MySQL 连接错误 : Every derived table must have its own alias

Question

我查看了围绕 S.O 的可能解决方案，但我的查询仍然无法正常工作。

这是查询:

SELECT `messages`.`from_id`, 
  `messages`.`to_id`, 
  `messages`.`message` 
FROM `messages`
JOIN 
(
  SELECT `users`.`profile_pic` 
  FROM `users` AS `from_pic` 
  WHERE `users`.`id` = `from_id`
)  
  ON `users`.`id` = `messages`.`from_id` 
JOIN 
(
  SELECT `users`.`profile_pic` 
  FROM `users` AS `to_pic` 
  WHERE `users`.`id` = `to_id`
)  
  ON `users`.`id` = `messages`.`to_id` 
WHERE  `messages`.`from_id` = 23 
  AND `messages`.`to_id` = 24 
ORDER BY `date_sent` DESC;

表用户

id | profile_pic | name

表消息

message_id | from_id | to_id | message

Answer 1

当你使用子查询时，你需要为它们提供一个别名:

SELECT 
  `messages`.`from_id`, 
  `messages`.`to_id`, 
  `messages`.`message` 
FROM `messages`
JOIN 
(
  SELECT 
    `from_pic`.`id`, 
    `from_pic`.`profile_pic` 
  FROM `users` AS `from_pic` 
)  u1 --- alias
  ON u1.`id` = `messages`.`from_id` 
JOIN 
(
  SELECT 
    `to_pic`.`id`,
    `to_pic`.`profile_pic` 
  FROM `users` AS `to_pic` 
)  u2 ---- alias
  ON u2.`id` = `messages`.`to_id` 
WHERE  `messages`.`from_id` = 23 
  AND `messages`.`to_id` = 24 
ORDER BY `date_sent` DESC;

这也可以不用子查询来写:

SELECT 
  `messages`.`from_id`, 
  `messages`.`to_id`, 
  `messages`.`message` 
FROM `messages`
JOIN `users` AS `from_pic` 
  ON `from_pic` = `messages`.`from_id` 
JOIN `users` AS `to_pic` 
  ON `to_pic`.`id` = `messages`.`to_id` 
WHERE  `messages`.`from_id` = 23 
  AND `messages`.`to_id` = 24 
ORDER BY `date_sent` DESC;