Question

我无法弄清楚这个SQL语句中的任何错误。任何人都可以请帮助我哪里有错误，我该如何解决这个问题？提前谢谢，对不起拼写或语法错误。

SQL：

SELECT count(*) as `total` FROM `pet_info` LEFT JOIN `lostpets` ON `lostpets`.`petid` = `pet_info`.`id` WHERE `lostpets`.`reunited`='No' AND (CASE WHEN `pet_info`.`pet_aggg` = 'birth' THEN `pet_info`.`pet_birthdate` <= '2011-05-12' WHEN `pet_info`.`pet_aggg` = 'age' THEN `pet_info`.`pet_age` >= '7');

Answer 1

只需添加＆＃39;结束＆＃39;在您的选择结束时：那么pet_info。pet_age＆gt; =＆＃39; 7＆＃39;端）;

Answer 2

这是您的查询：

SELECT count(*) as `total`
FROM `pet_info` LEFT JOIN
     `lostpets`
      ON `lostpets`.`petid` = `pet_info`.`id`
WHERE `lostpets`.`reunited`='No' AND
      (CASE WHEN `pet_info`.`pet_aggg` = 'birth'
            THEN `pet_info`.`pet_birthdate` <= '2011-05-12'
            WHEN `pet_info`.`pet_aggg` = 'age'
            THEN `pet_info`.`pet_age` >= '7'
      );

明显的语法问题是缺少END。但是你有其他问题。查询应该如下所示：

SELECT count(*) as `total`
FROM `pet_info` pi JOIN
     `lostpets` l
      ON l.`petid` = pi.`id`
WHERE l.`reunited`= 'No' AND
      ((pi.pet_aggg = 'birth' AND pi.pet_birthdate <= '2011-05-12') OR
       (pi.pet_aggg = 'age' AND pi.pet_age >= 7)
      );

注意：

您的WHERE子句会将LEFT JOIN变为INNER JOIN。
请勿在数值周围使用单引号。
CASE在WHERE条款中往往令人困惑。只需使用基本的布尔逻辑。

MYSQL SQL语法错误如何解决？

2 个答案: