如何將ID與mysql中不同表的名稱匹配？

Question

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   | 

我一直在嘗試內部聯接以嘗試將名稱與ID匹配，但未成功

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，這是php / mysql的新手

Answer 1

嘗試：

SELECT event_id, res_name FROM event_choices, restaurants WHERE event_choices.res_id = restaurants.res_id

享受^ _ ^

Answer 2

SELECT event.event_name, restaurants.res_name FROM event_choices 
    INNER JOIN restaurants ON event_choices.res_id = restaurants.res_id
    INNER JOIN event ON event_choices.event_id = event.event_id 

可以工作。

如果選擇*而不是event.event_name ，則應該獲取所有數據並能夠從中進行選擇。 您還可以選擇<table>.<field>形式的特定字段。

編輯：添加res_name

Answer 3

嘗試這個：

  select ec.event_id as id from event_choices ec inner join restaurants r on ec.res_id = r.res_id