Oracle数仓中判断时间连续性的几种SQL写法示例

oracle 判断时间 oracle判断时间连续性 oracle判断日期是否连续 2023-02-05 13:02:56 245人浏览独家记忆

摘要

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

零、需求介绍

现有一张表数据如下：

Oracle数仓中判断时间连续性的几种SQL写法示例

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;

Oracle数仓中判断时间连续性的几种SQL写法示例

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;

Oracle数仓中判断时间连续性的几种SQL写法示例

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;

Oracle数仓中判断时间连续性的几种SQL写法示例

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;

Oracle数仓中判断时间连续性的几种SQL写法示例

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

Oracle数仓中判断时间连续性的几种SQL写法示例

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;

Oracle数仓中判断时间连续性的几种SQL写法示例