我有两个数字作为用户的输入，例如1000和1050。
如何使用sql查询在单独的行中生成这两个数字之间的数字？我想要这个：
1000
1001
1002
1003
.
.
1050
答案 0 :(得分：117)
使用VALUES关键字选择非持久值。然后使用JOIN生成许多组合（可以扩展以创建数十万行甚至更多）。
SELECT ones.n + 10*tens.n + 100*hundreds.n + 1000*thousands.n
FROM (VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) ones(n),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) tens(n),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) hundreds(n),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) thousands(n)
WHERE ones.n + 10*tens.n + 100*hundreds.n + 1000*thousands.n BETWEEN @userinput1 AND @userinput2
ORDER BY 1
一个较短的选择，这不容易理解：
WITH x AS (SELECT n FROM (VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) v(n))
SELECT ones.n + 10*tens.n + 100*hundreds.n + 1000*thousands.n
FROM x ones, x tens, x hundreds, x thousands
ORDER BY 1
答案 1 :(得分：79)
另一种解决方案是递归CTE：
DECLARE @startnum INT=1000
DECLARE @endnum INT=1050
;
WITH gen AS (
SELECT @startnum AS num
UNION ALL
SELECT num+1 FROM gen WHERE num+1<=@endnum
)
SELECT * FROM gen
option (maxrecursion 10000)
答案 2 :(得分：26)
SELECT DISTINCT n = number
FROM master..[spt_values]
WHERE number BETWEEN @start AND @end
请注意，此表格最多为2048，因为这些数字有差距。
这是使用系统视图的一种稍好的方法（从SQL-Server 2005开始）：
;WITH Nums AS
(
SELECT n = ROW_NUMBER() OVER (ORDER BY [object_id])
FROM sys.all_objects
)
SELECT n FROM Nums
WHERE n BETWEEN @start AND @end
ORDER BY n;
或使用自定义数字表。致Aaron Bertrand的信用，我建议阅读整篇文章：Generate a set or sequence without loops
答案 3 :(得分：24)
我最近编写了这个内联表值函数来解决这个问题。除了内存和存储之外，它的范围不受限制。它不访问任何表，因此通常不需要磁盘读取或写入。它在每次迭代时以指数方式添加连接值，因此即使对于非常大的范围，它也非常快。它在我的服务器上在五秒内创建了一千万条记录。它也适用于负值。
CREATE FUNCTION [dbo].[fn_ConsecutiveNumbers]
(
@start int,
@end int
) RETURNS TABLE
RETURN
select
x268435456.X
| x16777216.X
| x1048576.X
| x65536.X
| x4096.X
| x256.X
| x16.X
| x1.X
+ @start
X
from
(VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12),(13),(14),(15)) as x1(X)
join
(VALUES (0),(16),(32),(48),(64),(80),(96),(112),(128),(144),(160),(176),(192),(208),(224),(240)) as x16(X)
on x1.X <= @end-@start and x16.X <= @end-@start
join
(VALUES (0),(256),(512),(768),(1024),(1280),(1536),(1792),(2048),(2304),(2560),(2816),(3072),(3328),(3584),(3840)) as x256(X)
on x256.X <= @end-@start
join
(VALUES (0),(4096),(8192),(12288),(16384),(20480),(24576),(28672),(32768),(36864),(40960),(45056),(49152),(53248),(57344),(61440)) as x4096(X)
on x4096.X <= @end-@start
join
(VALUES (0),(65536),(131072),(196608),(262144),(327680),(393216),(458752),(524288),(589824),(655360),(720896),(786432),(851968),(917504),(983040)) as x65536(X)
on x65536.X <= @end-@start
join
(VALUES (0),(1048576),(2097152),(3145728),(4194304),(5242880),(6291456),(7340032),(8388608),(9437184),(10485760),(11534336),(12582912),(13631488),(14680064),(15728640)) as x1048576(X)
on x1048576.X <= @end-@start
join
(VALUES (0),(16777216),(33554432),(50331648),(67108864),(83886080),(100663296),(117440512),(134217728),(150994944),(167772160),(184549376),(201326592),(218103808),(234881024),(251658240)) as x16777216(X)
