Question

我使用这种结构创建表约会：

  CREATE TABLE Appointments
    (
         [Id] bigint, 
         [Name] varchar(250),
         [DateInit] date
    );


INSERT INTO Appointments ([Id],  [Name],  [DateInit])
values 
(1000, 'Lorena', '03/06/2016'),
(1000, 'Lorena', '01/06/2016'),
(1000, 'Lorena', '08/06/2016'),
(1000, 'Lorena', '10/06/2016'),
(1000, 'Lorena', '02/06/2016'),
(1000, 'Lorena', '20/06/2016'),
(7000, 'Susan', '04/06/2016'),
(7000, 'Susan', '08/06/2016'),
(7000, 'Susan', '09/06/2016'),
(7000, 'Susan', '01/06/2016');

这是最终结果：

我需要在第二天和前一天得到结果，例如，如果今天是2016年06月06日＆＃39;我需要从今天开始在表格中插入最后一个约会的结果，从今天开始在表格中插入下一个约会，我需要的结果是这样的：

   Name    Last Visit    Next Visit
    -----   ----------    -----------
   Lorena   2016-06-02    2016-06-08
   Susan    2016-06-01    2016-06-04

我怎样才能得到这个结果？感谢

Answer 1

执行GROUP BY，使用case表达式选择最后一次约会，并将未来的约会分开：

select name,
       max(case when DateInit < CONVERT(DATE,GETDATE()) then DateInit end) as LastVisit,
       min(case when DateInit > CONVERT(DATE,GETDATE()) then DateInit end) as NextVisit
from Appointments
group by name

Answer 2

我这样做是为了加入上一次和下一次访问，就像这样;

SELECT DISTINCT
     a.ID
    ,a.NAME
    ,l.LastVisit
    ,n.NextVisit
FROM Appointments a
LEFT JOIN (
    SELECT ID
        ,MIN(DateInit) NextVisit
    FROM Appointments
    WHERE DateInit > GETDATE()
    GROUP BY ID
    ) n ON a.ID = n.ID
LEFT JOIN (
    SELECT ID
        ,MAX(DateInit) LastVisit
    FROM Appointments
    WHERE DateInit < GETDATE()
    GROUP BY ID
    ) l ON a.ID = l.ID

Answer 3

DECLARE  @Appointments TABLE
    (
         [Id] bigint, 
         [Name] varchar(250),
         [DateInit] date
    );


INSERT INTO @Appointments ([Id],  [Name],  [DateInit])
values 
(1000, 'Lorena','2016/06/03'),
(1000, 'Lorena','2016/06/01'),
(1000, 'Lorena','2016/06/08'),
(1000, 'Lorena','2016/06/10'),
(1000, 'Lorena','2016/06/02'),
(1000, 'Lorena','2016/06/20'),
(7000, 'Susan', '2016/06/04'),
(7000, 'Susan', '2016/06/08'),
(7000, 'Susan', '2016/06/09'),
(7000, 'Susan', '2016/06/01');

DECLARE @Today DATE = GETDATE();

WITH CTE
AS (
    SELECT A.NAME
        ,ROW_NUMBER() OVER (
            PARTITION BY ID ORDER BY ID
            ) RN
        ,(
            SELECT TOP 1 DateInit
            FROM @Appointments B
            WHERE B.ID = A.ID
                AND DateInit < @TODAY
            ORDER BY DateInit DESC
            ) [Last Visit]
        ,(
            SELECT TOP 1 DateInit
            FROM @Appointments B
            WHERE B.ID = A.ID
                AND DateInit > @TODAY
            ORDER BY DateInit
            ) [Next Visit]
    FROM @Appointments A
        --GROUP BY ID
    )
SELECT C.NAME
    ,C.[Last Visit]
    ,C.[Next Visit]
    ,RN
FROM CTE C
WHERE RN = 1

如何从日期和日期的最后一个记录日期获取下一个记录日期？

3 个答案: