SQL按月、按用户、按位置计算派驻天数

我正在查询一个康复组织，那里的租户/客户/患者在第一次到达时住在一栋楼里，当他们在治疗中取得进展时，他们会搬到另一栋楼里，当他们接近治疗结束时，他们会住在第三栋楼里

为了筹集资金，我们需要知道每个租户每月在每栋楼里住多少晚。 我可以使用DateDiff来获得总的夜数，但是如何获得每个建筑中每个月每个客户的总夜数呢

例如，约翰·史密斯正在建造一座9/12-11/3大楼；搬到B栋11/3-15；迁至C栋，至今仍在：11月15日-今天

什么查询返回一个结果，显示他在其中度过的夜晚数： 9月、10月和11月在A栋大楼。 11月建造B 十一月份的C大楼

两个表包含客户的名称、建筑名称和迁入日期和迁出日期

CREATE TABLE [dbo].[clients](
[ID] [nvarchar](50) NULL,
[First_Name] [nvarchar](100) NULL,
[Last_Name] [nvarchar](100) NULL
) ON [PRIMARY]

--populate w/ two records  
insert into clients (ID,First_name, Last_name)
values ('A2938', 'John', 'Smith')

insert into clients (ID,First_name, Last_name)
values ('A1398', 'Mary', 'Jones')




CREATE TABLE [dbo].[Buildings](
[ID_U] [nvarchar](50) NULL,
[Move_in_Date_Building_A] [datetime] NULL,
[Move_out_Date_Building_A] [datetime] NULL,
[Move_in_Date_Building_B] [datetime] NULL,
[Move_out_Date_Building_B] [datetime] NULL,
[Move_in_Date_Building_C] [datetime] NULL,
[Move_out_Date_Building_C] [datetime] NULL,
[Building_A] [nvarchar](50) NULL,
[Building_B] [nvarchar](50) NULL,
[Building_C] [nvarchar](50) NULL
) ON [PRIMARY]


-- Populate the tables with two records
insert into buildings (ID_U,Move_in_Date_Building_A,Move_out_Date_Building_A, Move_in_Date_Building_B,
Move_out_Date_Building_B, Move_in_Date_Building_C, Building_A, Building_B, Building_C)
VALUES ('A2938','2010-9-12', '2010-11-3','2010-11-3','2010-11-15', '2010-11-15', 'Kalgan', 'Rufus','Waylon')


insert into buildings (ID_U,Move_in_Date_Building_A,Building_A)
VALUES ('A1398','2010-10-6', 'Kalgan')

---

谢谢你的帮助

我会使用一个正确规范化的数据库模式，您的Buildings表没有这样的用处。在把它分开之后，我相信得到你的答案会很容易

编辑和更新：这里有一个CTE，它将采用这个奇怪的表结构，并将其拆分为一个更规范化的表单，显示用户id、建筑名称、迁入和迁出日期。通过对您想要的数据进行分组并使用DATEPART等，您应该能够获得所需的数据

WITH User_Stays AS (
    SELECT
        ID_U,
        Building_A Building,
        Move_in_Date_Building_A Move_In,
        COALESCE(Move_out_Date_Building_A, CASE WHEN ((Move_in_Date_Building_B IS NULL) OR (Move_in_Date_Building_C<Move_in_Date_Building_B)) AND (Move_in_Date_Building_C>Move_in_Date_Building_A) THEN Move_in_Date_Building_C WHEN Move_in_Date_Building_B>=Move_in_Date_Building_A THEN Move_in_Date_Building_B END, GETDATE()) Move_Out
    FROM dbo.Buildings 
    WHERE Move_in_Date_Building_A IS NOT NULL   
    UNION ALL
    SELECT
        ID_U, 
        Building_B,
        Move_in_Date_Building_B, 
        COALESCE(Move_out_Date_Building_B, CASE WHEN ((Move_in_Date_Building_A IS NULL) OR (Move_in_Date_Building_C<Move_in_Date_Building_A)) AND (Move_in_Date_Building_C>Move_in_Date_Building_B) THEN Move_in_Date_Building_C WHEN Move_in_Date_Building_A>=Move_in_Date_Building_B THEN Move_in_Date_Building_A END, GETDATE())
    FROM dbo.Buildings 
    WHERE Move_in_Date_Building_B IS NOT NULL
    UNION ALL
    SELECT
        ID_U, 
        Building_C,
        Move_in_Date_Building_C, 
        COALESCE(Move_out_Date_Building_C, CASE WHEN ((Move_in_Date_Building_B IS NULL) OR (Move_in_Date_Building_A<Move_in_Date_Building_B)) AND (Move_in_Date_Building_A>Move_in_Date_Building_C) THEN Move_in_Date_Building_A WHEN Move_in_Date_Building_B>=Move_in_Date_Building_C THEN Move_in_Date_Building_B END, GETDATE())
    FROM dbo.Buildings
    WHERE Move_in_Date_Building_C IS NOT NULL
)
SELECT *
FROM User_Stays
ORDER BY ID_U, Move_In

