Sea*_*ean 18
以下可以编码为函数.您需要修剪前导零以满足您的问题要求.
declare @intvalue int
set @intvalue=5
declare @vsresult varchar(64)
declare @inti int
select @inti = 64, @vsresult = ''
while @inti>0
begin
select @vsresult=convert(char(1), @intvalue % 2)+@vsresult
select @intvalue = convert(int, (@intvalue / 2)), @inti=@inti-1
end
select @vsresult
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Mat*_*ank 17
实际上,使用普通的旧SQL非常简单.只需使用按位AND.我有点惊讶的是,没有一个简单的解决方案在线发布(没有调用UDF).在我的情况下,我真的想检查位是打开还是关闭(数据来自dotnet eNums).
因此,这里有一个例子,它将单独和一起给你 - 位值和二进制字符串(大联盟只是一种产生数字的hacky方式,可以在DBs上工作:
select t.Number
, cast(t.Number & 64 as bit) as bit7
, cast(t.Number & 32 as bit) as bit6
, cast(t.Number & 16 as bit) as bit5
, cast(t.Number & 8 as bit) as bit4
, cast(t.Number & 4 as bit) as bit3
, cast(t.Number & 2 as bit) as bit2
,cast(t.Number & 1 as bit) as bit1
, cast(cast(t.Number & 64 as bit) as CHAR(1))
+cast( cast(t.Number & 32 as bit) as CHAR(1))
+cast( cast(t.Number & 16 as bit) as CHAR(1))
+cast( cast(t.Number & 8 as bit) as CHAR(1))
+cast( cast(t.Number & 4 as bit) as CHAR(1))
+cast( cast(t.Number & 2 as bit) as CHAR(1))
+cast(cast(t.Number & 1 as bit) as CHAR(1)) as binary_string
--to explicitly answer the question, on MSSQL without using REGEXP (which would make it simple)
,SUBSTRING(cast(cast(t.Number & 64 as bit) as CHAR(1))
+cast( cast(t.Number & 32 as bit) as CHAR(1))
+cast( cast(t.Number & 16 as bit) as CHAR(1))
+cast( cast(t.Number & 8 as bit) as CHAR(1))
+cast( cast(t.Number & 4 as bit) as CHAR(1))
+cast( cast(t.Number & 2 as bit) as CHAR(1))
+cast(cast(t.Number & 1 as bit) as CHAR(1))
,
PATINDEX('%1%', cast(cast(t.Number & 64 as bit) as CHAR(1))
+cast( cast(t.Number & 32 as bit) as CHAR(1))
+cast( cast(t.Number & 16 as bit) as CHAR(1))
+cast( cast(t.Number & 8 as bit) as CHAR(1))
+cast( cast(t.Number & 4 as bit) as CHAR(1))
+cast( cast(t.Number & 2 as bit) as CHAR(1))
+cast(cast(t.Number & 1 as bit) as CHAR(1) )
)
,99)
from (select 1 as Number union all select 2 union all select 3 union all select 4 union all select 5 union all select 6
union all select 7 union all select 8 union all select 9 union all select 10) as t
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产生这个结果:
num bit7 bit6 bit5 bit4 bit3 bit2 bit1 binary_string binary_string_trimmed
1 0 0 0 0 0 0 1 0000001 1
2 0 0 0 0 0 1 0 0000010 10
3 0 0 0 0 0 1 1 0000011 11
4 0 0 0 1 0 0 0 0000100 100
5 0 0 0 0 1 0 1 0000101 101
6 0 0 0 0 1 1 0 0000110 110
7 0 0 0 0 1 1 1 0000111 111
8 0 0 0 1 0 0 0 0001000 1000
9 0 0 0 1 0 0 1 0001001 1001
10 0 0 0 1 0 1 0 0001010 1010
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这是一个通用的基本转换器
http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html
你可以做
select reverse(dbo.ConvertToBase(5, 2)) -- 101
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