创建一个季度每天处方金额的日期表

时间：2013-10-02 19:35:29

标签： sql sql-server

我正在尝试开发一个查询，以确定个人在一个季度内每天所服用的药物数量。在某些日子里，没有规定的药物，对于其他人，可能有重叠，我需要一个总量（意思是，每个总和一天的力量）。药物数量，优势，供应天数等可能有所不同。这是一些数据：

create table #MemberInfo 
(ProgramName varchar(255), 
DateFilled datetime, 
DaySupply integer, 
MemberID varchar(255), 
Strength integer, 
Tradename varchar(255));

insert into #MemberInfo
Values ('InsureCo', '20130612', 30, 'MEM001', 10, 'Sedative')
, ('InsureCo', '20130429', 30, 'MEM001', 20, 'Sedative')
, ('InsureCo', '20130401', 30, 'MEM001', 20, 'Sedative')
, ('InsureCo', '20130529', 30, 'MEM001', 30, 'Sedative')

我真的不知道最佳方法可能是在一个季度内将某一天服用的药物加起来。如果可以的话，我想避免使用游标。我正在考虑创建一个临时表，其中包含一个季度的所有日期，然后以某种方式将这些日期加入药物的每一天（即，DateFilled +每隔一天直到DaySupply）。一旦我达到每个药物在一个季度中的日期和数量的点，我可以按天分组并获得每天的力量总和。我还需要能够获得超过四分之一的平均金额。

附加要求：

  1. 我有开始日期和天数。我想创建一行 每个成员每天都有处方（并且做了 所有处方都一样）。然后我会总结力量 每天所有的药物。如果它有助于任何，所有的药物 将属于同一类，力量将是相同的 剂量，这意味着我可以总结它们。

  2. 对于报告，我需要能够连续数天计算 金额大于某些截止值（假设为100）。这就是我的原因 试图每天获得金额。

    Desired output

MemberID    Date        SumStrength
MEM001     2013-04-29  40
MEM001     2013-04-30  40
MEM001     2013-05-01  20
ETC FOR EVERY DAY FOR THIS MEMBER

MEM002     2013-04-01  60
MEM002     2013-04-02  40
ETC FOR EVERY DAY FOR THIS MEMBER

3 个答案:

答案 0 :(得分：1)

我认为只是一个简单的小组。

create table #MemberInfo 
(ProgramName varchar(255), 
DateFilled datetime, 
DaySupply integer, 
MemberID varchar(255), 
Strength integer, 
Tradename varchar(255));

insert into #MemberInfo
Values ('InsureCo', '20130612', 30, 'MEM001', 10, 'Sedative')
, ('InsureCo', '20130429', 30, 'MEM001', 20, 'Sedative')
, ('InsureCo', '20130429', 30, 'MEM002', 25, 'Sedative')
, ('InsureCo', '20130515', 30, 'MEM002', 25, 'Sedative')
, ('InsureCo', '20130401', 30, 'MEM001', 20, 'Sedative')
, ('InsureCo', '20130529', 30, 'MEM001', 30, 'Sedative')
, ('InsureCo', '20130529', 30, 'MEM003', 35, 'Sedative')
, ('InsureCo', '20130529', 30, 'MEM003', 45, 'Sedative')

select memberid,datefilled,SUM(strength) as [Strength sum]
from #MemberInfo
where memberid = 'MEM003' -- or whatever, could be a parameter
group by memberid,DateFilled

order by Memberid,DateFilled

drop table #MemberInfo

答案 1 :(得分：1)

以下是如何使用CTE构建日历并使用OVER(PARTION BY)

进行聚合的示例

<强>查询： 

-- Declare a Start and End Date required to build a calendar
DECLARE @StartDate DATETIME = '2013-01-01'
DECLARE @EndDate   DATETIME = '2015-01-01'

-- Build out a Day/Quarter Calendar
;WITH Calendar ([Date], [Quarter]) AS (
    SELECT @StartDate, 1
    UNION ALL
    SELECT [Date] + 1, (DATEDIFF(m, @StartDate, [Date] + 1) / 3) + 1
    FROM Calendar 
    WHERE [Date] + 1 < @EndDate
)

-- Build Result Set
SELECT  ProgramName, 
        DateFilled, 
        DaySupply, 
        MemberID, 
        Strength,
        Quarter,
        SUM(Strength) OVER(PARTITION BY ProgramName, DaySupply, MemberID, Quarter) AS QuarterlyTotal,
        AVG(Strength) OVER(PARTITION BY ProgramName, DaySupply, MemberID, Quarter) AS QuarterlyAverage
FROM #MemberInfo MI
JOIN Calendar C ON MI.DateFilled = C.[Date]
ORDER BY MemberID, DateFilled
OPTION (MAXRECURSION 0)

测试数据： 

create table #MemberInfo 
(ProgramName varchar(255), 
DateFilled datetime, 
DaySupply integer, 
MemberID varchar(255), 
Strength integer, 
Tradename varchar(255));

