通过一个月的居住天的SQL计算数量，按用户，按位置
https://stackoverflow.com/questions/4258218
-
27-09-2019 - |
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我工作的康复机构，其中租户（客户端/例）生活在一个建筑物的时候，他们初来乍到的查询，因为他们在他们的治疗进展，他们移动到另一个建筑物和他们接近治疗结束时，他们是在第三建筑物。
有关资金的目的，我们需要知道一个租户中的每个建筑在每个月多少个夜晚度过的。 我可以使用则DateDiff得到晚上的总数，但我如何得到总的为每个客户每个月在每栋楼？
例如，约翰·史密斯是在建筑物A 9 / 12-11 / 3;移动到B楼11 / 3-15;移动到C楼上仍然存在：11/15 - 今天
查询返回什么结果，显示晚上他花的数量： A栋在Septmeber，十月和十一月。 十一月Buidling乙 中心C在十一月
两个表中保存了客户的名字，在布展日期和迁出日期建筑物的名称和
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。
解决方案
我会使用正确规范化的数据库架构，你的建筑物表不是这样有用。分开后它相信让你的答案将是相当容易的。
编辑（和更新的）：这是一种CTE将这个奇怪的表的结构和它分割成更规范化的形式，显示用户ID，在建筑物的名称，移动和迁出日期。通过对你想要的那些分组（使用DATEPART()等），你应该能够得到你需要的数据。
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
此在您的样本数据的查询的运行产生他下面的输出：
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)
你可以看到，从这里开始它会更容易每位患者或建筑物的天隔离，并找到具体个月的记录，并计算在这种情况下正确的停留时间。注意，CTE显示当前日期的患者，其仍然在建筑物。
编辑（再一次）：为了让所有月份，包括所有相关年份的开始和结束日期，你可以用一个CTE这样的：
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;
因此，既然我们现在拥有的所有日期范围的表格表示它可以是感兴趣的，我们只是一起加入这个：
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
这最终产生以下输出：
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
其他提示
- 设置哪个月你想
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，你可以创建一个像这样：
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)
只需修改初始日期人口，以满足您的需求（在我的测试情况下，上述产生日期从2000年1月2日至2015年10月26日）。随着日期表，查询是相当直接的，是这样的：
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
如果你不能重组的建表，您可以创建一个查询，将其归为您和以便更容易计算：
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
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