Oracle數(shù)倉中判斷時(shí)間連續(xù)性的幾種SQL寫法示例

更新時(shí)間：2023年02月04日 11:52:20 作者：趙延?xùn)|的一畝三分地

這篇文章主要給大家介紹了關(guān)于Oracle數(shù)倉中判斷時(shí)間連續(xù)性的幾種SQL寫法,文中通過實(shí)例代碼介紹的非常詳細(xì),對(duì)大家學(xué)習(xí)或者使用Oracle具有一定的參考學(xué)習(xí)價(jià)值,需要的朋友可以參考下

零、需求介紹

現(xiàn)有一張表數(shù)據(jù)如下：

此表是一張鏡像表，policyno列代表一個(gè)保單號(hào)，state列代表這個(gè)保單號(hào)在snapdate當(dāng)天的最后一次狀態(tài)（state每天可能會(huì)變很多次，鏡像表只保留snapdate時(shí)間點(diǎn)凌晨的最后一次狀態(tài)），snapdate代表當(dāng)天做鏡像的時(shí)間，現(xiàn)在有個(gè)需求，我們想取出來這個(gè)保單號(hào)連續(xù)保持某個(gè)狀態(tài)的起止時(shí)間，例如：

保單號(hào)sm1保持狀態(tài)1的起止時(shí)間為2021020120210202，然后在20210203時(shí)候變成了狀態(tài)2，又在20210204時(shí)候變成了狀態(tài)3，最終又在2021020520210209時(shí)間段保持在狀態(tài)1，然后鏡像表的程序可能期間出現(xiàn)過問題，在20210210開始到20210215日沒有鏡像成功，直到20210216日才恢復(fù)，20210216~20210219日保單號(hào)sm1的狀態(tài)一直保持為1，后續(xù)還有可能繼續(xù)變，那么，上面說的保單sm1的幾個(gè)狀態(tài)的連續(xù)時(shí)間，我們想要的結(jié)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
.........................

我這里提供5種寫法，可以歸結(jié)為兩大類：

一類：通過使用分析函數(shù)或自關(guān)聯(lián)獲取數(shù)據(jù)連續(xù)性，構(gòu)造一個(gè)分組字段進(jìn)行分組求最大最小值。

二類：通過樹形層次查詢獲取連續(xù)性，獲取起止時(shí)間。

一、通過使用lag分析函數(shù)獲取前后時(shí)間，根據(jù)當(dāng)前時(shí)間與前后時(shí)間的差值進(jìn)行判斷獲取時(shí)間連續(xù)性標(biāo)志，然后使用sum()over()對(duì)連續(xù)性標(biāo)志進(jìn)行累加，從而生成一個(gè)新的臨時(shí)分組字段，最終根據(jù)policyno,state,臨時(shí)分組字段進(jìn)行分組取最大最小值

這里為了好理解，每一個(gè)處理步驟都單獨(dú)寫出來了，實(shí)際使用中可以簡寫一下：

with t as--求出來每條數(shù)據(jù)當(dāng)天的前一天鏡像時(shí)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(dāng)天鏡像時(shí)間和前一天的鏡像時(shí)間+1是否相等，如果相等就置為0否則置為1，新增臨時(shí)字段lxzt意為：連續(xù)狀態(tài)標(biāo)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--根據(jù)lxzt字段進(jìn)行sum()over()求和，求出來一個(gè)新的用來做分組依據(jù)的字段，簡稱fzyj
 (select t1.*, sum(lxzt) over(order by policyno, snapdate) as fzyj from t1)
select policyno,--最后根據(jù)policyno,state,fzyj進(jìn)行分組求最大最小值即為狀態(tài)連續(xù)的開始結(jié)束時(shí)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(shù),通過自關(guān)聯(lián)也能判斷出來哪些天連續(xù),然后后面操作步驟同上，這個(gè)寫法算是對(duì)lag()over()函數(shù)的一個(gè)回寫，擺脫對(duì)分析函數(shù)的依賴

下面這種寫法，需要讀兩次表，上面lag的方式是對(duì)這個(gè)寫法的一種優(yōu)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(gòu)造樹形結(jié)構(gòu)，確定根節(jié)點(diǎn)和葉子節(jié)點(diǎn)來獲取狀態(tài)連續(xù)的開始和結(jié)束時(shí)間

先按照數(shù)據(jù)的連續(xù)性構(gòu)造顯示每層關(guān)系的樹狀結(jié)構(gòu)：

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(jié)構(gòu),
         level as 樹中層次,
         decode(level, 1, 1) 是否根節(jié)點(diǎn),
         decode(connect_by_isleaf, 1, 1) 是否葉子節(jié)點(diǎn),
         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(xù)狀態(tài)的開始日期作為每個(gè)樹的根，分支節(jié)點(diǎn)即樹杈和葉子節(jié)點(diǎn)的關(guān)系一步步拓展開來，分析上面數(shù)據(jù)我們能夠知道，如果我們想要獲取每個(gè)保單狀態(tài)連續(xù)時(shí)間范圍，以上面的數(shù)據(jù)現(xiàn)有分布方式，現(xiàn)在就可以：通過policyno,state,主根值進(jìn)行g(shù)roup by 取snapdate的最大最小值，類似前面兩個(gè)寫法的最終步驟；

接下來，我們這個(gè)第三種寫法就是按照這個(gè)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(jié)構(gòu),
         level as 樹中層次,
         decode(level, 1, 1) 是否根節(jié)點(diǎn),
         decode(connect_by_isleaf, 1, 1) 是否葉子節(jié)點(diǎn),
         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(jīng)獲取了每條數(shù)據(jù)的主根值和葉子節(jié)點(diǎn)的值，這就代表了我們知道了每個(gè)保單狀態(tài)的連續(xù)開始和結(jié)束時(shí)間，那直接取出來葉子節(jié)點(diǎn)數(shù)據(jù)，葉子節(jié)點(diǎn)主根值就是開始日期，葉子節(jié)點(diǎn)的值就是結(jié)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(jié)構(gòu),
         level as 樹中層次,
         decode(level, 1, 1) 是否根節(jié)點(diǎn),
         decode(connect_by_isleaf, 1, 1) 是否葉子節(jié)點(diǎn),
         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(jié)點(diǎn) = 1
 order by policyno, snapdate

五、在Oracle10g之前，上面樹狀查詢的關(guān)鍵函數(shù) connect_by_root還不支持，如果使用樹形結(jié)構(gòu)，可以通過sys_connect_by_path來實(shí)現(xiàn)

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，這里不再一一列舉！

總結(jié)

到此這篇關(guān)于Oracle數(shù)倉中判斷時(shí)間連續(xù)性的幾種SQL寫法的文章就介紹到這了,更多相關(guān)Oracle數(shù)倉判斷時(shí)間連續(xù)性內(nèi)容請(qǐng)搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持腳本之家！