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个具有id,名称和描述的对象。我可以使用join
关键字执行此操作吗?如果是这样的话......我怎么样,悄悄地研究几个小时。或者可能采用不同的方法?
如果我要创建一个我需要获得的输出数据表,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
答案 0 :(得分:5)
这不是join
的用例。由于您希望两个表中的行一个接一个,而不是并排,您应该使用union all
:
SELECT id, name, description
FROM browsers
UNION ALL
SELECT id, name, description
FROM cars