php – MySQL ON DUPLICATE KEY UPDATE问题
发布时间:2020-05-26 01:54:37 所属栏目:PHP 来源:互联网
导读:嗨,有人可以看看这个并告诉我哪里出错了. 我有一个SQL语句,当我使用 PHP回应时,我得到这个屏幕 INSERT INTO moviedb.genre SET GenreID = 18 , GenreName = Drama ON DUPLICATE KEY UPDATE GenreName = Drama WHERE GenreID = 18INSER
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嗨,有人可以看看这个并告诉我哪里出错了.
INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '18','GenreName' = 'Drama' ON DUPLICATE KEY UPDATE 'GenreName' = 'Drama' WHERE 'GenreID' = '18' INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '16','GenreName' = 'Animation' ON DUPLICATE KEY UPDATE 'GenreName' = 'Animation' WHERE 'GenreID' = '16' 这是声明 $sql="INSERT INTO 'moviedb'.'genre' SET 'GenreID' = '{$genresID[$i]}','GenreName' = '{$genreName[$i]}' ON DUPLICATE KEY UPDATE 'GenreName' = '{$genreName[$i]}' WHERE 'GenreID' = '{$genresID[$i]}'";
这是我收到的错误: 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 ''moviedb'.'genre' SET 'GenreID' = '18','GenreName' = 'Drama' ON DUPLICATE KEY ' at line 1 任何帮助将不胜感激,提前感谢. 您不能将WHERE与ON DUPLICATE KEY组合使用.删除WHERE子句,MySql只会更新导致重复键的行. 对于多行INSERT,使用VALUES()告诉MySql更新将插入的值,例如: INSERT INTO moviedb.genre (GenreID,GenreName)
VALUES ('18','Drama'),('16','Animation')
ON DUPLICATE KEY UPDATE
GenreName = VALUES(GenreName); (编辑:安卓应用网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
