SQL按月、按用户、按位置计算居住天数

Question

我正在查询一家康复机构，其中租户(客户/患者)在他们第一次到达时住在一栋楼里，随着他们治疗的进展，他们搬到另一栋楼，当他们接近治疗结束时，他们在第三栋楼里。

出于融资目的，我们需要知道租户每个月在每栋建筑中度过了多少个晚上。 我可以使用 DateDiff 获取总住宿天数，但如何获取每个客户每个月在每栋建筑中的总住宿天数？

例如，John Smith 在 A 楼 9/12-11/3；搬到 B 楼 11/3-15；搬到 C 楼并仍然在那里:11 月 15 日 - 今天

什么查询返回显示他度过的夜晚数的结果: 9 月、10 月和 11 月的 A 楼。 11月B栋 11月C栋

两张表保存了客户姓名、楼宇名称和搬入搬出日期

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。

Answer 1

我会使用适当规范化的数据库模式，您的 Buildings 表没有这样用。拆分后，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、搬入和搬出日期。通过根据需要分组(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

此查询对您的样本数据运行产生以下输出:

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)

如您所见，从这里开始，可以更轻松地隔离每个患者或建筑物的天数，还可以更轻松地查找特定月份的记录并计算这种情况下的正确停留时间。请注意，CTE 显示仍在建筑物中的患者的当前日期。

---

编辑(再次):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;

既然我们现在有一个可能感兴趣的所有日期范围的表格表示，我们只需将它们连接在一起:

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

最终产生这个输出:

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