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在仔细研究了一个类似的问题并发现它从未提供完整的解决方案之后，我终于找到了我无法解决的问题的核心。我正在寻找一个人可以服用一定数量药物的连续天数。由于处方开始和结束，一个人服用 X 种药物可能有多个不连续的时间间隔。以下 SQL 脚本生成我将立即发布的查询的结果集：另外，我没有 SQL Server 2012。

create table test
(pat_id int, cal_date date, grp_nbr int, drug_qty int,[ranking] int)
go
insert into test(pat_id,cal_date, grp_nbr,drug_qty,[ranking])
values
(1, '1/8/2007',7,2, 1),
(1, '1/9/2007',7,2, 1),
(1, '1/10/2007',7,  2,1),
(1, '1/11/2007',7,  2,1),
(1, '1/12/2007',7,  2,1),
(1, '1/13/2007',7,  2,1),
(1, '1/14/2007',7,  2,1),
(1, '1/15/2007',7,  2,1),
(1, '6/1/2007',7,2, 1),
(1, '6/2/2007',7,2, 1),
(1, '6/3/2007',7,2, 1)

请注意这里有两个不连续的时间间隔，此人同时服用两种药物。在省略的日子里，drug_qty 超过了两个。此示例中的最后一列是我尝试添加另一个可以分组以帮助解决问题的字段（没有用）。

查询创建表：

 CREATE TABLE [dbo].[rx](
            [pat_id] [int] NOT NULL,
            [fill_Date] [date] NOT NULL,
            [script_End_Date]  AS (dateadd(day,[dayssup],[filldate])),
            [drug_Name] [varchar](50) NULL,
            [days_Sup] [int] NOT NULL,
            [quantity] [float] NOT NULL,
            [drug_Class] [char](3) NOT  NULL,
            CHECK(fill_Date <=script_End_Date
PRIMARY KEY CLUSTERED 
(
            [clmid] ASC
)


CREATE TABLE [dbo].[Calendar](
             [cal_date] [date] PRIMARY KEY,
[Year] AS YEAR(cal_date) PERSISTED,
[Month] AS MONTH(cal_date) PERSISTED,
[Day] AS DAY(cal_date) PERSISTED,
             [julian_seq] AS 1+DATEDIFF(DD, CONVERT(DATE, CONVERT(varchar,YEAR(cal_date))+'0101'),cal_date),
     id int identity);

我用来生成结果集的查询：

;WITH x 
     AS (SELECT rx.pat_id, 
                c.cal_date, 
                Count(DISTINCT rx.drug_name) AS distinctDrugs 
         FROM   rx, 
                calendar AS c 
         WHERE  c.cal_date BETWEEN rx.fill_date AND rx.script_end_date 
                AND rx.ofinterest = 1 
         GROUP  BY rx.pat_id, 
                   c.cal_date 
         --the query example I used having count(1) =2, but to illustrate the non-contiguous intervals, in practice I need the below having statement
         HAVING Count(*) > 1), 
     y 
     AS (SELECT x.pat_id, 
                x.cal_date 
                --c2.id is the row number in the calendar table. 
                , 
                c2.id - Row_number() 
                          OVER( 
                            partition BY x.pat_id 
                            ORDER BY x.cal_date) AS grp_nbr, 
                distinctdrugs 
         FROM   x, 
                calendar AS c2 
         WHERE  c2.cal_date = x.cal_date) 
SELECT *, 
       Rank() 
         OVER( 
           partition BY pat_id, grp_nbr 
           ORDER BY distinctdrugs) AS [ranking] 
FROM   y 
WHERE  y.pat_id = 1604012867 
       AND distinctdrugs = 2

除了我不应该在日历表中有一个名为“id”的列之外，这种方法有什么严重的错误吗？我可以让查询向我显示 distinctDrugs=x 的不同间隔，但它只适用于该整数，而不适用于 >1。我的意思是我可以找到患者服用两种药物的单独间隔，但只有当我在有子句中使用 =2，而不是 >1 时。我不能做类似的事情

SELECT pat_id, 
       Min(cal_date), 
       Max(cal_date), 
       distinctdrugs 
FROM   y 
GROUP  BY pat_id, 
          grp_nbr

因为这将选择第二组不连续的日期。有谁知道这个问题的优雅解决方案？

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1 回答 1

1

关键是一个简单的观察。如果你有一个日期序列，那么它们和递增序列之间的差异是恒定的。以下是这样做的，假设您使用的是 SQL Server 2005 或更高版本：

select pat_id, MIN(cal_date), MAX(cal_date), MIN(drug_qty)
from (select t.*,
             cast(cal_date as datetime) - ROW_NUMBER() over (partition by pat_id, drug_qty order by cal_date) as grouping
      from #test t
     ) t
group by pat_id, grouping
于 2013-01-30T20:15:05.663 回答