如何将id与mysql中不同表的名称相匹配？

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   | 

我一直在尝试使用内部联接来尝试将名称与id匹配但不成功

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，非常新的php / mysql

Answer 1

尝试：

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 

可行。

如果您选择*而非event.event_name，则应获取所有数据并从中进行选择。您还可以选择<table>.<field>。

形式的特定字段

编辑：添加res_name

Answer 3

试试这个：

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