A. event_choices
event_id | res_id |
4 | 10 |
B. restaurants
res_id | res_name |
10 | xyz |
C. event
event_id | event_name |
4 | birthday |
我一直在试图内连接尝试匹配名字ID,但没有成功如何将id与来自mysql中不同表的名称进行匹配?
select event_id as id from event_choices inner join restaurants on res_id.id = res_name
任何帮助,将不胜感激,很新的PHP/MySQL的
你试图得到什么数据? – 2012-07-30 17:48:52