正如您所看到的，从这里开始，隔离每个患者或建筑物的天数，以及查找特定月份的记录和计算这种情况下的正确住院时间将变得更加容易。请注意，CTE显示仍在建筑物内的患者的当前日期

再次编辑：为了获得所有月份，包括所有相关年份的开始和结束日期，您可以使用如下CTE：

WITH User_Stays AS (             
        [...see above...]
    )
,
    Months AS (          
        SELECT  m.IX,
                y.[Year], dateadd(month,(12*y.[Year])-22801+m.ix,0) StartDate, dateadd(second, -1, dateadd(month,(12*y.[Year])-22800+m.ix,0)) EndDate
                FROM    (            
                    SELECT  1 IX UNION ALL 
                    SELECT  2 UNION ALL 
                    SELECT  3 UNION ALL 
                    SELECT  4 UNION ALL 
                    SELECT  5 UNION ALL 
                    SELECT  6 UNION ALL 
                    SELECT  7 UNION ALL 
                    SELECT  8 UNION ALL 
                    SELECT  9 UNION ALL 
                    SELECT  10 UNION ALL 
                    SELECT  11 UNION ALL 
                    SELECT  12 
                )
        m 
            CROSS JOIN (             
                    SELECT  Datepart(YEAR, us.Move_In) [Year] 
                    FROM    User_Stays us UNION 
                    SELECT  Datepart(YEAR, us.Move_Out) 
                    FROM    User_Stays us 
                )
        y 
    )
SELECT  * 
FROM    months;

select c.First_name, c.Last_name,
    b.Building_A BuildingName, dA.year, dA.month, count(distinct dA.day) daysInBuilding
from clients c
    join Buildings b on c.ID = b.ID_U
    left join Dates dA on dA.date between b.Move_in_Date_Building_A and isnull(b.Move_out_Date_Building_A, getDate())
group by c.First_name, c.Last_name,
    b.Building_A, dA.year, dA.month
UNION
select c.First_name, c.Last_name,
    b.Building_B, dB.year, dB.month, count(distinct dB.day)
from clients c
    join Buildings b on c.ID = b.ID_U
    left join Dates dB on dB.date between b.Move_in_Date_Building_B and isnull(b.Move_out_Date_Building_B, getDate())
group by c.First_name, c.Last_name,
    b.Building_B, dB.year, dB.month
UNION
select c.First_name, c.Last_name,
    b.Building_C, dC.year, dC.month, count(distinct dC.day)
from clients c
    join Buildings b on c.ID = b.ID_U
    left join Dates dC on dC.date between b.Move_in_Date_Building_C and isnull(b.Move_out_Date_Building_C, getDate())
group by c.First_name, c.Last_name,
    b.Building_C, dC.year, dC.month

因此，由于我们现在有了所有感兴趣的日期范围的表格表示，我们只需将其连接在一起：

WITH User_Stays AS ([...]),
Months AS ([...])
SELECT  m.[Year],
    DATENAME(MONTH, m.StartDate) [Month],
    us.ID_U,
    us.Building,
    DATEDIFF(DAY, CASE WHEN us.Move_In>m.StartDate THEN us.Move_In ELSE m.StartDate END, CASE WHEN us.Move_Out<m.EndDate THEN us.Move_Out ELSE DATEADD(DAY, -1, m.EndDate) END) Days 
FROM    Months m 
JOIN User_Stays us ON (us.Move_In < m.EndDate) AND (us.Move_Out >= m.StartDate)
ORDER BY m.[Year],
    us.ID_U,
    m.Ix,
    us.Move_In

---

-设置所需月份的日期

Declare @startDate datetime
declare @endDate datetime

set @StartDate = '09/01/2010'
set @EndDate = '09/30/2010'


select 
-- determine if the stay occurred during this month
    Case When @StartDate <= Move_out_Date_Building_A and @EndDate >= Move_in_Date_Building_A
         Then 
                  (DateDiff(d, @StartDate , @enddate+1) 
                   )
-- drop the days off the front
                - (Case When @StartDate <  Move_in_Date_Building_A
                       Then datediff(d, @StartDate, Move_in_Date_Building_A)
                       Else 0
                  End)
--drop the days of the end
                - (Case When @EndDate > Move_out_Date_Building_A
                       Then datediff(d, @EndDate,  Move_out_Date_Building_A)
                       Else 0
                  End)
        Else 0
    End AS Building_A_Days_Stayed
from Clients c 
inner join Buildings b
on c.id = b.id_u

