如何MySQL的UPDATE fieldINT!=“4”不起作用[关闭]
-
11-09-2019 - |
题
林试图更新queueStatusINT WHERE statusINT是8和queueStatusINT不等于2和类型为$类型。 不过,我不断收到错误:
ERROR 1064 (42000) at line 1: 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 ‘ queueStatusINT!='2’, type=’int'' at line 1
林使用该SQL查询来执行更新:
UPDATE $mysqlTable SET queueStatusINT='2’ WHERE statusINT='8’, queueStatusINT!='2’, type=‘$type’;
我也注意到,我可以在SELECT命令做一个不等于...
SELECT nameTXT FROM $mysqlTable WHERE queueStatusINT!='2' ORDER BY queueStatusINT DESC, priorityINT DESC, id ASC LIMIT 7;
解决方案
您需要使用与您的标准相结合,不只是逗号。
例如
UPDATE $mysqlTable
SET queueStatusINT = '2'
WHERE statusINT = '8'
AND queueStatusINT != '2'
AND type = '$type'
其他提示
您更新更改为:
UPDATE $mysqlTable
SET queueStatusINT='2’
WHERE statusINT=8
AND queueStatusINT !=2
AND type=‘$type’;
我认为queueStatusINT是一个整数(顾名思义) - 你应该离开了',因为它们象征着一个字符串/字符
祝, 费边
不隶属于 StackOverflow