on x16777216.X <= @end-@start
join
(VALUES (0),(268435456),(536870912),(805306368),(1073741824),(1342177280),(1610612736),(1879048192)) as x268435456(X)
on x268435456.X <= @end-@start
WHERE @end >=
x268435456.X
| isnull(x16777216.X, 0)
| isnull(x1048576.X, 0)
| isnull(x65536.X, 0)
| isnull(x4096.X, 0)
| isnull(x256.X, 0)
| isnull(x16.X, 0)
| isnull(x1.X, 0)
+ @start
GO
SELECT X FROM fn_ConsecutiveNumbers(5, 500);
它也适用于日期和时间范围：
SELECT DATEADD(day,X, 0) DayX
FROM fn_ConsecutiveNumbers(datediff(day,0,'5/8/2015'), datediff(day,0,'5/31/2015'))
SELECT DATEADD(hour,X, 0) HourX
FROM fn_ConsecutiveNumbers(datediff(hour,0,'5/8/2015'), datediff(hour,0,'5/8/2015 12:00 PM'));
您可以在其上使用交叉应用连接，以根据表中的值拆分记录。因此，例如，为了在表格中的时间范围内创建每分钟的记录，您可以执行以下操作：
select TimeRanges.StartTime,
TimeRanges.EndTime,
DATEADD(minute,X, 0) MinuteX
FROM TimeRanges
cross apply fn_ConsecutiveNumbers(datediff(hour,0,TimeRanges.StartTime),
datediff(hour,0,TimeRanges.EndTime)) ConsecutiveNumbers
答案 4 :(得分：19)
我使用的最佳选择如下：
DECLARE @min bigint, @max bigint
SELECT @Min=919859000000 ,@Max=919859999999
SELECT TOP (@Max-@Min+1) @Min-1+row_number() over(order by t1.number) as N
FROM master..spt_values t1
CROSS JOIN master..spt_values t2
我已经使用它生成了数百万条记录，而且效果很好。
答案 5 :(得分：8)
它对我有用！
select top 50 ROW_NUMBER() over(order by a.name) + 1000 as Rcount
from sys.all_objects a
答案 6 :(得分：6)
没什么新东西，但我改写了Brian Pressler解决方案，以使其更容易使用，这对某人可能有用（即使只是为了我）：
alter function [dbo].[fn_GenerateNumbers]
(
@start int,
@end int
) returns table
return
with
b0 as (select n from (values (0),(0x00000001),(0x00000002),(0x00000003),(0x00000004),(0x00000005),(0x00000006),(0x00000007),(0x00000008),(0x00000009),(0x0000000A),(0x0000000B),(0x0000000C),(0x0000000D),(0x0000000E),(0x0000000F)) as b0(n)),
b1 as (select n from (values (0),(0x00000010),(0x00000020),(0x00000030),(0x00000040),(0x00000050),(0x00000060),(0x00000070),(0x00000080),(0x00000090),(0x000000A0),(0x000000B0),(0x000000C0),(0x000000D0),(0x000000E0),(0x000000F0)) as b1(n)),
b2 as (select n from (values (0),(0x00000100),(0x00000200),(0x00000300),(0x00000400),(0x00000500),(0x00000600),(0x00000700),(0x00000800),(0x00000900),(0x00000A00),(0x00000B00),(0x00000C00),(0x00000D00),(0x00000E00),(0x00000F00)) as b2(n)),
b3 as (select n from (values (0),(0x00001000),(0x00002000),(0x00003000),(0x00004000),(0x00005000),(0x00006000),(0x00007000),(0x00008000),(0x00009000),(0x0000A000),(0x0000B000),(0x0000C000),(0x0000D000),(0x0000E000),(0x0000F000)) as b3(n)),
b4 as (select n from (values (0),(0x00010000),(0x00020000),(0x00030000),(0x00040000),(0x00050000),(0x00060000),(0x00070000),(0x00080000),(0x00090000),(0x000A0000),(0x000B0000),(0x000C0000),(0x000D0000),(0x000E0000),(0x000F0000)) as b4(n)),
b5 as (select n from (values (0),(0x00100000),(0x00200000),(0x00300000),(0x00400000),(0x00500000),(0x00600000),(0x00700000),(0x00800000),(0x00900000),(0x00A00000),(0x00B00000),(0x00C00000),(0x00D00000),(0x00E00000),(0x00F00000)) as b5(n)),
