php - 如何将 id 与 mysql 中不同表的名称匹配？

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   | 

我一直在尝试内部连接以尝试将名称与 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，对 php/mysql 来说非常新

Answer 1

尝试：

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

欣赏^_^

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。

如果您选择*而不是event.event_name您应该获取所有数据并能够从中进行选择。您还可以选择 形式的特定字段<table>.<field>。

编辑：添加res_name

Answer 3

尝试这个：

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