Address 为Innodb表，AppUser 为FEDERATED表，下面的连接查询如果用INNER JOIN 就正常，用left join连接的结果Name全为null，即使有匹配的行，怀疑是mysql的bug
select a.*,b.Name from Address AS aLEFT JOIN AppUser AS b ON b.AID = a.CheckAdminIDwhere a.AID=1006948
暂时解决办法：将AppUser改名AppUser1，建个视图AppUser引用AppUser1，结果正常。
转载于:https://www.cnblogs.com/seawwh/archive/2011/10/28/2227403.html
