mysql题目，求大神帮忙做一下 顺便给讲解一下，小白怎么做也有错误？

CREATE TABLE user (
id int(11) NOT NULL,
name VARCHAR(100),
no varchar(6),
id_number varchar(20),
birth_day datetime
)
COLLATE='utf8_general_ci'
ENGINE=InnoDB
ROW_FORMAT=DEFAULT

insert into user values(3,'张三','180202','322724189809723','1990-07-03');
select * from user a where a.id=3
update user a set a.name='李四' where a.id=3
delete from user where id=3
select * from user a where a.name like '%李四%'

校验

1、创建表
CREATE TABLE user (
id int(11) NOT NULL,
name VARCHAR(100),
no varchar(6),
id_number varchar(20),
birth_day datetime
)
COLLATE='utf8_general_ci'
ENGINE=InnoDB
ROW_FORMAT=DEFAULT

2、插入
insert into user values(3,'张三','180202','322724189809723','1990-07-03');
3、查询
select * from user a where a.id=3
4、更新
update user a set a.name='李四' where a.id=3
5、删除
delete from user where id=3
select * from user a where a.name like '%李四%'

这是数据库基础，涉及建表，添加数据，查询，修改，删除等等基本操作
已经基础到不能再基础了。你必须自己完成。

CREATE TABLE IF NOT EXISTS `user`(
`id` INT UNSIGNED AUTO_INCREMENT,
`name` VARCHAR(100) NOT NULL,
`student` VARCHAR(100) NOT NULL,
`idnumer` VARCHAR(100) NOT NULL,
`birthday` VARCHAR(100) NOT NULL,
PRIMARY KEY ( `id` )

)ENGINE=InnoDB DEFAULT CHARSET=utf8;

INSERT INTO `user` ( `name`, `student`,`idnumer`, `birthday`)
VALUES
( '张三', 182012,'372925199904024XX','1992-07-45');

SELECT * from `user`



UPDATE `user` SET `name`='李四' WHERE id=1

DELETE from `user` WHERE id=1

SELECT * FROM `user` WHERE name LIKE '%李四%';