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列代表一個保單號，state列代表這個保單號在snapdate當天的最后一次狀態(tài)（state每天可能會變很多次，鏡像表只保留snapdate時間點凌晨的最后一次狀態(tài)），snapdate代表當天做鏡像的時間，現(xiàn)在有個需求，我們想取出來這個保單號連續(xù)保持某個狀態(tài)的起止時間，例如：

保單號sm1保持狀態(tài)1的起止時間為2021020120210202，然后在20210203時候變成了狀態(tài)2，又在20210204時候變成了狀態(tài)3，最終又在2021020520210209時間段保持在狀態(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，20210216~20210219日保單號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的幾個狀態(tài)的連續(xù)時間，我們想要的結(jié)果為：

POLICYNO	STATE	START_DATE	END_DATEsm1		1	20210201	20210202sm1		2	20210203	20210203sm1		3	20210204	20210204sm1		1	20210205	20210209sm1     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)或自關聯(lián)獲取數(shù)據(jù)連續(xù)性，構造一個分組字段進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，獲取起止時間。

一、通過使用lag分析函數(shù)獲取前后時間，根據(jù)當前時間與前后時間的差值進行判斷獲取時間連續(xù)性標志，然后使用sum()over()對連續(xù)性標志進行累加，從而生成一個新的臨時分組字段，最終根據(jù)policyno,state,臨時分組字段進行分組取最大最小值

這里為了好理解，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--求出來每條數(shù)據(jù)當天的前一天鏡像時間 (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--判斷當天鏡像時間和前一天的鏡像時間+1是否相等，如果相等就置為0否則置為1，新增臨時字段lxzt意為：連續(xù)狀態(tài)標志 (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字段進行sum()over()求和，求出來一個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進行分組求最大最小值即為狀態(tài)連續(xù)的開始結(jié)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(shù),通過自關聯(lián)也能判斷出來哪些天連續(xù),然后后面操作步驟同上，這個寫法算是對lag()over()函數(shù)的一個回寫，擺脫對分析函數(shù)的依賴

下面這種寫法，需要讀兩次表，上面lag的方式是對這個寫法的一種優(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、通過構造樹形結(jié)構，確定根節(jié)點和葉子節(jié)點來獲取狀態(tài)連續(xù)的開始和結(jié)束時間

先按照數(shù)據(jù)的連續(xù)性構造顯示每層關系的樹狀結(jié)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(jié)構, level as 樹中層次, decode(level, 1, 1) 是否根節(jié)點, decode(connect_by_isleaf, 1, 1) 是否葉子節(jié)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(xù)狀態(tài)的開始日期作為每個樹的根，分支節(jié)點即樹杈和葉子節(jié)點的關系一步步拓展開來，分析上面數(shù)據(jù)我們能夠知道，如果我們想要獲取每個保單狀態(tài)連續(xù)時間范圍，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(jié)構, level as 樹中層次, decode(level, 1, 1) 是否根節(jié)點, decode(connect_by_isleaf, 1, 1) 是否葉子節(jié)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(jīng)獲取了每條數(shù)據(jù)的主根值和葉子節(jié)點的值，這就代表了我們知道了每個保單狀態(tài)的連續(xù)開始和結(jié)束時間，那直接取出來葉子節(jié)點數(shù)據(jù)，葉子節(jié)點主根值就是開始日期，葉子節(jié)點的值就是結(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, level as 樹中層次, decode(level, 1, 1) 是否根節(jié)點, decode(connect_by_isleaf, 1, 1) 是否葉子節(jié)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(jié)點 = 1 order by policyno, snapdate

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

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