Question

我的第一个问题，我觉得我没有清楚解释，但是示例数据应该如此。

如何找到一个“类型”的最早日期早于另一“类型”的最早日期，说明不应该包含的“相同”类型的日期，然后为系列中的其他顺序日期确定相同的日期？

表格中的示例数据：

PERSON_ID  SERVICE_DATE SERVICE_TYPE
ABC        10/4/2018    INTAKE
ABC        10/8/2018    INTAKE
ABC        10/19/2018   DISCHARGE
ABC        10/25/2018   DISCHARGE
ABC        11/21/2018   INTAKE
ABC        12/3/2018    INTAKE
ABC        12/6/2018    INTAKE
ABC        12/26/2018   DISCHARGE

我想退货：

PERSON_ID  INTAKE_DATE DISCHARGE_DATE
ABC        10/4/2018   10/19/2018
ABC        11/21/2018  12/26/2018

编辑：第二个例子。如果进出日期在同一日期，我也想记录下来。

PERSON_ID  SERVICE_DATE  SERVICE_TYPE
DEF        10/1/2018     INTAKE
DEF        10/1/2018     DISCHARGE
DEF        11/5/2018     INTAKE
DEF        12/31/2018    DISCHARGE

我想退货：

PERSON_ID  INTAKE_DATE DISCHARGE_DATE
DEF        10/1/2018   10/1/2018
DEF        11/5/2018   12/31/2018

Answer 1

您可以使用MATCH_RECOGNIZE轻松匹配多行中的模式，如以下查询所示：

WITH test_vals AS (
    SELECT 'ABC' as PERSON_ID,TO_DATE('10/4/2018','mm/dd/yyyy') as SERVICE_DATE,'INTAKE' as SERVICE_TYPE FROM DUAL
    UNION SELECT 'ABC',TO_DATE('10/8/2018','mm/dd/yyyy'),'INTAKE' FROM DUAL
    UNION SELECT 'ABC',TO_DATE('10/19/2018','mm/dd/yyyy'),'DISCHARGE' FROM DUAL
    UNION SELECT 'ABC',TO_DATE('10/25/2018','mm/dd/yyyy'),'DISCHARGE' FROM DUAL
    UNION SELECT 'ABC',TO_DATE('11/21/2018','mm/dd/yyyy'),'INTAKE' FROM DUAL
    UNION SELECT 'ABC',TO_DATE('12/3/2018','mm/dd/yyyy'),'INTAKE' FROM DUAL
    UNION SELECT 'ABC',TO_DATE('12/6/2018','mm/dd/yyyy'),'INTAKE' FROM DUAL
    UNION SELECT 'ABC',TO_DATE('12/26/2018','mm/dd/yyyy'),'DISCHARGE' FROM DUAL
)

SELECT m.person_id,
       m.earliest_intake_date,
       m.earliest_discharge_date
FROM test_vals t
match_recognize (
    PARTITION BY person_id
    ORDER BY service_date, service_type DESC /* Order such that INTAKE comes before DISCHARGE if two items have the same service date */
    MEASURES
        MIN(service_date) AS earliest_intake_date,
        MAX(service_date) AS earliest_discharge_date
    ONE ROW PER match
    pattern (
        intake+ /* Match one or more INTAKE codes, followed by a DISCHARGE */
        discharge
    )
    define 
        intake AS service_type = 'INTAKE',
        discharge AS service_type = 'DISCHARGE'
) m

Answer 2

，test_table为（选择DUAL UNION ALL中的'ABC'PERSON_ID，to_date（'10 / 04/2018'，'MM / DD / YYYY'）SERVICE_DATE，'INTAKE'SERVICE_TYPE 选择DUAL UNION ALL中的'ABC'PERSON_ID，to_date（'10 / 08/2018'，'MM / DD / YYYY'）SERVICE_DATE，'INTAKE'SERVICE_TYPE 选择DUAL UNION ALL中的'ABC'PERSON_ID，to_date（'10 / 19/2018'，'MM / DD / YYYY'）SERVICE_DATE，'DISCHARGE'SERVICE_TYPE 选择DUAL UNION ALL中的'ABC'PERSON_ID，to_date（'10 / 25/2018'，'MM / DD / YYYY'）SERVICE_DATE，'DISCHARGE'SERVICE_TYPE 选择DUAL UNION ALL中的'ABC'PERSON_ID，to_date（'11 / 21/2018'，'MM / DD / YYYY'）SERVICE_DATE，'INTAKE'SERVICE_TYPE 从DUAL UNION ALL中选择“ ABC” PERSON_ID，to_date（'12 / 03/2018'，'MM / DD / YYYY'）SERVICE_DATE，'INTAKE'SERVICE_TYPE 选择DUAL UNION ALL中的'ABC'PERSON_ID，to_date（'12 / 06/2018'，'MM / DD / YYYY'）SERVICE_DATE，'INTAKE'SERVICE_TYPE 选择'ABC'PERSON_ID，to_date（'12 / 26/2018'，'MM / DD / YYYY'）SERVICE_DATE，'DISCHARGE'SERVICE_TYPE FROM DUAL ），参数为（ SELECT T. ，行数rn 来自（选择t1。，LAG（SERVICE_TYPE，1，'DISCHARGE'）OVER（ORDER BY SERVICE_DATE）BEFORE_ROW 从test_table t1 ）吨哪里 T.SERVICE_TYPE ='INTAKE'并且 T.BEFORE_ROW ='DISCHARGE' ）选择 tt1.person_id PERSON_ID， tt1.service_date INTAKE，（从TEST_TABLE T3的TEST.TABLE T3.SERVICE_TYPE ='DISCHARGE'中选择MAX（T3.SERVICE_DATE）并且（T3.SERVICE_DATE TT1.SERVICE_DATE ORDER BY tt1.SERVICE_DATE;

查找同一系列中多次出现的最早和最晚日期

2 个答案: