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)
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