---

尝试使用日期表。例如，您可以这样创建一个：

CREATE TABLE Dates
(
  [date]    datetime,
  [year]    smallint,
  [month]   tinyint,
  [day]     tinyint
)

INSERT INTO Dates(date)
SELECT dateadd(yy, 100, cast(row_number() over(order by s1.object_id) as datetime))
FROM sys.objects s1
  CROSS JOIN sys.objects s2

UPDATE Dates
SET [year] = year(date),
    [month] = month(date),
    [day] = day(date)

只需修改初始日期填充以满足您对我的测试实例的需求，上述生成日期为2000-01-02到2015-10-26。对于日期表，查询非常简单，如下所示：

WITH User_Stays AS (             
        [...see above...]
    )
,
    Months AS (          
        SELECT  m.IX,
                y.[Year], dateadd(month,(12*y.[Year])-22801+m.ix,0) StartDate, dateadd(second, -1, dateadd(month,(12*y.[Year])-22800+m.ix,0)) EndDate
                FROM    (            
                    SELECT  1 IX UNION ALL 
                    SELECT  2 UNION ALL 
                    SELECT  3 UNION ALL 
                    SELECT  4 UNION ALL 
                    SELECT  5 UNION ALL 
                    SELECT  6 UNION ALL 
                    SELECT  7 UNION ALL 
                    SELECT  8 UNION ALL 
                    SELECT  9 UNION ALL 
                    SELECT  10 UNION ALL 
                    SELECT  11 UNION ALL 
                    SELECT  12 
                )
        m 
            CROSS JOIN (             
                    SELECT  Datepart(YEAR, us.Move_In) [Year] 
                    FROM    User_Stays us UNION 
                    SELECT  Datepart(YEAR, us.Move_Out) 
                    FROM    User_Stays us 
                )
        y 
    )
SELECT  * 
FROM    months;

select c.First_name, c.Last_name,
    b.Building_A BuildingName, dA.year, dA.month, count(distinct dA.day) daysInBuilding
from clients c
    join Buildings b on c.ID = b.ID_U
    left join Dates dA on dA.date between b.Move_in_Date_Building_A and isnull(b.Move_out_Date_Building_A, getDate())
group by c.First_name, c.Last_name,
    b.Building_A, dA.year, dA.month
UNION
select c.First_name, c.Last_name,
    b.Building_B, dB.year, dB.month, count(distinct dB.day)
from clients c
    join Buildings b on c.ID = b.ID_U
    left join Dates dB on dB.date between b.Move_in_Date_Building_B and isnull(b.Move_out_Date_Building_B, getDate())
group by c.First_name, c.Last_name,
    b.Building_B, dB.year, dB.month
UNION
select c.First_name, c.Last_name,
    b.Building_C, dC.year, dC.month, count(distinct dC.day)
from clients c
    join Buildings b on c.ID = b.ID_U
    left join Dates dC on dC.date between b.Move_in_Date_Building_C and isnull(b.Move_out_Date_Building_C, getDate())
group by c.First_name, c.Last_name,
    b.Building_C, dC.year, dC.month

---

如果无法重新构造建筑表，则可以创建一个查询，以便对其进行规范化，并简化计算：

SELECT "A" as Building, BuidlingA as Name, Move_in_Date_Building_A as MoveInDate, 
Move_out_Date_Building_A As MoveOutDate
UNION
SELECT "B", BuidlingB, Move_in_Date_Building_B, Move_out_Date_Building_B 
 UNION
SELECT "C", BuidlingC, Move_in_Date_Building_C, Move_out_Date_Building_C

---

规范化数据库的唯一挑战是它不是我的数据库。不过我会努力的。如果有其他关于如何继续我的想法，我很乐意听到。我添加了一个带有CTE的查询，应该可以让您开始。这个解决方案的好处是它不需要更改数据库，您可以运行一个查询，使其正常化，然后将该查询用作源。我喜欢这样，虽然我不知道如何获得每个月的天数，但客户在每栋楼里。例如，我需要证明A1398在10月25日和11月23日在卡根。如果你喜欢，那就投票吧！我编辑了答案以包含完整的解决方案，只是因为编写代码很有趣。但不要指望别人在这里做你的工作-