SQL Oracle 中的滚动 90 天日期窗口

Question

SELECT *
  FROM (SELECT date'2021-05-06' AS tran_dt FROM dual UNION
        SELECT date'2021-05-24' FROM dual UNION
        SELECT date'2021-06-25' FROM dual UNION
        SELECT date'2021-07-02' FROM dual UNION
        SELECT date'2021-07-27' FROM dual UNION
        SELECT date'2021-08-16' FROM dual UNION
        SELECT date'2021-08-23' FROM dual UNION
        SELECT date'2021-10-01' FROM dual UNION
        SELECT date'2021-12-31' FROM dual)

你好

我正在使用 SQL Oracle 中的日期字段。

我正在尝试创建滚动窗口，该窗口返回第一行并在自上次返回行以来的日期 >90 时返回另一行。

在下面的示例中，我想计算并返回警报字段 = 是的位置。

如果有人有任何想法，不胜感激。

谢谢

Shee7

TRAN_DT    Alert    Days Since Last Alert
06/05/2021  Yes     0
24/05/2021  No      18
25/06/2021  No      50
02/07/2021  No      57
27/07/2021  No      82
16/08/2021  Yes     102
23/08/2021  No      7
01/10/2021  No      46
31/12/2021  Yes     137

Answer 1

你可以使用递归

with firstr as (
  -- a row to start with
  select TRAN_DT
  from (
    select t.*
     , row_number() over(order by TRAN_DT) rn
    from trans t
  )
  where rn=1
),
cte(TRAN_DT) as (
  select * 
  from firstr
  union all
  select t.*
  from cte c
  join trans t on t.TRAN_DT >= c.TRAN_DT + 90 and not exists (
               select 1 
               from trans t2 
               where t2.TRAN_DT >= c.TRAN_DT + 90 and t2.TRAN_DT < t.TRAN_DT)
)
select * 
from cte;

Answer 2

也许会有第三个答案

with tab1 as (
SELECT TRAN_DT
  FROM (SELECT TO_DATE('20210506','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210524','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210625','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210702','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210727','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210816','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210823','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20211001','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20211231','YYYYMMDD') TRAN_DT FROM DUAL)
)
, tab2 as (
select t1.tran_dt dt,
       row_number() over(order by t1.tran_dt) rn
  from tab1 t1
)
, t0(dt, rn, temp) as (
select t1.dt, rn, t1.dt from tab2 t1 where t1.rn = 1
union all
select t2.dt, 
       t2.rn, 
       case when t2.dt - t0.temp > 90 then t2.dt else t0.temp end 
  from tab2 t2, t0 
 where t0.rn + 1 = t2.rn
)
select t1.dt, 
       case when t1.dt = t1.temp then 'yes' else 'no' end, 
       nvl(t1.dt - lag(t1.temp) over(order by t1.rn), 0)
  from t0 t1

第三个答案，也许还有第四个

with tab1 as (
SELECT TRAN_DT
  FROM (SELECT TO_DATE('20210506','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210524','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210625','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210702','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210727','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210816','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20210823','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20211001','YYYYMMDD') TRAN_DT FROM DUAL UNION
        SELECT TO_DATE('20211231','YYYYMMDD') TRAN_DT FROM DUAL)
)
, tab2 as (
select t1.tran_dt dt
  from tab1 t1
 order by t1.tran_dt
 fetch next 1 row only
)
, t0(dt) as (
select t1.dt from tab2 t1
union all
select (select min(v1.TRAN_DT) from tab1 v1 where t1.dt + 90 < v1.TRAN_DT)
  from t0 t1
 where t1.dt is not null
)
select t1.dt
  from t0 t1
 where t1.dt is not null

Answer 3

从 Oracle 12 开始，这是 MATCH_RECOGNIZE 用于的查询类型：

SELECT tran_dt,
       alert,
       tran_dt
         - LAG(CASE alert WHEN 'Yes' THEN tran_dt END, 1, tran_dt)
             IGNORE NULLS OVER (ORDER BY tran_dt)
         AS days
FROM   table_name
MATCH_RECOGNIZE (
  ORDER BY tran_dt
  MEASURES
    CLASSIFIER() AS alert
  ALL ROWS PER MATCH
  PATTERN ( "Yes" "No"* )
  DEFINE
    "No" AS tran_dt <= "Yes".tran_dt + INTERVAL '90' DAY
)

在 Oracle 12 之前，您可以使用递归查询：

WITH dates (tran_dt, rn) AS (
  SELECT tran_dt,
         ROW_NUMBER() OVER (ORDER BY tran_dt) AS rn
  FROM   table_name
),
rolling_dates (tran_dt, alert, rn, days, last_alert) AS (
  SELECT tran_dt, 'Yes', rn, 0, tran_dt
  FROM   dates
  WHERE  rn = 1
UNION ALL
  SELECT d.tran_dt,
         CASE
         WHEN d.tran_dt <= r.last_alert + INTERVAL '90' DAY
         THEN 'No'
         ELSE 'Yes'
         END,
         d.rn,
         d.tran_dt - r.last_alert,
         CASE
         WHEN d.tran_dt <= r.last_alert + INTERVAL '90' DAY
         THEN r.last_alert
         ELSE d.tran_dt
         END
  FROM   rolling_dates r
         INNER JOIN dates d
         ON (r.rn + 1 = d.rn)
)
SELECT tran_dt,
       alert,
       days
FROM   rolling_dates;

这两个输出：

TRAN_DT	ALERT	DAYS
2021-05-06 00:00:00	Yes	0
2021-05-24 00:00:00	No	18
2021-06-25 00:00:00	No	50
2021-07-02 00:00:00	No	57
2021-07-27 00:00:00	No	82
2021-08-16 00:00:00	Yes	102
2021-08-23 00:00:00	No	7
2021-10-01 00:00:00	No	46
2021-12-31 00:00:00	Yes	137

db小提琴here