depart_id);
    查询出那些薪资比所有部门的平均薪资都低的员工=》薪资在所有部门平均线以下的无产阶级劳苦大众
    select * from employee where salary < all (
    select avg(salary) from employee group by depart_id);
    
    查询出那些薪资比任意一个部门的平均薪资低的员工=》薪资在任一部门平均线以下的员工select * from employee where salary < any ( select avg(salary) from employee group by depart_id); 
    查询出那些薪资比任意一个部门的平均薪资高的员工=》薪资在任一部门平均线以上的员工
    select * from employee where salary > any (
    select avg(salary) from employee group by depart_id);
    

    四:带比较运算的子查询

    #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;
    

    五: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

    #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)
    
    一:in与exists
    !!!!!!当in和exists在查询效率上比较时,in查询的效率快于exists的查询效率!!!!!!
    ==============================exists==============================
    # exists
    exists后面一般都是子查询,后面的子查询被称做相关子查询(即与主语句相关),当子查询返回行数时,exists条件返回true,
    否则返回false,exists是不返回列表的值的,exists只在乎括号里的数据能不能查找出来,是否存在这样的记录。
    
    # 例
    查询出那些班级里有学生的班级
    select * from class where exists (select * from stu where stu.cid=class.id)
    
    # exists的执行原理为:
    1、依次执行外部查询:即select * from class 
    2、然后为外部查询返回的每一行分别执行一次子查询:即(select * from stu where stu.cid=class.cid)
    3、子查询如果返回行,则exists条件成立,条件成立则输出外部查询取出的那条记录
    
    ==============================in==============================
    # in
    in后跟的都是子查询,in()后面的子查询 是返回结果集的
    
    # 例
    查询和所有女生年龄相同的男生
    select * from stu where sex='男' and age in(select age from stu where sex='女')
    
    # in的执行原理为:
    in()的执行次序和exists()不一样,in()的子查询会先产生结果集,
    然后主查询再去结果集里去找符合要求的字段列表去.符合要求的输出,反之则不输出.
    
    二:not in与 not exists
    !!!!!!not exists查询的效率远远高与not in查询的效率。!!!!!!
    
    ==============================not in==============================
    not in()子查询的执行顺序是:
    为了证明not in成立,即找不到,需要一条一条地查询表,符合要求才返回子查询的结果集,不符合的就继续查询下一条记录,直到把表中的记录查询完,只能查询全部记录才能证明,并没有用到索引。
                    
    ==============================not exists==============================
    not exists:
    如果主查询表中记录少,子查询表中记录多,并有索引。
    例如:查询那些班级中没有学生的班级
    select * from class
    
    where not exists
    
    (select * from student where student.cid = class.cid)
    
    not exists的执行顺序是:
    在表中查询,是根据索引查询的,如果存在就返回true,如果不存在就返回false,不会每条记录都去查询。
    
    数据准备
    create database db13;
    
    use db13
    
    create table student(
        id int primary key auto_increment,
        name varchar(16)
    );
    
    create table course(
        id int primary key auto_increment,
        name varchar(16),
        comment varchar(20)
    );
    
    create table student2course(
        id int primary key auto_increment,
        sid int,
        cid int,
        foreign key(sid) references student(id),
        foreign key(cid) references course(id)
    );
    
    
    insert into student(name) values
    ("egon"),
    ("lili"),
    ("jack"),
    ("tom");
    
    insert into course(name,comment) values
    ("数据库","数据仓库"),
    ("数学","根本学不会"),
    ("英语","鸟语花香");
    
    
    insert into student2course(sid,cid) values
    (1,1),
    (1,2),
    (1,3),
    (2,1),
    (2,2),
    (3,2);
    
    示例
    # 1、查询选修了所有课程的学生id、name:(即该学生根本就不存在一门他没有选的课程。)
    select * from student s where not exists
        (select * from course c where not exists
            (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));
    
    
    select s.name from student as s
    inner join student2course as sc
    on s.id=sc.sid
    group by s.name 
    having count(sc.id) = (select count(id) from course);
    
    # 2、查询没有选择所有课程的学生,即没有全选的学生。(存在这样的一个学生,他至少有一门课没有选)
    select * from student s where exists
        (select * from course c where not exists
            (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));
    
    # 3、查询一门课也没有选的学生。(不存这样的一个学生,他至少选修一门课程)
    select * from student s where not exists
        (select * from course c where exists
            (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));
    
    # 4、查询至少选修了一门课程的学生。
    select * from student s where exists
        (select * from course c where exists
            (select * from student2course sc where sc.sid=s.id and sc.cid=c.id));
    
    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
    

    答案

    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(链表)
    

    答案二

    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

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  • 原文地址:https://www.cnblogs.com/wait59/p/13634287.html
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