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 |
我一直在試圖內連接嘗試匹配名字ID,但沒有成功如何將id與來自mysql中不同表的名稱進行匹配?
select event_id as id from event_choices inner join restaurants on res_id.id = res_name
任何幫助,將不勝感激,很新的PHP/MySQL的
你試圖得到什麼數據? – 2012-07-30 17:48:52