MYSQL 连接错误。如何解决？

Question

这个问题已经有答案了:

Mysql CASE WHEN JOIN Statement Error (1 个回答)

已关闭 6 年前。

MYSQL查询:

SELECT * FROM `pet_info` LEFT JOIN `lostpets` ON `pet_info`.`id` = `lostpets`.`petid` LEFT JOIN `pet_images` ON `pet_info`.`id` = `pet_images`.`petid` LEFT JOIN ON `pet_info`.`pet_cat` CASE WHEN `pet_info`.`pet_cat` = 1 THEN `dog_breeds`.`id` WHEN `pet_info`.`pet_cat` = 2 THEN `cat_breeds`.`id` END WHERE `pet_info`.`pet_user_id` = 581

错误:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'ON `pet_info`.`pet_cat` CASE WHEN `pet_info`.`pet_cat` = 1 THEN `do' at line 7

为什么出现错误，请帮助我。我可以解决这个错误吗？

Answer 1

这是您的查询的格式化版本，突出显示了一些错误。

SELECT * 
  FROM pet_info p
  LEFT 
  JOIN lostpets l
    ON p.id = l.petid 
  LEFT 
  JOIN pet_images i
    ON p.id = i.petid 
  LEFT 
  JOIN  !! ERROR 1
    ON p.pet_cat !! ERROR 2
  CASE WHEN p.pet_cat = 1 
       THEN d.id !! ERROR 3
       WHEN p.pet_cat = 2 
       THEN c.id END !! ERROR 4
 WHERE p.pet_user_id = 581