零、需求介绍
现有一张表数据如下：
此表是一张镜像表，policyno列代表一个保单号，state列代表这个保单号在snapdate当天的最后一次状态（state每天可能会变很多次，镜像表只保留snapdate时间点凌晨的最后一次状态），
snapdate代表当天做镜像的时间，现在有个需求，我们想取出来这个保单号连续保持某个状态的起止时间，例如：
保单号sm1保持状态1的起止时间为20210201~20210202，然后在20210203时候变成了状态2，又在20210204时候变成了状态3，最终又在20210205~20210209时间段保持在状态1，
然后镜像表的程序可能期间出现过问题，在20210210开始到20210215日没有镜像成功，直到20210216日才恢复，20210216~20210219日保单号sm1的状态一直保持为1，
后续还有可能继续变，那么，上面说的保单sm1的几个状态的连续时间，我们想要的结果为：
POLICYNO STATE START_DATE END_DATE
sm1 1 20210201 20210202
sm1 2 20210203 20210203
sm1 3 20210204 20210204
sm1 1 20210205 20210209
sm1 1 20210216 20210219
.........................我这里提供5种写法，可以归结为两大类：
一类：通过使用分析函数或自关联获取数据连续性，构造一个分组字段进行分组求最大最小值。
二类：通过树形层次查询获取连续性，获取起止时间。
一、通过使用lag分析函数获取前后时间，根据当前时间与前后时间的差值进行判断获取时间连续性标志，然后使用sum()over()对连续性标志进行累加，从而生成一个新的临时分组字段，最终根据policyno,state,临时分组字段进行分组取最大最小值
这里为了好理解，每一个处理步骤都单独写出来了，实际使用中可以简写一下：
with t as--求出来每条数据当天的前一天镜像时间
(select a.policyno,
a.state,
a.snapdate,
lag(a.snapdate) over(partition by a.policyno, a.state order by a.snapdate) as lag_tim
from zyd.temp_0430 a
order by a.policyno, a.snapdate),
t1 as--判断当天镜像时间和前一天的镜像时间+1是否相等，如果相等就置为0否则置为1，新增临时字段lxzt意为：连续状态标志
(select t.*,
case
when t.snapdate = t.lag_tim + 1 then
0
else
1
end as lxzt
from t
order by policyno, snapdate),
t2 as--根据lxzt字段进行sum()over()求和，求出来一个新的用来做分组依据的字段，简称fzyj
(select t1.*, sum(lxzt) over(order by policyno, snapdate) as fzyj from t1)
select policyno,--最后根据policyno,state,fzyj进行分组求最大最小值即为状态连续的开始结束时间
state,
-- fzyj,
min(snapdate) as start_snap,
max(snapdate) as end_snap
from t2
group by policyno, state, fzyj
order by fzyj;
二、不使用lag分析函数,通过自关联也能判断出来哪些天连续,然后后面操作步骤同上，这个写法算是对lag()over()函数的一个回写，摆脱对分析函数的依赖
下面这种写法，需要读两次表，上面lag的方式是对这个写法的一种优化:
with t as
(select a.policyno, a.state, a.snapdate, b.snapdate as snap2
from zyd.temp_0430 a, zyd.temp_0430 b
where a.policyno = b.policyno(+)
and a.state = b.state(+)
and a.snapdate - 1 = b.snapdate(+)
order by policyno, snapdate),
t1 as
(select t.*,
case
when snap2 is null then
1
else
0
end as lxzt
from t
order by policyno, snapdate),
t2 as
(select t1.*, sum(lxzt) over(order by policyno, snapdate) as fzyj
from t1
order by policyno, snapdate)
select policyno,
state,
fzyj,
min(snapdate) as start_snap,
max(snapdate) as end_snap
from t2
group by policyno, state, fzyj
order by fzyj;
三、通过构造树形结构，确定根节点和叶子节点来获取状态连续的开始和结束时间
先按照数据的连续性构造显示每层关系的树状结构：
with t as
(select a.policyno,
a.state,
a.snapdate,
lag(a.snapdate) over(partition by a.policyno, a.state order by a.snapdate) as lag_tim
from zyd.temp_0430 a --where policyno='sm1'
order by a.policyno, a.snapdate),
t1 as
(select t.*,
case
when t.snapdate = t.lag_tim + 1 then
0
else
1
end as lxzt
from t
order by policyno, snapdate),
t2 as
(select t1.*,
lpad('->', (level - 1) * 2, '->') || snapdate as 树状结构,
level as 树中层次,
decode(level, 1, 1) 是否根节点,
decode(connect_by_isleaf, 1, 1) 是否叶子节点,
case
when (connect_by_isleaf = 0 and level > 1) then
1
end 是否树杈,
(prior snapdate) as 根值,
connect_by_root snapdate 主根值
from t1
start with (lxzt = 1)
connect by (prior snapdate = snapdate - 1
and prior state = state and
prior policyno = policyno)
order by policyno, snapdate)
select * from t2;
