Oracle數倉中判斷時間連續性的幾種SQL寫法範例

2023-02-06 06:03:55

零、需求介紹

現有一張表資料如下：

此表是一張映象表，policyno列代表一個保單號，state列代表這個保單號在snapdate當天的最後一次狀態（state每天可能會變很多次，映象表只保留snapdate時間點凌晨的最後一次狀態），snapdate代表當天做映象的時間，現在有個需求，我們想取出來這個保單號連續保持某個狀態的起止時間，例如：

保單號sm1保持狀態1的起止時間為2021020120210202，然後在20210203時候變成了狀態2，又在20210204時候變成了狀態3，最終又在2021020520210209時間段保持在狀態1，然後映象表的程式可能期間出現過問題，在20210210開始到20210215日沒有映象成功，直到20210216日才恢復，20210216~20210219日保單號sm1的狀態一直保持為1，後續還有可能繼續變，那麼，上面說的保單sm1的幾個狀態的連續時間，我們想要的結果為：

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
.........................

我這裡提供5種寫法，可以歸結為兩大類：

一類：通過使用分析函數或自關聯獲取資料連續性，構造一個分組欄位進行分組求最大最小值。

二類：通過樹形層次查詢獲取連續性，獲取起止時間。

一、通過使用lag分析函數獲取前後時間，根據當前時間與前後時間的差值進行判斷獲取時間連續性標誌，然後使用sum()over()對連續性標誌進行累加，從而生成一個新的臨時分組欄位，最終根據policyno,state,臨時分組欄位進行分組取最大最小值

這裡為了好理解，每一個處理步驟都單獨寫出來了，實際使用中可以簡寫一下：

with t as--求出來每條資料當天的前一天映象時間
 (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--判斷當天映象時間和前一天的映象時間+1是否相等，如果相等就置為0否則置為1，新增臨時欄位lxzt意為：連續狀態標誌
 (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--根據lxzt欄位進行sum()over()求和，求出來一個新的用來做分組依據的欄位，簡稱fzyj
 (select t1.*, sum(lxzt) over(order by policyno, snapdate) as fzyj from t1)
select policyno,--最後根據policyno,state,fzyj進行分組求最大最小值即為狀態連續的開始結束時間
       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分析函數,通過自關聯也能判斷出來哪些天連續,然後後面操作步驟同上，這個寫法算是對lag()over()函數的一個回寫，擺脫對分析函數的依賴

下面這種寫法，需要讀兩次表，上面lag的方式是對這個寫法的一種優化:

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、通過構造樹形結構，確定根節點和葉子節點來獲取狀態連續的開始和結束時間

先按照資料的連續性構造顯示每層關係的樹狀結構：

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 樹狀結構,
         level as 樹中層次,
         decode(level, 1, 1) 是否根節點,
         decode(connect_by_isleaf, 1, 1) 是否葉子節點,
         case
           when (connect_by_isleaf = 0 and level > 1) then
            1
         end  是否樹杈,
         (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;

從上面能清晰的看出來，每一次連續狀態的開始日期作為每個樹的根，分支節點即樹杈和葉子節點的關係一步步拓展開來，分析上面資料我們能夠知道，如果我們想要獲取每個保單狀態連續時間範圍，以上面的資料現有分佈方式，現在就可以：通過policyno,state,主根值進行group by 取snapdate的最大最小值，類似前面兩個寫法的最終步驟；

接下來，我們這個第三種寫法就是按照這個方式寫：

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 樹狀結構,
         level as 樹中層次,
         decode(level, 1, 1) 是否根節點,
         decode(connect_by_isleaf, 1, 1) 是否葉子節點,
         case
           when (connect_by_isleaf = 0 and level > 1) then
            1
         end  是否樹杈,
         (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、參照過程三，既然已經獲取了每條資料的主根值和葉子節點的值，這就代表了我們知道了每個保單狀態的連續開始和結束時間，那直接取出來葉子節點資料，葉子節點主根值就是開始日期，葉子節點的值就是結束日期，這樣我們就不需再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 樹狀結構,
         level as 樹中層次,
         decode(level, 1, 1) 是否根節點,
         decode(connect_by_isleaf, 1, 1) 是否葉子節點,
         case
           when (connect_by_isleaf = 0 and level > 1) then
            1
         end 是否樹杈,
         (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 是否葉子節點 = 1
 order by policyno, snapdate

五、在Oracle10g之前，上面樹狀查詢的關鍵函數 connect_by_root還不支援，如果使用樹形結構，可以通過sys_connect_by_path來實現

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;

還有好多其他寫法，這裡不再一一列舉！

總結

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Oracle數倉中判斷時間連續性的幾種SQL寫法範例

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