如何在springboot中实现简单的草稿箱

建表sql语句相关代码

建表

我们只需要给表加一个状态的字段来进行判断他是不是草稿即可 例如

CREATE TABLE `recipe` (

`id` int NOT NULL AUTO_INCREMENT,

`recipe_name` varchar(220) CHARACTER SET utf8 COLLATE utf8_general_ci NOT NULL,

`author` int NOT NULL,

`time` varchar(220) CHARACTER SET utf8 COLLATE utf8_general_ci DEFAULT NULL,

`recipe_picture` varchar(220) CHARACTER SET utf8 COLLATE utf8_general_ci NOT NULL,

`introduction` varchar(1000) CHARACTER SET utf8 COLLATE utf8_general_ci NOT NULL,

`quick` int(1) unsigned zerofill DEFAULT '0',

`difficulty` varchar(255) CHARACTER SET utf8 COLLATE utf8_general_ci NOT NULL,

`need_time` varchar(220) CHARACTER SET utf8 COLLATE utf8_general_ci NOT NULL,

`collectNums` int DEFAULT '0',

`status` int DEFAULT '0',

PRIMARY KEY (`id`) USING BTREE

) ENGINE=InnoDB AUTO_INCREMENT=187 DEFAULT CHARSET=utf8mb3

sql语句

1.将内容填入表中并且指定状态

insert into recipe( recipe_name, author, time, recipe_picture, introduction

, difficulty, need_time, status)

values (#{recipe_name}, #{author}, #{time}, #{recipe_picture}, #{introduction}, #{difficulty},

#{need_time}, 3)

2.如果对草稿箱有改动

update recipe set recipe_name = #{recipe_name},

author = #{author},

time = #{time},

recipe_picture = #{recipe_picture},

introduction = #{introduction},

difficulty = #{difficulty},

need_time = #{need_time},

status = #{status}

这时可以指定是保存还是上传草稿箱的内容

相关代码

1.dao层 2.service层 3.impl层 之后在controller层接收数据即可。