How should i store data like users's gender, religion, political views which is selecting from a list of 2-8 max values like 'male', 'female' or 'orthodox', 'muslim','judaism','catholic' etc? Also this values is constant, even admin cannot change 'female' to something else. In a Database it looks wierd to store a similar tables with only this 2-8 values and make JOIN with a parent table on foreign key. Second way - special object inside program code - but it's always bad to mix program logic with a data.
2条回答 默认 最新
- douyao1994 2015-04-27 14:25关注
Whether or not something is looking "weird" depends on personal preferences or design structures. However, it is entirely logical to store anything in a database that has to do with, well, data. Even a given set of options can change in the distant or not so distant future. I can't count the times a client asked me to change a set of options a day, a week, or even a few years after having ensured me that the set wouldn't change, ever.
Storing a list of options in a separate table is part of a relational database design. Relational database designs make it easy to get a set of data which includes or even excludes the options in any way in my opinion.
I'd recommend doing it the good, old fashioned way, for example:
- Table
user
(id, user_name) - Table
option
(id, option_label) - Table
user_option
(id, user_id, option_id)
A user that is both
male
andcatholic
would have a relation with two options:Table user Table option Table user_option +----+-----------+ +----+--------------+ +----+---------+-----------+ | id | user_name | | id | option_label | | id | user_id | option_id | +----+-----------+ +----+--------------+ +----+---------+-----------+ | 1 | john | | 1 | male | | 1 | 1 | 1 | | 2 | melody | | 2 | female | | 2 | 1 | 6 | | 3 | gerald | | 3 | orthodox | +----+---------+-----------+ +----+-----------+ | 4 | muslim | | 5 | judaism | | 6 | catholic | +----+--------------+
Showing all selected options per user can be done with the following query:
SELECT `u`.*, GROUP_CONCAT( `o`.`option_label` SEPARATOR ', ' ) AS `options` FROM `user` AS `u` LEFT JOIN `user_option` AS `uo` ON `uo`.`user_id` = `u`.`id` LEFT JOIN `option` AS `o` ON `uo`.`option_id` = `o`.`id`
本回答被题主选为最佳回答 , 对您是否有帮助呢?解决 无用评论 打赏 举报 - Table
悬赏问题
- ¥15 banner广告展示设置多少时间不怎么会消耗用户价值
- ¥16 mybatis的代理对象无法通过@Autowired装填
- ¥15 可见光定位matlab仿真
- ¥15 arduino 四自由度机械臂
- ¥15 wordpress 产品图片 GIF 没法显示
- ¥15 求三国群英传pl国战时间的修改方法
- ¥15 matlab代码代写,需写出详细代码,代价私
- ¥15 ROS系统搭建请教(跨境电商用途)
- ¥15 AIC3204的示例代码有吗,想用AIC3204测量血氧,找不到相关的代码。
- ¥20 CST怎么把天线放在座椅环境中并仿真