列出在另一组记录的x小时内发生的所有SQL记录

我已经接近了吗？
我试图列出所有包含“猿”的记录。。。以及所有的 在任何“Ape”记录之前x天出现的记录。 我想我需要一个自联接表

-- Doesn't work:
SELECT tblA.MyDate, tblA.MyPet  FROM TestTable As tblA                      WHERE tblA.MyPet='Ape'
UNION
SELECT tblA.MyDate, tblA.MyPet  FROM TestTable AS tblA, TestTable AS tblB   WHERE tblA.MyPet='Ape' AND tblA.MyDate>tblB.MyDate-0.5 AND tblA.MyDate<tblB.MyDate
ORDER BY tblA.MyDate ASC

-- Doesn't work:
SELECT * FROM TestTable As tblA     INNER JOIN TestTable As tblB ON tblA.MyPet =         'Ape'     AND tblA.MyDate>tblB.MyDate-1  AND tblA.MyKey>tblB.MyKey

-- Doesn't work:
SELECT * FROM TestTable WHERE MyDate IN (SELECT MyDate-1 FROM TestTable WHERE MyPet='Ape')

-- Doesn't work
SELECT id, uid, date FROM orders current
WHERE EXISTS 
(
   SELECT * from orders future 
   WHERE future.date < DateAdd(DAYS, 1, current.date)
   AND future.date > getdate()
   AND future.uid = current.uid
)

-- Doesn't work
SELECT * FROM TestTable AS tblA
WHERE EXISTS
( 
   SELECT * FROM TestTable AS tblB
   WHERE tblB < DateAdd(DAYS, 1, tblA.MyDate)
   AND tblB.MYDate > GetDate()
   AND tblA.MyKey = tblB.MyKey
)

过程：分别编写查询的每一半，并进行测试以确保其正常工作。然后将它们与一个联盟结合在一起

如果要删除重复项，请使用UNION ALL而不是UNION

这就是你想要的：

SELECT DISTINCT
    tblA.*
FROM TestTable AS tblA
    INNER JOIN TestTable AS tblB
        ON (tblA.MyDate >= DATEADD(day, -1, tblB.MyDate)) AND (tblA.MyDate <= tblB.Mydate) AND (tblB.MyPet = 'Ape');

选择DISTINCT
tblA*
作为tblA从测试表中删除
内部联接测试表作为tblB
在（tblA.MyDate>=DATEADD（day，-1，tblB.MyDate））和（tblA.MyDate

查询1：

-- Doesn't work:
SELECT tblA.MyDate, tblA.MyPet  
FROM TestTable As tblA
WHERE tblA.MyPet='Ape'
UNION all
SELECT tblA.MyDate, tblA.MyPet  
FROM TestTable AS tblA
WHERE dateadd(day, 1, tblA.MyDate) < (select min(MyDate) from TestTable where MyPet = 'Ape')
ORDER BY tblA.MyDate ASC

|                          MYDATE |  MYPET |
--------------------------------------------
| December, 01 2012 06:12:10+0000 |    Cat |
| December, 01 2012 10:11:10+0000 |    Dog |
| December, 01 2012 14:13:10+0000 |   Fish |
| December, 01 2012 16:14:10+0000 |   Duck |
| December, 01 2012 17:15:10+0000 |   Bird |
| December, 01 2012 20:16:10+0000 | Kitten |
| December, 02 2012 01:17:10+0000 |    Dog |
| December, 02 2012 12:19:10+0000 |   Fish |
| December, 02 2012 13:20:10+0000 |   Duck |
| December, 02 2012 14:21:10+0000 |   Bird |
| December, 03 2012 14:25:10+0000 |    Ape |
| December, 06 2012 16:40:10+0000 |    Ape |

编辑

如果您需要在第一次Ape记录之前的x天内发生的记录
那么查询应该是（在示例x=1中）

我认为您正在寻找的公式是：

SELECT *
FROM TestTable tblA
WHERE EXISTS (SELECT *
              FROM TestTable tblB
              WHERE tblB.mydate between tblA.MyDate - X and tblA.MyDate and
                    tblB.MyPet = 'Ape'
             )

两个注意事项。首先，这使用了语法“date-x”，当date是
datetime
而不是
date
时，该语法起作用。它相当于
dateadd（day，-x，tbla.MyDate）

其次，这将返回所有带有“Ape”的记录，因为
between
处理相等

Jesse的查询使用外部连接和distinct实现了本质上相同的功能。
我不知道如何“编写两个部分”。（“然后用UNION连接它们”是最简单的部分。）代码+结果列出的记录比任何“ape”记录的前一天还要早（我看到12月3日的ape…然后是12月1日的记录。）…我对此进行了投票，但准确地说，它应该是
选择distinct tblA.
，因为当多个记录匹配时，您将获得重复的记录。您不需要
外部联接（您没有），因为原始记录将始终匹配。@GordonLinoff我将使用
distinct
更新答案。谢谢！
SELECT tblA.MyDate, tblA.MyPet  
FROM TestTable As tblA
WHERE tblA.MyPet='Ape'
UNION
SELECT tblA.MyDate, tblA.MyPet  
FROM TestTable AS tblA
WHERE (select min(MyDate) from TestTable where MyPet = 'Ape') between tblA.MyDate and dateadd(day, 1, tblA.MyDate)
ORDER BY tblA.MyDate ASC

SELECT *
FROM TestTable tblA
WHERE EXISTS (SELECT *
              FROM TestTable tblB
              WHERE tblB.mydate between tblA.MyDate - X and tblA.MyDate and
                    tblB.MyPet = 'Ape'
             )