Sql 根据不同的ID范围创建一组ID

我有一组缺少文档ID范围的表。开始范围是开始ID，结束范围是结束ID，缺少是该范围内缺少的行数，包括其计数中的开始ID和结束ID。我想知道如何将单个ID解析到一个新表中，而不是基于缺少的列的实际范围

以下是数据的呈现方式：

+-------------+-----------+---------+-----------+
| Start Range | End Range | Missing | Date      |
+-------------+-----------+---------+-----------+
| 184         | 186       | 3       | 1/9/1979  |
| 204         | 207       | 4       | 1/9/1979  |
| 209         | 212       | 4       | 1/9/1979  |
| 223         | 224       | 2       | 1/9/1979  |
| 240         | 241       | 2       | 1/10/1979 |
| 243         | 243       | 1       | 1/10/1979 |
| 248         | 249       | 2       | 1/10/1979 |
| 261         | 265       | 5       | 1/11/1979 |
+-------------+-----------+---------+-----------+

我希望获得如下成果：

+-----+-----------+
| ID  | Date      |
+-----+-----------+
| 184 | 1/9/1979  |
| 185 | 1/9/1979 |
| 186 | 1/9/1979 |
| 204 | 1/9/1979 |
| 205 | 1/9/1979 |
| 206 | 1/9/1979 |
| 207 | 1/9/1979 |
| 209 | 1/9/1979 |
| 210 | 1/9/1979 |
| 211 | 1/9/1979 |
| 212 | 1/9/1979 |
| 223 | 1/9/1979 |
| 224 | 1/9/1979 |
| 240 | 1/10/1979 |
| 241 | 1/10/1979 |
| 243 | 1/10/1979 |
| 248 | 1/10/1979 |
| 249 | 1/10/1979 |
| 261 | 1/11/1979 |
| 262 | 1/11/1979 |
| 263 | 1/11/1979 |
| 264 | 1/11/1979 |
| 265 | 1/11/1979 |
+-----+-----------+

---

实现这一目标的最佳方法是什么？感谢您的帮助。

我使用自定义项生成范围，但数字表或计数表也可以

Declare @Table table (StartRange int,EndRange int,Date Date)
Insert into @Table values
(184,186,'1979-01-09'),
(204,207,'1979-01-09')


Select B.ID
      ,A.Date
 From @Table A
 Join (Select ID=cast(RetVal as int) from [dbo].[udf-Create-Range-Number](1,9999,1)) B
   on B.ID between A.StartRange and A.EndRange
 Order by B.ID,Date

返回

ID  Date
184 1979-01-09
185 1979-01-09
186 1979-01-09
204 1979-01-09
205 1979-01-09
206 1979-01-09
207 1979-01-09

UDF

CREATE FUNCTION [dbo].[udf-Create-Range-Number] (@R1 money,@R2 money,@Incr money)

-- Syntax Select * from [dbo].[udf-Create-Range-Number](0,100,2)

Returns 
@ReturnVal Table (RetVal money)

As
Begin
    With NumbTable as (
        Select NumbFrom = @R1
        union all
        Select nf.NumbFrom + @Incr
        From NumbTable nf
        Where nf.NumbFrom < @R2
    )
    Insert into @ReturnVal(RetVal)

    Select NumbFrom from NumbTable Option (maxrecursion 32767)

    Return
End

---

我使用UDF生成范围，但数字表或计数表也可以

Declare @Table table (StartRange int,EndRange int,Date Date)
Insert into @Table values
(184,186,'1979-01-09'),
(204,207,'1979-01-09')


Select B.ID
      ,A.Date
 From @Table A
 Join (Select ID=cast(RetVal as int) from [dbo].[udf-Create-Range-Number](1,9999,1)) B
   on B.ID between A.StartRange and A.EndRange
 Order by B.ID,Date

返回

ID  Date
184 1979-01-09
185 1979-01-09
186 1979-01-09
204 1979-01-09
205 1979-01-09
206 1979-01-09
207 1979-01-09

UDF

CREATE FUNCTION [dbo].[udf-Create-Range-Number] (@R1 money,@R2 money,@Incr money)

-- Syntax Select * from [dbo].[udf-Create-Range-Number](0,100,2)

Returns 
@ReturnVal Table (RetVal money)

As
Begin
    With NumbTable as (
        Select NumbFrom = @R1
        union all
        Select nf.NumbFrom + @Incr
        From NumbTable nf
        Where nf.NumbFrom < @R2
    )
    Insert into @ReturnVal(RetVal)

    Select NumbFrom from NumbTable Option (maxrecursion 32767)

    Return
End

---

建立理货台或CTE。

这只是一列递增的整数n

Select tally.n as Id,  missingRange.date
from tally
 inner join missingRange
On tally.n >= beginRange 
And tally.n <= endRange

---

建立理货台或CTE。

这只是一列递增的整数n

Select tally.n as Id,  missingRange.date
from tally
 inner join missingRange
On tally.n >= beginRange 
And tally.n <= endRange

---

数字或理货台将是最简单的方法数字或理货台将是最简单的方法完美-这完全符合我的需要。谢谢你教我一些新东西，我非常感激。很高兴能帮上忙。记录在案的是，迈克的帖子也很完美——这正是我所需要的。谢谢你教我一些新东西，我非常感激。很高兴能帮上忙。记录在案的是，迈克的帖子也很中肯