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mysql阶段总结

mysql阶段总结

作者: 明日计划 | 来源:发表于2018-07-16 20:22 被阅读0次

    title: 漂亮的mysql
    tags: database
    notebook: Database


    mysql总结:database篇

    #:代表错误对比
    1. create database data_name charset utf8;
    2. create database data_name charact set=utf8;
       create database if not exists data_name charset utf8;
    3. show create database data_name;
    4. #show database status; 查看整个数据库的参数状态;
    5. drop database data_name;
       #drop database if not exists data_name;
    6. alter database data_name charset utf8;
    7. #alter database if not exists data_name charact utf8;
    

    mysql总结:table篇

    1. create table table_name(
    id int(50) not null auto_increment primary key comment '身份证号',
    name varchar(100) not null comment '姓名',
    birthday date not null comment '出生日期'
    ) comment '表明';
    2. alter table table_name convert to charset utf8;
    3. rename table old_table_name to new_table_name;
    4. drop table table_name;
    
    table字段篇
    1. alter table table_name add 字段;
    2. alter table table_name modify 字段; 该字段用的是旧名,设置新的格式
    3. alter table table_name change old_column_name new_column_name 字段格式;
    4. alter table table_name drop column column_name; 
    
    table数据篇
    1. insert into table_name (id,name) values (23,'r3r3'),(232,'sisi');
    2. updata table_name set column_name='haha' where id=23;
    3. delete from table_name where id=32; id:标示要删的行数据
    4. truncate table_name; 删除表格,重建列表
    
    table查询篇
    1. select * from table_name where id=23;
    2. select id from data_name.table_name where id=43;
    3. select * from table_name where id=43 and name='haha';#and左右链接的两个条件同时满足
    4. select * from table_name where id=43 or name='haha';#or左右两个条件都作为要挑选的分段,即满足左条件select数据、满足右条件select数据,select出来的两种数据并存不冲突
       select name,age from table_name where (id=32 or id=43) and age >= 23;
    5. select name,age from table_name order by age desc,name;
    6. select name,age from table_name order by age asc;
    7. select name.age from table_name where age=32 and order by name;
    8. select name,age from table_name where age!=19 and age!=23;
    9. #select name,age from table_name where age!=19 or age!=23; 该语句不起作用
    10. select name,age from table_name where age not in (18,21,22);
    11. #select name,age from table_name where height=null;
    12. select name,age from table_name where height is null;
    13. select name,age from table_name where birthday between '1990/2/2' and '2000/3/2';
    14. select name,age from table_name order by age limit 3,4;
    15. select name from table_name order by age limit 4 offset 3;
    

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