b6 as (select n from (values (0),(0x01000000),(0x02000000),(0x03000000),(0x04000000),(0x05000000),(0x06000000),(0x07000000),(0x08000000),(0x09000000),(0x0A000000),(0x0B000000),(0x0C000000),(0x0D000000),(0x0E000000),(0x0F000000)) as b6(n)),
b7 as (select n from (values (0),(0x10000000),(0x20000000),(0x30000000),(0x40000000),(0x50000000),(0x60000000),(0x70000000)) as b7(n))
select s.n
from (
select
b7.n
| b6.n
| b5.n
| b4.n
| b3.n
| b2.n
| b1.n
| b0.n
+ @start
n
from b0
join b1 on b0.n <= @end-@start and b1.n <= @end-@start
join b2 on b2.n <= @end-@start
join b3 on b3.n <= @end-@start
join b4 on b4.n <= @end-@start
join b5 on b5.n <= @end-@start
join b6 on b6.n <= @end-@start
join b7 on b7.n <= @end-@start
) s
where @end >= s.n
GO
答案 7 :(得分：6)
最好的方法是使用递归ctes。
declare @initial as int = 1000;
declare @final as int =1050;
with cte_n as (
select @initial as contador
union all
select contador+1 from cte_n
where contador <@final
) select * from cte_n option (maxrecursion 0)
saludos。
答案 8 :(得分：6)
如果您在服务器中安装CLR程序集时遇到问题，那么在.NET中编写一个表值函数是个不错的选择。通过这种方式，您可以使用简单的语法，轻松加入其他查询，并且因为结果是流式传输而不会浪费内存。
创建包含以下类的项目：
using System;
using System.Collections;
using System.Data;
using System.Data.Sql;
using System.Data.SqlTypes;
using Microsoft.SqlServer.Server;
namespace YourNamespace
{
public sealed class SequenceGenerator
{
[SqlFunction(FillRowMethodName = "FillRow")]
public static IEnumerable Generate(SqlInt32 start, SqlInt32 end)
{
int _start = start.Value;
int _end = end.Value;
for (int i = _start; i <= _end; i++)
yield return i;
}
public static void FillRow(Object obj, out int i)
{
i = (int)obj;
}
private SequenceGenerator() { }
}
}
将程序集放在服务器上的 并运行：
USE db;
CREATE ASSEMBLY SqlUtil FROM 'c:\path\to\assembly.dll'
WITH permission_set=Safe;
CREATE FUNCTION [Seq](@start int, @end int)
RETURNS TABLE(i int)
AS EXTERNAL NAME [SqlUtil].[YourNamespace.SequenceGenerator].[Generate];
现在你可以运行：
select * from dbo.seq(1, 1000000)
答案 9 :(得分：5)
2年后，但我发现我有同样的问题。这是我如何解决它。 （编辑为包含参数）
DECLARE @Start INT, @End INT
SET @Start = 1000
SET @End = 1050
SELECT TOP (@End - @Start+1) ROW_NUMBER() OVER (ORDER BY S.[object_id])+(@Start - 1) [Numbers]
FROM sys.all_objects S WITH (NOLOCK)
答案 10 :(得分：3)
slartidan's answer可以通过消除对笛卡尔积的所有引用并使用ROW_NUMBER()代替（execution plan compared）来改善性能：
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS n FROM
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x1(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x2(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x3(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x4(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x5(x)
ORDER BY n
将其包裹在CTE中并添加where子句以选择所需的数字：
DECLARE @n1 AS INT = 100;
DECLARE @n2 AS INT = 40099;
WITH numbers AS (
SELECT ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) AS n FROM
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x1(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x2(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x3(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x4(x),
