【发布时间】:2015-12-08 22:40:30
【问题描述】:
考虑以下数据库:
浏览器表:
id | name | description | stuff different from cars table
-------------------------------------------------------------------------------------
1 | Chrome | Some description
2 | Firefox | Some other description
3 | Vivaldi | Even more description
汽车表:
id | name | description | stuff different from browsers table
-------------------------------------------------------------------------------------
1 | Hyundai | Some korean description
2 | Ford | Some ford ther description
3 | Ferrari | Even ferrari more description
我需要在 PHP 中获得的输出是 6 个带有 id、名称和描述的对象。我可以用join 关键字来做到这一点吗?如果是这样...如何,我已经悄悄地研究了几个小时。或者也许是不同的方法?
如果我要制作一张我需要获取的输出数据的表格,那就是:
id | name | description
------------------------------------------------
1 | Hyundai | Some korean description
2 | Ford | Some ford ther description
3 | Ferrari | Even ferrari more description
1 | Chrome | Some description
2 | Firefox | Some other description
3 | Vivaldi | Even more description
【问题讨论】: