sql

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 
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 
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') 

如果表数据量大，千万不要用union all等操作