时间:2021-07-01 10:21:17 帮助过:5人阅读
UPDATE school_more_info SET comments = replace( comments, '�', "'" ) WHERE school_id IN (SELECT school_id FROM school_more_info WHERE comments LIKE '%�%')
这个sql的意思是将comments字段中的�替换更新成‘, 更新的条件是只有含有�的行才更新,没有�的行则不更新。咋一看,应该没有错啊,我们经常这么写。最后查资料,mysql中不能这么用,那串英文错误提示就是说,不能先select出同一表中的某些值,再update这个表(在同一语句中)。
改写后的sql,这样是可以正确执行的。如下:
UPDATE school_more_info SET comments = replace( comments, '�', "'" ) WHERE school_id IN (SELECT school_id FROM (SELECT *FROM school_more_info WHERE comments LIKE '%�%') AS a)