Oracle数仓中判断时间连续性的几种SQL写法示例

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2021020120210202，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2021020520210209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！

总结

到此这篇关于Oracle数仓中判断时间连续性的几种SQL写法的文章就介绍到这了,更多相关Oracle数仓判断时间连续性内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！