SELECT * FROM time WHERE time_id <> ALL( SElECT time_id FROM time AS a WHERE (time_date > ( SELECT MIN(time_date) FROM time WHERE DATE(time_date) = DATE(a.time_date) AND time_name = '2' ) AND time_date < ( SELECT MAX(time_date) FROM time WHERE DATE(time_date) = DATE(a.time_date) AND time_name = '2' ) ) AND time_name = '2' ) AND time_name = '2';
虽然能找出同一日期中多余的指纹登陆时间!但是我这样写法有问题 查询数据库很慢需要1.5秒左右!知道问题出在 DATE(a.time_date) AND time_name = '2' 但是因为学习SQL没多久 不知道该怎么搞!求个大神答疑解惑!!小弟在此感谢
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