数据库--续

13、数据CRUD–初级
13.0 增加数据
insert into customer values(0,’yonghuming123’,’password123’,’sexboy’,0);
insert into customer values(0,’yonghuming212’,’password234’,’sexgirl’,1);
需要按字段的顺序填写，否则就会报错
13.1 查询语句
select * from customer; 获得表中所有的数据
select name as ‘姓名’,gender from customer; 获得一个新的表格 name 和gender 并且将name改名为姓名。
13.2 按列插入数据
insert into customer(user_name,password) values(‘老张’,’123456’);
13.3 按列插入多行数据
insert into customer(user_name,password,email) values(‘用户名asd’,’123456’,’[email protected]’),(‘用户名12313’,’123456’,’[email protected]’);
13.4 修改数据
update customer set email=’[email protected]’ where id=3;

    13.5 删除数据
    delete from customer where id=1;
    13.6 逻辑删除
     update customer set is_delete=1 where id=5;
     select * from customer where is_delete=0;

14、数据CRUD–中级（条件）

    14.2 比较运算符的问题
    select * from customer where id > 7;
    select * from customer where id !=7;
    select * from customer where id <>7;(> >= < <= != <>)

    14.3逻辑运算符
    select * from customer where id > 7 and user_name='老王';（and or not）

    14.4 模糊查询
    select * from customer where user_name like '王%';
    %代表匹配任意的多个字符
    select * from customer where user_name like '王%' or user_name like '_王%' or user_name like '%王';

    14.5 范围查询
    不连续范围查询
    select * from customer where id in(3,7,9,100);
    连续范围
    select * from customer where id between 3 and 9;(包括序号3和9)

    14.6 null
    select * from customer where gender is null;

    14.7 not null
    select * from customer where gender is not null and user_name like '%王';

    14.8 顺序的问题
    括号 > not >比较运算符 > 逻辑运算符

    14.9 排序的问题
    desc降序 asc升序
    select * from customer order by user_name desc ,id desc;
    语句顺序的问题，如果第一个条件出现同值的情况，按照第二个语句来进行排序

15、数据CRUD–高级（查询结果）
14.1 删除重复行
select distinct user_name from customer;
15.1聚合函数
sum() max() min() avg() count()

    select avg(id) from customer where id >7 ;
    select sum(id)/count(id) as '均值' from customer where id >7 ;
    select count(id) from customer where id >7 ;

    15.2 分组的问题
    select gender from customer group by gender;
    select gender,group_concat(user_name) from customer group by gender;
    select gender,group_concat(user_name) from customer group by gender having id>3;
    select gender,group_concat(user_name,id) from customer group by gender having group_concat(id)>3;

    having 和 where 功能是一样的
    select * from customer having id>3;

    select gender,avg(id) from customer group by gender;
     15.3 分页的问题（解决数据量大的问题）
    select * from customer limit 0,5;

猜你喜欢