我正在研究一个康复组织的问题，其中租户（客户/病人）在他们第一次到达时住在一幢建筑物中，因为他们在治疗中取得进展，他们搬到另一幢建筑物，当他们接近治疗结束时在第三栋楼里。
出于资助目的，我们需要知道每个月每个建筑物中租户花了多少个夜晚。 我可以使用DateDiff来获取总夜数，但是如何获得每个建筑物中每个月的每个客户的总数？
例如，John Smith在A楼9 / 12-11 / 3;搬到B楼11 / 3-15;搬到C楼，现在仍然在那里：11月15日 - 今天
什么查询返回显示他花费的夜数的结果： 在Septmeber，10月和11月建造A. 11月的Buidling B. 11月C楼
两个表格包含客户的名称，建筑物名称和搬入日期以及搬出日期
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')
感谢您的帮助。
答案 0 :(得分：2)
我使用正确规范化的数据库架构，您的Buildings表没有这样的用处。分手后，我相信得到你的答案会很容易。
编辑（和更新）：这是一个CTE，它将采用这种奇怪的表结构并将其拆分为更规范化的形式，显示用户ID，建筑物名称，移入和移出日期。通过分组您想要的（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
对您的样本数据运行此查询会产生以下输出：
ID_U Building Move_In Move_Out
-------- ----------- ----------------------- -----------------------
A1398 Kalgan 2010-10-06 00:00:00.000 2010-11-23 18:35:59.050
A2938 Kalgan 2010-09-12 00:00:00.000 2010-11-03 00:00:00.000
A2938 Rufus 2010-11-03 00:00:00.000 2010-11-15 00:00:00.000
A2938 Waylon 2010-11-15 00:00:00.000 2010-11-23 18:35:59.050
(4 row(s) affected)
正如您所看到的，从这里可以更容易地隔离每个患者或建筑物的天数，并且还可以查找特定月份的记录并计算该情况下的正确停留持续时间。请注意，CTE显示仍在建筑物内的患者的当前日期。
编辑（再次）：为了获得所有相关年份的所有月份，包括其开始日期和结束日期，您可以使用这样的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;
因为我们现在有一个感兴趣的所有日期范围的表格表示，我们只是将它们加在一起：
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
最终产生此输出：
Year Month ID_U Building Days
----------- ------------ -------- ---------- -----------
2010 October A1398 Kalgan 25
2010 November A1398 Kalgan 22
2010 September A2938 Kalgan 18
2010 October A2938 Kalgan 30
2010 November A2938 Kalgan 2
2010 November A2938 Rufus 12
2010 November A2938 Waylon 8
答案 1 :(得分：0)
- 设置您想要的月份的日期
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
答案 2 :(得分：0)
尝试使用日期表。例如，你可以像这样创建一个：
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)
只需修改初始日期数量即可满足您的需求（在我的测试实例中，上述产生的日期为2000-01-02至2015-10-26）。使用日期表，查询非常简单，如下所示：
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
答案 3 :(得分：0)
如果您无法重构Building表，您可以创建一个查询，为您规范化它并允许更容易的计算：
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