如下A表 记录用户每次动作
mobile time action re_value（返回值）
13631521739 20120915 1（开通） 0(成功）
15143728915 20120915 1（开通） 0(成功）
15143728915 20120915 1（开通） 2(重复开通）
15143728915 20120918 2（注销） 0(成功）
13631521739 20120925 2（注销） 0(成功）
1. 查询在开通四天之内又注销的用户
2.查询在开通四天后又注销的用户
------解决方案--------------------
select a.mobile, a.time as timea, nvl(b.time,20990101) as timeb
from (select mobile, time
from ttt
where action = 1
and re_value = 0) a
left join (select mobile, time
from ttt
where action = 2
and re_value = 0) b on a.mobile = b.mobile；
至于4天内外 自己外面包
------解决方案--------------------
select ta.mobile from, ta.maxt, tb.maxt from
(select mobile, max(time) maxt, action from a group by mobile where action=1) ta,
(select mobile, max(time) maxt, action from a group by mobile where action=2) tb
where ta.mobile=tb.mobile and round(to_date(tb.maxt,'yyyymmdd')-to_date(ta.maxt,'yyyymmdd')) = 4;
不知道行不行，反正我没试。
------解决方案--------------------
with t as(
select '13631521739' as moble,TO_DATE('20120915','YYYYMMDD') as time,'1' as action,'0' as re_value from dual
union
select '15143728915' as moble,TO_DATE('20120915','YYYYMMDD') as time,'1' as action,'0' as re_value from dual
union
select '15143728915' as moble,TO_DATE('20120915','YYYYMMDD') as time,'1' as action,'2' as re_value from dual
union
select '15143728915' as moble,TO_DATE('20120918','YYYYMMDD') as time,'2' as action,'0' as re_value from dual
union
select '13631521739' as moble,TO_DATE('20120925','YYYYMMDD') as time,'2' as action,'0' as re_value from dual
)
--四天以及之内的
SELECT A.MOBLE,A.TIME,B.TIME,TRUNC(B.TIME-A.TIME) AS DAY
FROM T A, T B
WHERE A.MOBLE = B.MOBLE
AND A.RE_VALUE != 2 AND B.RE_VALUE != 2
AND A.ACTION = 1 AND B.ACTION= 2
AND TRUNC(B.TIME-A.TIME)<=4
--四天之后的