Question

我有两张桌子 - 活动和与会者。

我正在尝试使用php通过events.id点击与会者列表，并显示由id链接的与会者表中列出的与会者。

包含的字段是events.id和attendees.id，并显示该事件的与会者。

这就是我现在所拥有的。

SELECT events.*, attendees.* 
FROM attendees ON event.id = attendees.id 
WHERE event.id = attendees.id

Answer 1

您的连接语法是隐式和显式JOIN语法的错误组合，缺少JOIN关键字，而是复制WHERE子句中的连接条件。

SELECT
  events.*, 
  attendees.* 
FROM
  attendees 
  JOIN events ON event.id = attendees.id
WHERE
  event.id = <event to find attendees for>

请注意，在PHP中使用events.*, attendees.* 不可取，因为您将拥有PHP无法访问的重复列名。相反，要明确：

SELECT
  /* Be explicit about the columns you select in a JOIN query */
  events.id AS event_id,
  events.name AS event_name,
  events.someothercol,
  attendees.id AS attendee_id,
  attendees.name AS attendee_name
FROM
  attendees 
  JOIN events ON event.id = attendees.id
WHERE
  event.id = <event to find attendees for>

如果您仍想获取活动详细信息，即使没有与会者，请改为使用LEFT JOIN：

SELECT
  /* Be explicit about the columns you select in a JOIN query */
  events.id AS event_id,
  events.name AS event_name,
  events.someothercol,
  attendees.id AS attendee_id,
  attendees.name AS attendee_name
FROM
  events
  /* LEFT JOIN will return event details even when there are no attendees */
  LEFT JOIN attendees ON event.id = attendees.id
WHERE
  event.id = <event to find attendees for>

Answer 2

如果您有两个这样的表：

events:  
  ID  
  ...  

attendees:  
  event  
  ...

您需要获取与其活动相关的与会者列表，请尝试以下操作：

SELECT * FROM attendees LEFT OUTER JOIN events ON attendees.event = events.ID

如果没有，请澄清你的问题。

使用Id显示链接表中的MySql数据

2 个答案: