PL/SQL：在由開始和結束定義的重疊日期範圍中查找島嶼

Question

我有一個非常大的數據集，其中包含人員以及他們保險的開始和結束日期。每個人可以有多個重疊日期的記錄。我需要爲每個人找到每個「島」的開始和結束。

例如：

SKP_PERSON DATE_INSURANCE_START DATE_INSURANCE_END SKP_INSURANCE 
1 1   7.11.2015    1.1.3000    1 
2 1   7.11.2015    1.1.3000    2 
3 2   10.4.2015    1.8.2016 23:59:59 3 
4 3   28.3.2016    1.1.3000    4 
5 4   5.12.2015    31.12.2015 23:59:59 5 
6 4   5.12.2015    1.5.2016 23:59:59 6 
7 4   1.2.2016    1.5.2016 23:59:59 7 
8 5   15.1.2016    2.3.2016 23:59:59 8 
9 5   15.3.2016    2.6.2016 23:59:59 9 

結果我需要的是這樣的：

SKP_PERSON DATE_INSURANCE_START DATE_INSURANCE_END 
1 1   7.11.2015    1.1.3000    
2 2   10.4.2015    1.8.2016 23:59:59  
3 3   28.3.2016    1.1.3000    
4 4   5.12.2015    1.5.2016 23:59:59  
5 5   15.1.2016    2.3.2016 23:59:59  
6 5   15.3.2016    2.6.2016 23:59:59  

我設法找到通過連接所有可能的時間（從分鐘（開始溶液）最大值（結束）），併爲每一天找到滯後和主導值 - 但是記錄太多，日期範圍太大，因此需要很長時間。有沒有更有效的解決方案使用PL/SQL？

編輯： 我試過（簡體）查詢：

WITH table1 AS (
SELECT d.dtime_day, COUNT(i.dkp_insurance), i.skp_person 
FROM date d --a date table, contains a record for every day 
JOIN insurance i ON d.dtime_day BETWEEN i.DATE_INSURANCE_START AND i.DATE_INSURANCE_END  
GROUP BY d.dtime_day, i.skp_person 
) 
SELECT * FROM 
(
SELECT distinct skp_person, 
CASE WHEN LAG(dtime_day) OVER (PARTITION BY skp_person ORDER BY dtime_day) <> dtime_day -1 THEN dtime_day END AS start, 
CASE WHEN LEAD(dtime_day) OVER (PARTITION BY skp_person ORDER BY dtime_day) <> dtime_day +1 THEN dtime_day END AS end 
FROM table1 t1) 
WHERE start IS NOT NULL OR end IS NOT NULL 
ORDER BY skp_person 
; 

Answer 1

我改編了我以前的解決方案以適應這種情況（請參閱原始問題的評論）。需要愚蠢的+1/86400（增加一秒）來處理表中奇怪的結束日期/時間。

with 
    inputs (skp_person, date_insurance_start, date_insurance_end) as (
     select 1, to_date('7.11.2015', 'dd.mm.yyyy'), to_date('1.1.3000'   , 'dd.mm.yyyy')    from dual union all 
     select 1, to_date('7.11.2015', 'dd.mm.yyyy'), to_date('1.1.3000'   , 'dd.mm.yyyy')    from dual union all 
     select 2, to_date('10.4.2015', 'dd.mm.yyyy'), to_date('1.8.2016 23:59:59' , 'dd.mm.yyyy hh24:mi:ss') from dual union all 
     select 3, to_date('28.3.2016', 'dd.mm.yyyy'), to_date('1.1.3000'   , 'dd.mm.yyyy')    from dual union all 
     select 4, to_date('5.12.2015', 'dd.mm.yyyy'), to_date('31.12.2015 23:59:59', 'dd.mm.yyyy hh24:mi:ss') from dual union all 
     select 4, to_date('5.12.2015', 'dd.mm.yyyy'), to_date('1.5.2016 23:59:59' , 'dd.mm.yyyy hh24:mi:ss') from dual union all 
     select 4, to_date('1.2.2016' , 'dd.mm.yyyy'), to_date('1.5.2016 23:59:59' , 'dd.mm.yyyy hh24:mi:ss') from dual union all 
     select 5, to_date('15.1.2016', 'dd.mm.yyyy'), to_date('2.3.2016 23:59:59' , 'dd.mm.yyyy hh24:mi:ss') from dual union all 
     select 5, to_date('15.3.2016', 'dd.mm.yyyy'), to_date('2.6.2016 23:59:59' , 'dd.mm.yyyy hh24:mi:ss') from dual 
    ), 
    m (skp_person, date_insurance_start, m_date) as (
     select skp_person, date_insurance_start, 
       max(date_insurance_end + 1/86400) 
       over (partition by skp_person order by date_insurance_start 
        rows between unbounded preceding and 1 preceding) 
     from inputs 
     union all 
     select skp_person, null, max(date_insurance_end + 1/86400) 
     from inputs 
     group by skp_person 
    ), 
    f (skp_person, date_insurance_start, e_date) as (
     select skp_person, date_insurance_start, 
       lead(m_date) over 
       (partition by skp_person order by date_insurance_start) 
     from m 
     where date_insurance_start > m_date 
      or date_insurance_start is null or m_date is null 
    ) 
select skp_person, date_insurance_start, e_date - 1/86400 as date_insurance_end 
from f where date_insurance_start is not null 
; 

輸出：（使用我NLS_DATE_FORMAT設置）

SKP_PERSON DATE_INSURANCE_STAR DATE_INSURANCE_END 
---------- ------------------- ------------------- 
     1 07.11.2015 00:00:00 01.01.3000 00:00:00 
     2 10.04.2015 00:00:00 01.08.2016 23:59:59 
     3 28.03.2016 00:00:00 01.01.3000 00:00:00 
     4 05.12.2015 00:00:00 01.05.2016 23:59:59 
     5 15.01.2016 00:00:00 02.03.2016 23:59:59 
     5 15.03.2016 00:00:00 02.06.2016 23:59:59 

Answer 2

這裏的理念是：

• 使用lag()或一些其他方法來確定當一個島嶼開始
• 構建當島開始時爲1的標誌
• 運行累積和
• Reaggregate

結果查詢是這樣的：

select skp_person, 
     min(date_insurance_start) as date_insurance_start, 
     min(date_insurance_end) as date_insurance_end 
from (select t.*, 
      sum(isIslandFlag) over (partition by skp_person order by date_insurance_start) as grp 
     from (select t.*, 
        (case when exists (select 1 
             from t t2 
             where t2.skp_person = t.skp_person and 
              t2.date_insurance_start between t.date_insurance_start and t.date_insurance_end 
            ) 
         then 0 else 1 
        end) as IsIslandFlag 
      from t 
      ) t 
    ) t 
group by skp_person, grp; 

注意：此方法不是傻瓜證明，但它在大多數現實世界的情況下工作。例如，如果您有多個策略在同一天開始，那麼它需要稍微調整一下。