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
-- www.jbxue.com
-- ----------------------------
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。
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