INSERT INTO #MemberInfo
Values
    --MEM001
    --Q1
     ('InsureCo', '20130112', 30, 'MEM001', 10, 'Sedative')
    ,('InsureCo', '20130129', 30, 'MEM001', 20, 'Sedative')
    ,('InsureCo', '20130401', 30, 'MEM001', 20, 'Sedative')
    --Q2
    ,('InsureCo', '20130529', 30, 'MEM001', 30, 'Sedative')
    ,('InsureCo', '20130429', 30, 'MEM001', 20, 'Sedative')
    ,('InsureCo', '20130401', 30, 'MEM001', 20, 'Sedative')
    --Q3
    ,('InsureCo', '20130829', 30, 'MEM001', 30, 'Sedative')

    --MEM002
    --Q1
    ,('InsureCo', '20130112', 30, 'MEM002', 10, 'Sedative')
    ,('InsureCo', '20130129', 30, 'MEM002', 20, 'Sedative')
    ,('InsureCo', '20130401', 30, 'MEM002', 20, 'Sedative')
    --Q2
    ,('InsureCo', '20130529', 30, 'MEM002', 30, 'Sedative')
    ,('InsureCo', '20130429', 30, 'MEM002', 20, 'Sedative')
    ,('InsureCo', '20130401', 30, 'MEM002', 20, 'Sedative')
    --Q3
    ,('InsureCo', '20130829', 30, 'MEM002', 30, 'Sedative')
    --Q4
    ,('InsureCo', '20131129', 30, 'MEM002', 30, 'Sedative')

<强>结果： 

ProgramName DateFilled  DaySupply   MemberID    Strength    Quarter QuarterlyTotal  QuarterlyAverage
InsureCo    2013-01-12  30          MEM001      10          1       30              15
InsureCo    2013-01-29  30          MEM001      20          1       30              15
InsureCo    2013-04-01  30          MEM001      20          2       90              22
InsureCo    2013-04-01  30          MEM001      20          2       90              22
InsureCo    2013-04-29  30          MEM001      20          2       90              22
InsureCo    2013-05-29  30          MEM001      30          2       90              22
InsureCo    2013-08-29  30          MEM001      30          3       30              30
InsureCo    2013-01-12  30          MEM002      10          1       30              15
InsureCo    2013-01-29  30          MEM002      20          1       30              15
InsureCo    2013-04-01  30          MEM002      20          2       90              22
InsureCo    2013-04-01  30          MEM002      20          2       90              22
InsureCo    2013-04-29  30          MEM002      20          2       90              22
InsureCo    2013-05-29  30          MEM002      30          2       90              22
InsureCo    2013-08-29  30          MEM002      30          3       30              30
InsureCo    2013-11-29  30          MEM002      30          4       30              30

答案 2 :(得分：0)

今晚我做了一些游戏，我离得更近了：

一些数据：

create TABLE dateranges (drug VARCHAR(5), date_begin DATETIME, numdays integer,strength        integer)
INSERT into dateranges values ('DrugA', '2010-01-01', 5, 10);
INSERT into dateranges values ('DrugB', '2008-02-27', 10, 20);
INSERT into dateranges values ('DrugC', '2010-04-26', 3, 20);
INSERT into dateranges values ('DrugD', '2000-02-01', 5, 30);

CTE：

WITH cte (id, d, s)
     AS (SELECT tbl.drug AS id
                ,tbl.date_begin AS d
                ,tbl.strength AS s
           FROM dateranges tbl
          WHERE DATEDIFF(DAY, tbl.date_begin, tbl.date_begin+numdays-1) <= 100
         UNION ALL
         SELECT tbl.drug AS id
                ,DATEADD(DAY, 1, cte.d) AS d
                ,tbl.strength as s
           FROM cte
                INNER JOIN dateranges tbl
                  ON cte.id = tbl.drug
          WHERE cte.d < tbl.date_begin+numdays-1)
SELECT id AS drug
       ,d AS dates
       ,s AS strength
  FROM cte
 ORDER BY id, d, s

结果：

DRUG    DATES   STRENGTH
DrugA   January, 01 2010 00:00:00+0000  10
DrugA   January, 02 2010 00:00:00+0000  10
DrugA   January, 03 2010 00:00:00+0000  10
DrugA   January, 04 2010 00:00:00+0000  10
DrugA   January, 05 2010 00:00:00+0000  10
DrugB   February, 27 2008 00:00:00+0000 20
DrugB   February, 28 2008 00:00:00+0000 20
DrugB   February, 29 2008 00:00:00+0000 20
DrugB   March, 01 2008 00:00:00+0000    20
DrugB   March, 02 2008 00:00:00+0000    20
DrugB   March, 03 2008 00:00:00+0000    20
DrugB   March, 04 2008 00:00:00+0000    20
DrugB   March, 05 2008 00:00:00+0000    20
DrugB   March, 06 2008 00:00:00+0000    20
DrugB   March, 07 2008 00:00:00+0000    20
DrugC   April, 26 2010 00:00:00+0000    20
DrugC   April, 27 2010 00:00:00+0000    20
DrugC   April, 28 2010 00:00:00+0000    20
DrugD   February, 01 2000 00:00:00+0000 30
DrugD   February, 02 2000 00:00:00+0000 30
DrugD   February, 03 2000 00:00:00+0000 30
DrugD   February, 04 2000 00:00:00+0000 30
DrugD   February, 05 2000 00:00:00+0000 30

从这里开始，我计划按药物，日期，力量（总和力量）进行分组。我应该能够将这些结果扔到临时表中，然后计算超过我提到的阈值的天数。

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