(VALUES(0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x5(x)
)
SELECT numbers.n
FROM numbers
WHERE n BETWEEN @n1 and @n2
ORDER BY n
答案 11 :(得分：3)
我知道我已经熬了4年了，但我偶然发现了另一个解决这个问题的方法。速度问题不仅仅是预过滤，而且还阻止了排序。可以强制连接顺序以笛卡尔积作为连接的结果实际计数的方式执行。使用slartidan的答案作为跳跃点：
WITH x AS (SELECT n FROM (VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) v(n))
SELECT ones.n + 10*tens.n + 100*hundreds.n + 1000*thousands.n
FROM x ones, x tens, x hundreds, x thousands
ORDER BY 1
如果我们知道我们想要的范围，我们可以通过@Upper和@Lower指定它。通过将连接提示REMOTE和TOP组合在一起，我们只能计算出我们想要的值的子集而没有浪费。
WITH x AS (SELECT n FROM (VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) v(n))
SELECT TOP (1+@Upper-@Lower) @Lower + ones.n + 10*tens.n + 100*hundreds.n + 1000*thousands.n
FROM x thousands
INNER REMOTE JOIN x hundreds on 1=1
INNER REMOTE JOIN x tens on 1=1
INNER REMOTE JOIN x ones on 1=1
连接提示REMOTE强制优化器首先在连接的右侧进行比较。通过将每个连接指定为从最高到最低有效值的REMOTE，连接本身将正确向上计数一个。无需使用WHERE进行过滤，也无需使用ORDER BY进行排序。
如果要增加范围，只要在FROM子句中从大多数到最不重要的顺序排序，就可以继续添加其他连接数量级逐渐增加的连接数。
请注意，这是特定于SQL Server 2008或更高版本的查询。
答案 12 :(得分：3)
以下是几个非常优化且兼容的解决方案：
USE master;
declare @min as int; set @min = 1000;
declare @max as int; set @max = 1050; --null returns all
-- Up to 256 - 2 048 rows depending on SQL Server version
select isnull(@min,0)+number.number as number
FROM dbo.spt_values AS number
WHERE number."type" = 'P' --integers
and ( @max is null --return all
or isnull(@min,0)+number.number <= @max --return up to max
)
order by number
;
-- Up to 65 536 - 4 194 303 rows depending on SQL Server version
select isnull(@min,0)+value1.number+(value2.number*numberCount.numbers) as number
FROM dbo.spt_values AS value1
cross join dbo.spt_values AS value2
cross join ( --get the number of numbers (depends on version)
select sum(1) as numbers
from dbo.spt_values
where spt_values."type" = 'P' --integers
) as numberCount
WHERE value1."type" = 'P' --integers
and value2."type" = 'P' --integers
and ( @max is null --return all
or isnull(@min,0)+value1.number+(value2.number*numberCount.numbers)
<= @max --return up to max
)
order by number
;
答案 13 :(得分：3)
declare @start int = 1000
declare @end int =1050
;with numcte
AS
(
SELECT @start [SEQUENCE]
UNION all
SELECT [SEQUENCE] + 1 FROM numcte WHERE [SEQUENCE] < @end
)
SELECT * FROM numcte
答案 14 :(得分：2)
这也可以
DECLARE @startNum INT = 1000;
DECLARE @endNum INT = 1050;
INSERT INTO dbo.Numbers
( Num
)
SELECT CASE WHEN MAX(Num) IS NULL THEN @startNum
ELSE MAX(Num) + 1
END AS Num
FROM dbo.Numbers
GO 51
答案 15 :(得分：2)
运行查询时的最佳速度
DECLARE @num INT = 1000
WHILE(@num<1050)
begin
INSERT INTO [dbo].[Codes]
( Code
)
VALUES (@num)
SET @num = @num + 1
end
答案 16 :(得分：2)
指数大小的递归CTE（即使默认值为100递归，也可以构建最多2 ^ 100个数字）：
// Create response et hydrate headers
$response = new Response();
$response->setContent(file_get_contents('page.html'));
$response->headers->set('Content-Type', 'text/html');
