你如何解决这个 SQL 查询？

Question

在这个难题中，我们必须根据患者入院日期和出院日期对数据进行分组。如果任何患者出院日期 + 1 = 入院日期，那么我们将两行分组为一行，并对两行的成本求和。请查看示例输入和预期输出以了解详细信息。

Sample Input
PatientID   AdmissionDate   DischargeDate   Cost
1009        27-07-2014      31-07-2014      1050
1009        01-08-2014      23-08-2014      1070
1009        31-08-2014      31-08-2014      1900
1009        01-09-2014      14-09-2014      1260
1009        01-12-2014      31-12-2014      2090
1024        07-06-2014      28-06-2014      1900
1024        29-06-2014      31-07-2014      2900
1024        01-08-2014      02-08-2014      1800

Expected Output
PatientId   AdminssionDate  DischargeDate   Cost
1009        27-07-2014      23-08-2014      2120
1009        31-08-2014      14-09-2014      3160
1009        01-12-2014      31-12-2014      2090
1024        07-06-2014      02-08-2014      6600

我想不出解决办法。

用它来生成 table:

CREATE TABLE PatientProblem
(
PatientID integer,
AdmissionDate date,
DischargeDate date,
Cost numeric(20,2)
);
 
--Insert Data
INSERT INTO PatientProblem(PatientID,AdmissionDate,DischargeDate
,Cost)
VALUES
(1009,'2014-07-27','2014-07-31',1050.00),
(1009,'2014-08-01','2014-08-23',1070.00),
(1009,'2014-08-31','2014-08-31',1900.00),
(1009,'2014-09-01','2014-09-14',1260.00),
(1009,'2014-12-01','2014-12-31',2090.00),
(1024,'2014-06-07','2014-06-28',1900.00),
(1024,'2014-06-29','2014-07-31',2900.00),
(1024,'2014-08-01','2014-08-02',1800.00)

Answer 1

这称为间隙和孤岛问题。我们通常用 window 函数来解决这个问题。使用 LAG 您可以看到上一行的值。这有助于我们标记每个患者开始新日期范围的行。然后，通过构建标记数量的运行计数，我们可以获得患者日期范围的组号。

select
  patientid,
  min(admissiondate) as range_start,
  max(dischargedate) as range_end,
  sum(cost) as total_cost
from
(
  select
    patientid, admissiondate, dischargedate, cost,
    count(marker) over (partition by patientid order by admissiondate) as grp
  from
  (
    select
      patientid, admissiondate, dischargedate, cost,
      case when admissiondate > lag(dischargedate) over (partition by patientid order by admissiondate) + interval '1 day' then
        1
      end as marker
    from mytable
  ) marked
) grouped
group by patientid, grp
order by patientid, grp;

演示：https://dbfiddle.uk/?rdbms=postgres_14&fiddle=d2ee8fede9bb999ada79310047e5ae27

你如何解决这个 SQL 查询？

How do you solve this SQL query?

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