SQL Server 递归查询

Question

几天前我问了一个关于help on a recursive query的问题。

那个问题的问题是“如何获取一个人的约会历史”。

现在我遇到了一个类似的问题，但它应该回答一个稍微不同的问题：

How to get an Appointment history?

例如，如果 ID = 5 的约会被推迟了一次，并且是另一个约会的推迟，我如何得到以下结果？

AppointmentID         PrevAppointmentID
-----------------    ----------------------
1                     NULL
5                     1
12                    5

感谢您的帮助

更新：

这些脚本将有助于为您的试验创建表

CREATE TABLE [dbo].[Appointments](
    [AppointmentID] [int] IDENTITY(1,1) NOT NULL,
    [IssueID] [int] NOT NULL,
    [Location] [varchar](255) NOT NULL,
    [Description] [varchar](255) NOT NULL,
    [AppointmentDate] [datetime] NOT NULL,
    [AppointmentHour] [datetime] NOT NULL,
    [Done] [bit] NOT NULL,
    [PrevAppointmentID] [int] NULL,
 CONSTRAINT [PK_Appointments] PRIMARY KEY CLUSTERED 
(
    [AppointmentID] ASC
)WITH (PAD_INDEX  = OFF, STATISTICS_NORECOMPUTE  = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS  = ON, ALLOW_PAGE_LOCKS  = ON) ON [PRIMARY]
) ON [PRIMARY]

Answer 1

我不想完全劫持他的答案，所以：

使用布拉德的答案

这是获取完整历史记录的查询：

WITH    cte
          AS ( SELECT   AppointmentId ,
                        PrevAppointmentId
               FROM     Appointments
               WHERE    AppointmentId = @AppointmentId
               UNION ALL
               SELECT   prev.AppointmentId ,
                        prev.PrevAppointmentId
               FROM     Appointments prev
                        INNER JOIN cte curr ON prev.AppointmentId = curr.PrevAppointmentId
             ),
        cte1
          AS ( SELECT   AppointmentId ,
                        PrevAppointmentId
               FROM     Appointments
               WHERE    AppointmentId = @AppointmentId
               UNION ALL
               SELECT   prev.AppointmentId ,
                        prev.PrevAppointmentId
               FROM     Appointments prev
                        INNER JOIN cte1 curr ON prev.PrevAppointmentId = curr.AppointmentId
             )
    SELECT  *
    FROM    cte
    UNION
    SELECT  *
    FROM    cte1

Answer 2

逻辑：

1. 获取相关约会
2. 重复加入父应用程序
3. 选择所有结果

查询：

DECLARE @appointmentId INT
SET @appointmentId = 3

--
;
WITH  past
      AS ( SELECT   AppointmentId ,
                    PrevAppointmentId
           FROM     Appointments
           WHERE    AppointmentId = @AppointmentId
           UNION ALL
           SELECT   prev.AppointmentId ,
                    prev.PrevAppointmentId
           FROM     Appointments prev
                    INNER JOIN cte curr ON prev.AppointmentId = curr.PrevAppointmentId
         ),
    future
      AS ( SELECT   AppointmentId ,
                    PrevAppointmentId
           FROM     Appointments
           WHERE    AppointmentId = @AppointmentId
           UNION ALL
           SELECT   prev.AppointmentId ,
                    prev.PrevAppointmentId
           FROM     Appointments prev
                    INNER JOIN cte1 curr ON prev.PrevAppointmentId = curr.AppointmentId
         )
SELECT  *
FROM    past OPTION(MAXRECURSION 500)
UNION
SELECT  *
FROM    future OPTION(MAXRECURSION 500)

Answer 3

回答 OP 的问题：

有效约会

有效日期是当您将DateTime 列添加到控制记录何时变为“有效”的表时。然后将该列添加到表的PK，使每个条目成为在给定时间点（生效日期）“有效”内容的记录。使用有效约会，您永远不会DELETE 或UPDATE，只有INSERT 意味着您始终拥有该对象的完整历史记录。

要查找最有效的记录，请选择具有最大有效但不在未来的行

SELECT *
FROM   Appointments a1
WHERE  EffectiveDate = (SELECT MAX(EffectiveDate)
                        FROM   Appointments a2
                        WHERE  a1.AppointmentId = a2.AppointmentId
                               AND a2.EffectiveDate <= ISNULL(@asOfDate, GETDATE()
                       )

这意味着您还可以提前记录。例如，您今天获准加薪，但在 2 周内不会生效。

因此，要查找约会的历史记录，您只需：

SELECT *
FROM   Appointments
WHERE  AppointmentId = @appointmentId
ORDER BY EffectiveDate