$response->headers->set('Content-disposition', 'filename=page.html');
// Return response
return $response;
答案 17 :(得分：2)
SQL 2017及更高版本的更新： 如果您想要的序列是<8k，那么它将起作用：
Declare @start_num int = 1000
, @end_num int = 1050
Select [number] = @start_num + ROW_NUMBER() over (order by (Select null))
from string_split(replicate(' ',@end_num-@start_num-1),' ')
答案 18 :(得分：1)
我必须使用类似的方法将图片文件路径插入数据库。以下查询工作正常：
DECLARE @num INT = 8270058
WHILE(@num<8270284)
begin
INSERT INTO [dbo].[Galleries]
(ImagePath)
VALUES
('~/Content/Galeria/P'+CONVERT(varchar(10), @num)+'.JPG')
SET @num = @num + 1
end
您的代码是：
DECLARE @num INT = 1000
WHILE(@num<1051)
begin
SELECT @num
SET @num = @num + 1
end
答案 19 :(得分：1)
阅读此线程后，我做了以下功能。简单快捷：
go
create function numbers(@begin int, @len int)
returns table as return
with d as (
select 1 v from (values(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0)) d(v)
)
select top (@len) @begin -1 + row_number() over(order by (select null)) v
from d d0
cross join d d1
cross join d d2
cross join d d3
cross join d d4
cross join d d5
cross join d d6
cross join d d7
go
select * from numbers(987654321,500000)
答案 20 :(得分：1)
;WITH u AS (
SELECT Unit FROM (VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) v(Unit)
),
d AS (
SELECT
(Thousands+Hundreds+Tens+Units) V
FROM
(SELECT Thousands = Unit * 1000 FROM u) Thousands
,(SELECT Hundreds = Unit * 100 FROM u) Hundreds
,(SELECT Tens = Unit * 10 FROM u) Tens
,(SELECT Units = Unit FROM u) Units
WHERE
(Thousands+Hundreds+Tens+Units) <= 10000
)
SELECT * FROM d ORDER BY v
答案 21 :(得分：0)
这是一种通用且相对较快的解决方案，它输出从1到@n的整数。它适用于@n的任何正整数（非常大的数将导致算术溢出），而无需添加或删除表连接。不需要使用系统表，也不需要更改最大递归。
declare @n int = 10000
;with d as (select * from (values (0),(1),(2),(3),(4),(5),(6),(7),(8),(9)) x (d)),
n as (
select d x from d where d > 0 and d <= @n
union all
select x * 10 + d from n, d where x * 10 + d <= @n
)
select x from n
您可以添加order by子句对数字进行排序。
答案 22 :(得分：0)
我已经开发并使用了很长一段时间的解决方案（在其他人的共享作品中进行骑乘）与至少一个发布的方案有点类似。它不引用任何表，并且返回最多1048576个值（2 ^ 20）的未排序范围，并且可以根据需要包含负数。您当然可以根据需要对结果进行排序。它运行非常快，尤其是在较小范围内。
Select value from dbo.intRange(-500, 1500) order by value -- returns 2001 values
create function dbo.intRange
(
@Starting as int,
@Ending as int
)
returns table
as
return (
select value
from (
select @Starting +
( bit00.v | bit01.v | bit02.v | bit03.v
| bit04.v | bit05.v | bit06.v | bit07.v
| bit08.v | bit09.v | bit10.v | bit11.v
| bit12.v | bit13.v | bit14.v | bit15.v
| bit16.v | bit17.v | bit18.v | bit19.v
) as value
from (select 0 as v union ALL select 0x00001 as v) as bit00
cross join (select 0 as v union ALL select 0x00002 as v) as bit01
cross join (select 0 as v union ALL select 0x00004 as v) as bit02
cross join (select 0 as v union ALL select 0x00008 as v) as bit03
cross join (select 0 as v union ALL select 0x00010 as v) as bit04
cross join (select 0 as v union ALL select 0x00020 as v) as bit05
cross join (select 0 as v union ALL select 0x00040 as v) as bit06
