要求：
计算用户距上次访问的天数，根据imei号区分不同的用户，如果时间段内只有一次访问则为0。
初始化数据：
复制代码 代码示例:
CREATE TABLE `pd` (
`imei` varchar(32) NOT NULL DEFAULT '',
`dat` datetime DEFAULT NULL
) ENGINE=InnoDB DEFAULT CHARSET=utf8;
-- ----------------------------
-- Records of pd
-- ----------------------------
INSERT INTO `pd` VALUES ('1', '2013-07-25 00:00:01');
INSERT INTO `pd` VALUES ('1', '2013-07-26 00:00:02');
INSERT INTO `pd` VALUES ('2', '2013-07-23 00:00:04');
INSERT INTO `pd` VALUES ('2', '2013-07-26 00:00:03');
INSERT INTO `pd` VALUES ('3', '2013-07-26 00:00:01');
脚本，使用@特殊变量：
复制代码 代码示例:
select * from (
select imei user_id, max(max_dd) , max(max_dd_2), to_days( max(max_dd)) - to_days(max(max_dd_2)) days from (
select imei, max_dd, max_dd_2 from (
select tmp.imei, tmp.dates,
if(@imei=tmp.imei, @rank:=@rank+1,@rank:=1) as rank,
if(@rank = 1, @max_d := tmp.dates, @max_d := null) as max_dd,
if(@rank = 2, @max_d_2 := tmp.dates, @max_d_2 := null) as max_dd_2,
@imei:=tmp.imei
from (select imei, dates from pb order by imei asc ,dates desc ) tmp ,
(select @rownum :=0 , @imei := null ,@rank:=0, @max_d :=null, @max_d_2 := null) a
) result
) t
group by imei
having count(*) > 1
) x where x.days >= 1 and EXISTS (select 'x' from pb where dates > '2013-07-26 00:00:00')
注意：
表数据量较大时，使用union all等操作将会有悲剧性的结果。
