SQL中的数据点整数按位值 [英] Pivot Integer Bitwise Values in SQL

问题描述

使用:SQL Server 2008 R2

Using: SQL Server 2008 R2

我的目标是取一个给定的整数(n)，并在2列中选择n位整数的每个置换，如下所示.

My goal is to take a given integer (n) and select every permutation of integers with n bits in 2 columns, as below.

Given: n=4

Output:

ColA  ColB
0     0
0     0
0     0
0     0
1     0
1     0
1     0
1     1
2     0
2     0
2     1
2     0
3     0
3     0
3     1
3     1
...
14    1
14    1
14    1
14    0
15    1
15    1
15    1
15    1

输出不会不是，重复不会，需要进行排序，这可能会或可能不会使此操作更容易.但是，我确实需要能够将n设置为任何整数...但是为了运行时，我们假设它将在10以下.

The output does not, repeat does NOT, need to be ordered, which may or may not make this easier. I do, however, need to be able to set n to any integer... but for the sake of runtime, let's say it'll be under 10.

提前谢谢.

推荐答案

您可能想尝试一下:

declare @bits int
set @bits = 4

;with AllTheNumbers as (
    select cast (POWER(2, @Bits) as int) - 1 Number
    union all
    select Number - 1
    from AllTheNumbers
    where Number > 0
),
Bits as (
    select @Bits - 1 Bit
    union all
    select  Bit - 1
    from Bits
    where Bit > 0
)
select *, case when (Number & cast (POWER(2, Bit) as int)) != 0 then 1 else 0 end
from AllTheNumbers cross join Bits
order by Number, Bit desc

AllTheNumbers产生从0到2 ^ n-1的数字，Bits产生从0到@Bits-1的位数，并且主要部分连接它们并计算每个位置的位数.

AllTheNumbers produces the numbers from 0 to 2^n-1, Bits produces bit numbers from 0 to @Bits - 1 and main part connects them and calculates bit value per position.

这篇关于SQL中的数据点整数按位值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！