从上面能清晰的看出来，每一次连续状态的开始日期作为每个树的根，分支节点即树杈和叶子节点的关系一步步拓展开来，分析上面数据我们能够知道，如果我们想要获取
每个保单状态连续时间范围，以上面的数据现有分布方式，现在就可以：通过policyno,state,主根值进行group by 取snapdate的最大最小值，类似前面两个写法的最终步骤；
接下来，我们这个第三种写法就是按照这个方式写：
with t as
(select a.policyno,
a.state,
a.snapdate,
lag(a.snapdate) over(partition by a.policyno, a.state order by a.snapdate) as lag_tim
from zyd.temp_0430 a --where policyno='sm1'
order by a.policyno, a.snapdate),
t1 as
(select t.*,
case
when t.snapdate = t.lag_tim + 1 then
0
else
1
end as lxzt
from t
order by policyno, snapdate),
t2 as
(select t1.*,
lpad('->', (level - 1) * 2, '->') || snapdate as 树状结构,
level as 树中层次,
decode(level, 1, 1) 是否根节点,
decode(connect_by_isleaf, 1, 1) 是否叶子节点,
case
when (connect_by_isleaf = 0 and level > 1) then
1
end 是否树杈,
(prior snapdate) as 根值,
connect_by_root snapdate 主根值
from t1
start with (lxzt = 1)
connect by (prior snapdate = snapdate - 1
and prior state = state and
prior policyno = policyno)
order by policyno, snapdate)
select policyno,
state,
min(snapdate) as start_date,
max(snapdate) as end_date
from t2
group by policyno, state, 主根值
order by policyno, state;
四、参照过程三，既然已经获取了每条数据的主根值和叶子节点的值，这就代表了我们知道了每个保单状态的连续开始和结束时间，那直接取出来叶子节点数据，叶子节点主根值就是开始日期，叶子节点的值就是结束日期，这样我们就不需再group by了
with t as
(select a.policyno,
a.state,
a.snapdate,
lag(a.snapdate) over(partition by a.policyno, a.state order by a.snapdate) as lag_tim
from zyd.temp_0430 a --where policyno='sm1'
order by a.policyno, a.snapdate),
t1 as
(select t.*,
case
when t.snapdate = t.lag_tim + 1 then
0
else
1
end as lxzt
from t
order by policyno, snapdate),
t2 as
(select t1.*,
lpad('->', (level - 1) * 2, '->') || snapdate as 树状结构,
level as 树中层次,
decode(level, 1, 1) 是否根节点,
decode(connect_by_isleaf, 1, 1) 是否叶子节点,
case
when (connect_by_isleaf = 0 and level > 1) then
1
end 是否树杈,
(prior snapdate) as 根值,
connect_by_root snapdate 主根值
from t1
start with (lxzt = 1)
connect by (prior snapdate = snapdate - 1 and prior state = state and
prior policyno = policyno)
order by policyno, snapdate)
select policyno, state, 主根值 as start_date, snapdate as end_date
from t2
where 是否叶子节点 = 1
order by policyno, snapdate
五、在Oracle10g之前，上面树状查询的关键函数 connect_by_root还不支持，如果使用树形结构，可以通过sys_connect_by_path来实现
with t as
(select a.policyno,
a.state,
a.snapdate,
lag(a.snapdate) over(partition by a.policyno, a.state order by a.snapdate) as lag_tim
--case when lag(a.snapdate) over(partition by a.policyno, a.state order by a.snapdate) is null then snapdate else lag(a.snapdate) over(partition by a.policyno, a.state order by a.snapdate) end as lag_tim
from zyd.temp_0430 a
order by a.policyno, a.snapdate),
t1 as
(select t.*,
case
when t.snapdate = t.lag_tim + 1 then
0
else
1
end as lxzt
from t
order by policyno, snapdate),
t2 as
(select t1.*,
sys_connect_by_path(snapdate, ',') as pt,
level,
connect_by_isleaf as cb
from t1
start with (lxzt = 1)
connect by (prior snapdate = snapdate - 1 and prior state = state and
prior policyno = policyno))
select t2.*,
regexp_substr(pt, '[^,]+', 1, 1) as start_date,
regexp_substr(pt, '[^,]+', 1, regexp_count(pt, ',')) as end_date
from t2
where cb = 1
order by policyno, state;
还有好多其他写法，这里不再一一列举，有兴趣的可以评论下面写出其他方案奥！