cross join (select 0 as v union ALL select 0x00080 as v) as bit07
cross join (select 0 as v union ALL select 0x00100 as v) as bit08
cross join (select 0 as v union ALL select 0x00200 as v) as bit09
cross join (select 0 as v union ALL select 0x00400 as v) as bit10
cross join (select 0 as v union ALL select 0x00800 as v) as bit11
cross join (select 0 as v union ALL select 0x01000 as v) as bit12
cross join (select 0 as v union ALL select 0x02000 as v) as bit13
cross join (select 0 as v union ALL select 0x04000 as v) as bit14
cross join (select 0 as v union ALL select 0x08000 as v) as bit15
cross join (select 0 as v union ALL select 0x10000 as v) as bit16
cross join (select 0 as v union ALL select 0x20000 as v) as bit17
cross join (select 0 as v union ALL select 0x40000 as v) as bit18
cross join (select 0 as v union ALL select 0x80000 as v) as bit19
) intList
where @Ending - @Starting < 0x100000
and intList.value between @Starting and @Ending
)
答案 23 :(得分：0)
我们的DEV服务器在36秒内完成了这项工作。就像Brian的回答一样，在查询中重点关注过滤到范围是很重要的;即使它不需要，BETWEEN仍试图在下限之前生成所有初始记录。
declare @s bigint = 10000000
, @e bigint = 20000000
;WITH
Z AS (SELECT 0 z FROM (VALUES (0),(1),(2),(3),(4),(5),(6),(7),(8),(9),(10),(11),(12),(13),(14),(15)) T(n)),
Y AS (SELECT 0 z FROM Z a, Z b, Z c, Z d, Z e, Z f, Z g, Z h, Z i, Z j, Z k, Z l, Z m, Z n, Z o, Z p),
N AS (SELECT ROW_NUMBER() OVER (PARTITION BY 0 ORDER BY z) n FROM Y)
SELECT TOP (1+@e-@s) @s + n - 1 FROM N
请注意 ROW_NUMBER 是 bigint ，因此我们无法通过任何方法查看2 ^^ 64（== 16 ^^ 16）生成的记录使用它。因此，此查询遵循生成值的相同上限。
答案 24 :(得分：0)
这就是我的工作，它非常快速，灵活，没有很多代码。
DECLARE @count int = 65536;
DECLARE @start int = 11;
DECLARE @xml xml = REPLICATE(CAST('<x/>' AS nvarchar(max)), @count);
; WITH GenerateNumbers(Num) AS
(
SELECT ROW_NUMBER() OVER (ORDER BY @count) + @start - 1
FROM @xml.nodes('/x') X(T)
)
SELECT Num
FROM GenerateNumbers;
请注意，（ORDER BY @count）是虚拟的。它什么都不做，但是ROW_NUMBER（）需要ORDER BY。
修改： 我意识到最初的问题是要获得从x到y的范围。我的脚本可以像这样修改以获取范围：
DECLARE @start int = 5;
DECLARE @end int = 21;
DECLARE @xml xml = REPLICATE(CAST('<x/>' AS nvarchar(max)), @end - @start + 1);
; WITH GenerateNumbers(Num) AS
(
SELECT ROW_NUMBER() OVER (ORDER BY @end) + @start - 1
FROM @xml.nodes('/x') X(T)
)
SELECT Num
FROM GenerateNumbers;
答案 25 :(得分：0)
Oracle 12c；快速但有限：
select rownum+1000 from all_objects fetch first 50 rows only;
注意：仅限于all_objects视图的行数；
答案 26 :(得分：0)
这将使用过程代码和表值函数。缓慢，但容易且可预测。
snapshot用法：
CREATE FUNCTION [dbo].[Sequence] (@start int, @end int)
RETURNS
@Result TABLE(ID int)
AS
begin
declare @i int;
set @i = @start;
while @i <= @end
begin
insert into @result values (@i);
set @i = @i+1;
end
return;
end
这是一个表，因此您可以在与其他数据的联接中使用它。我最经常使用此函数作为对GROUP BY小时，天等联接的左侧，以确保时间值的连续序列。
SELECT * FROM dbo.Sequence (3,7);
ID
3
4
5
6
7
性能令人鼓舞（百万行记录时间为16秒），但对于许多用途而言却足够好。
SELECT DateAdd(hh,ID,'2018-06-20 00:00:00') as HoursInTheDay FROM dbo.Sequence (0,23) ;
HoursInTheDay
2018-06-20 00:00:00.000
2018-06-20 01:00:00.000
2018-06-20 02:00:00.000
2018-06-20 03:00:00.000
2018-06-20 04:00:00.000
(...)
答案 27 :(得分：0)
这是我想出的：
create or alter function dbo.fn_range(@start int, @end int) returns table
return
with u2(n) as (
select n
from (VALUES (0),(1),(2),(3)) v(n)
),
u8(n) as (
select
x0.n | x1.n * 4 | x2.n * 16 | x3.n * 64 as n
from u2 x0, u2 x1, u2 x2, u2 x3
)
select
@start + s.n as n
from (
select
x0.n | isnull(x1.n, 0) * 256 | isnull(x2.n, 0) * 65536 as n
from u8 x0
left join u8 x1 on @end-@start > 256
left join u8 x2 on @end-@start > 65536
) s
where s.n < @end - @start
生成最多2 ^ 24个值。加入条件可以快速保持小值。
答案 28 :(得分：0)
只有某些应用程序表具有行时，这仅适用于序列。假设我想要序列来自1..100，并且应用程序表dbo.foo具有列（数字或字符串类型）foo.bar：
select
top 100
row_number() over (order by dbo.foo.bar) as seq
from dbo.foo
尽管它存在于order by子句中，但dbo.foo.bar不必具有不同的甚至非空值。
当然，SQL Server 2012具有序列对象，因此该产品是一个自然的解决方案。
答案 29 :(得分：0)
-- Generate Numeric Range
-- Source: http://www.sqlservercentral.com/scripts/Miscellaneous/30397/
CREATE TABLE #NumRange(
n int
)
DECLARE @MinNum int
DECLARE @MaxNum int
DECLARE @I int
SET NOCOUNT ON
SET @I = 0
WHILE @I <= 9 BEGIN
INSERT INTO #NumRange VALUES(@I)
SET @I = @I + 1
END
SET @MinNum = 1
SET @MaxNum = 1000000
SELECT num = a.n +
(b.n * 10) +
(c.n * 100) +
(d.n * 1000) +
(e.n * 10000)
FROM #NumRange a
CROSS JOIN #NumRange b
CROSS JOIN #NumRange c
CROSS JOIN #NumRange d
CROSS JOIN #NumRange e
WHERE a.n +
(b.n * 10) +
(c.n * 100) +
(d.n * 1000) +
(e.n * 10000) BETWEEN @MinNum AND @MaxNum
ORDER BY a.n +
(b.n * 10) +
(c.n * 100) +
(d.n * 1000) +
(e.n * 10000)
DROP TABLE #NumRange
答案 30 :(得分：-1)
DECLARE @a int=1000, @b int=1050
SELECT @a-1+ROW_NUMBER() OVER(ORDER BY y.z.value('(/n)[1]', 'int') ) rw
FROM (
SELECT CAST('<m>'+REPLICATE('<n>1</n>', @b-@a+1)+'</m>' AS XML ) x ) t
CROSS APPLY t.x.nodes('//m/n') y(z)
答案 31 :(得分：-1)
CREATE OR ALTER FUNCTION [dbo].[_ICAN_TF_Nums2](@a INT, @b INT)
-------------------------------------------------------------------------------------------------------------------
--INVENTIVE:Keyvan ARYAEE-MOEEN
-------------------------------------------------------------------------------------------------------------------
RETURNS @_ICAN_TF_Nums2 TABLE
(
num int
)
AS
BEGIN
------------------------------------------------------------------------------------------------------------------
WITH nums AS
(SELECT @a AS value
UNION ALL
SELECT value + 1 AS value
FROM nums
WHERE nums.value < @b)
INSERT @_ICAN_TF_Nums2
SELECT *
FROM nums
ORDER BY 1
option ( MaxRecursion 0 );
RETURN
END
-------------------------------------------------------------------------------------------------------------------
-- SELECT * FROM dbo._ICAN_TF_Nums2(1000, 1050)
-------------------------------------------------------------------------------------------------------------------