SQL 如何在两个数字之间生成一系列数字？
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How to generate a range of numbers between two numbers?
提问by user3211705
I have two numbers as input from the user, like for example 1000and 1050.
我有两个数字作为用户的输入，例如 1000和1050。
How do I generate the numbers between these two numbers, using a sql query, in seperate rows? I want this:
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？我要这个：
1000
1001
1002
1003
.
.
1050
回答by slartidan
Select non-persisted values with the VALUESkeyword. Then use JOINs to generate lots and lots of combinations (can be extended to create hundreds of thousands of rows and beyond).
使用VALUES关键字选择非持久值。然后使用JOINs 生成大量组合（可以扩展以创建数十万行及更多行）。
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
A shorter alternative, that is not as easy to understand:
一个较短的替代方案，这并不容易理解：
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
回答by Jayvee
an alternative solution is recursive CTE:
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)
回答by Tim Schmelter
SELECT DISTINCT n = number
FROM master..[spt_values]
WHERE number BETWEEN @start AND @end
Note that this table has a maximum of 2048 because then the numbers have gaps.
请注意，此表的最大值为 2048，因为数字之间存在差距。
Here's a slightly better approach using a system view(since from SQL-Server 2005):
这是使用系统视图（从 SQL-Server 2005 开始）的稍微好一点的方法：
;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;
or use a custom a number-table. Credits to Aaron Bertrand, i suggest to read the whole article: Generate a set or sequence without loops
或使用自定义数字表。感谢 Aaron Bertrand，我建议阅读整篇文章：Generate a set or sequence without loops
回答by Brian Pressler
I recently wrote this inline table valued function to solve this very problem. It's not limited in range other than memory and storage. It accesses no tables so there's no need for disk reads or writes generally. It adds joins values exponentially on each iteration so it's very fast even for very large ranges. It creates ten million records in five seconds on my server. It also works with negative values.
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，它不受范围限制。它不访问表，因此通常不需要磁盘读取或写入。它在每次迭代中以指数方式添加连接值，因此即使对于非常大的范围也非常快。它在我的服务器上五秒钟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);
It's handy for date and time ranges as well:
它对于日期和时间范围也很方便：
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'));
You could use a cross apply join on it to split records based on values in the table. So for example to create a record for every minute on a time range in a table you could do something like:
您可以在其上使用交叉应用连接来根据表中的值拆分记录。因此，例如要在表中的时间范围内为每分钟创建一条记录，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
回答by Habeeb
The best option I have used is as follows:
我使用过的最佳选择如下：
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
I have generated millions of records using this and it works perfect.
我已经使用它生成了数百万条记录，并且效果很好。
回答by Nguyen Son Tung
It work for me !
它对我有用！
select top 50 ROW_NUMBER() over(order by a.name) + 1000 as Rcount
from sys.all_objects a
回答by Sowbarani Karthikeyan
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
回答by cesargroening
The best way is using recursive ctes.
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.
致敬。
回答by Guillaume86
Nothing new but I rewrote Brian Pressler solution to be easier on the eye, it might be useful to someone (even if it's just future me):
没有什么新鲜事，但我重写了 Brian Pressler 的解决方案以使其更容易理解，它可能对某人有用（即使它只是未来的我）：
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
回答by AlexDev
If you don't have a problem installing a CLR assembly in your server a good option is writing a table valued function in .NET. That way you can use a simple syntax, making it easy to join with other queries and as a bonus won't waste memory because the result is streamed.
如果在服务器中安装 CLR 程序集没有问题，一个不错的选择是在 .NET 中编写表值函数。这样你就可以使用简单的语法，很容易加入其他查询，而且不会浪费内存，因为结果是流式的。
Create a project containing the following class:
创建一个包含以下类的项目：
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() { }
}
}
Put the assembly somewhere on the serverand run:
将程序集放在服务器上的某个位置并运行：
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];
Now you can run:
现在你可以运行：
select * from dbo.seq(1, 1000000)
