标签: php mysqli prepared-statement inner-join
所以我尝试使用select使用预准备语句进行inner join查询,它看起来像这样:
select
inner join
if ( !$prepared = $link->prepare('SELECT * FROM ? INNER JOIN Members ON ?.Member_UID = Members.Member_UID') ) { exit('Could not prepare the query! :S'); }
那将返回false并退出脚本,因为当然无法准备查询,我做错了什么。?
false