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一个有趣的SQL:根据登录日志,求系统无人登录时间

2015-09-12 01:20 381 查看
今天MySQL群里有人发了这个问题:



建立测试数据和样表:

SET FOREIGN_KEY_CHECKS=0;

-- ----------------------------
-- Table structure for `adtimelog`
-- ----------------------------
DROP TABLE IF EXISTS `adtimelog`;
CREATE TABLE `adtimelog` (
  `Aname` varchar(12) CHARACTER SET utf8 DEFAULT NULL,
  `LITime` timestamp NULL DEFAULT NULL,
  `LOTime` timestamp NULL DEFAULT NULL,
  `DTime` timestamp NULL DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

-- ----------------------------
-- Records of adtimelog
-- ----------------------------
INSERT INTO `adtimelog` VALUES ('甲', '2015-08-13 08:30:00', '2015-08-13 09:30:00', '2015-08-13 00:00:00');
INSERT INTO `adtimelog` VALUES ('甲', '2015-08-13 09:40:00', '2015-08-13 10:00:00', '2015-08-13 00:00:00');
INSERT INTO `adtimelog` VALUES ('甲', '2015-08-13 08:30:00', '2015-08-13 11:30:00', '2015-08-13 00:00:00');
INSERT INTO `adtimelog` VALUES ('丁', '2015-08-13 00:30:00', '2015-08-13 01:30:00', '2015-08-13 00:00:00');
INSERT INTO `adtimelog` VALUES ('丁', '2015-08-13 02:30:00', '2015-08-13 07:30:00', '2015-08-13 00:00:00');


SQL:

select '2015-08-13 00:00:00',min(LITime) from ADTimeLog where DTime='2015/8/13' 
 UNION ALL
select distinct a.LOTime,min(b.LITime) from 
(select LITime,LOTime from ADTimeLog where DTime='2015/8/13' order by LITime ) a,
(select LITime,LOTime from ADTimeLog where DTime='2015/8/13' order by LITime limit 1,99999) b
where a.LOTime<b.LITime GROUP BY a.LOTime
 UNION ALL
select max(LOTime),'2015-08-13 23:59:59' from ADTimeLog where DTime='2015/8/13'
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