sqlite删除与当前时间差值在10秒以上的记录

概述sqlite> delete from test1 where julianday('now')*86400 -julianday(a_date)*86400>10; sqlite的最小时间差的单位是天，当然是小数类型的，不是整型的，所以可以乘以86400（=24*3600）来计算相差的秒数 sqlite> select julianday('2016-05-31') - jul

sqlite> delete    from   test1  where julianday('Now')*86400 -julianday(a_date)*86400>10;

sqlite的最小时间差的单位是天，当然是小数类型的，不是整型的，所以可以乘以86400（=24*3600）来计算相差的秒数

sqlite> select   julianday('2016-05-31') - julianday('2016-06-01');-1.0sqlite> select   julianday('2016-05-31') - julianday('2016-06-02');-2.0sqlite> select   julianday('2016-06-11') - julianday('2016-06-02');9.0sqlite> select   julianday('2016-06-01 13:00:00') - julianday('2016-06-01 12:00:00');0.0416666665114462sqlite> select   julianday('2016-06-01 12:00:20') - julianday('2016-06-01 12:00:00');0.000231481622904539sqlite> select  *  from   test1 order  by   a_date;2016-04-23 17:43:352016-04-31 17:40:352016-05-26 19:40:352016-05-29 19:40:352016-05-30 19:40:352016-05-31 17:30:002016-05-31 17:30:012016-05-31 17:30:022016-05-31 17:30:032016-05-31 17:30:052016-05-31 17:30:152016-05-31 17:30:352016-05-31 17:40:352016-05-31 17:40:352016-05-31 19:40:352016-06-01 19:40:352016-12-01 19:40:35sqlite> select  *  from   test1  where julianday('Now') -julianday(a_date)>1;2016-05-26 19:40:352016-05-29 19:40:352016-04-31 17:40:352016-04-23 17:43:35sqlite> select  *  from   test1  where julianday('Now')*86400 -julianday(a_date)*86400>10;2016-05-30 19:40:352016-05-26 19:40:352016-05-29 19:40:352016-04-31 17:40:352016-04-23 17:43:35sqlite> delete    from   test1  where julianday('Now')*86400 -julianday(a_date)*86400>10;sqlite> select  *  from   test1 order  by   a_date;2016-05-31 17:30:002016-05-31 17:30:012016-05-31 17:30:022016-05-31 17:30:032016-05-31 17:30:052016-05-31 17:30:152016-05-31 17:30:352016-05-31 17:40:352016-05-31 17:40:352016-05-31 19:40:352016-06-01 19:40:352016-12-01 19:40:35sqlite>

sqlite> create  table   test1(a_date date);

这是建表语句 有一点很重要：在真机上测试，无法这样删除记录！！！！！！！！！ 总结

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