SQL Server将整数转换为二进制string
我想知道是否有一个简单的方法在SQL中将整数转换为其二进制表示，然后将其存储为一个varchar。
例如5将被转换为“101”并存储为一个varchar。
以下可以被编码成一个函数。 您将需要削减前导零来满足您的问题的要求。
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 其实这是真正简单的使用普通的旧SQL。 只要使用按位与。 我有点惊讶，没有一个简单的解决scheme在网上发布（没有调用UDF）。 在我的情况下，我真的想检查位是否打开或closures（数据来自dotnet eNums）。
因此，这里是一个例子，它会给你单独的位值和二进制string（大的联合只是一个生成数字的方法，
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
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
这是一个通用的基地转换器
http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html
你可以做
select reverse(dbo.ConvertToBase(5, 2)) -- 101
declare @i int /* input */ set @i = 42 declare @result varchar(32) /* SQL Server int is 32 bits wide */ set @result = '' while 1 = 1 begin select @result = convert(char(1), @i % 2) + @result, @i = convert(int, @i / 2) if @i = 0 break end select @result
请看这个博客文章， 整数转换为二进制string ，我张贴了一段时间。
declare @intVal Int set @intVal = power(2,12)+ power(2,5) + power(2,1); With ComputeBin (IntVal, BinVal,FinalBin) As ( Select @IntVal IntVal, @intVal %2 BinVal , convert(nvarchar(max),(@intVal %2 )) FinalBin Union all Select IntVal /2, (IntVal /2) %2, convert(nvarchar(max),(IntVal /2) %2) + FinalBin FinalBin From ComputeBin Where IntVal /2 > 0 ) select FinalBin from ComputeBin where intval = ( select min(intval) from ComputeBin);
这是对Sean公认的答案的一点改变，因为我发现它只限于在输出中允许硬编码的数字。 在我的日常使用中，我发现它只能得到最高的1位数，或者指定我期待的数字。 它会自动填充0的一侧，所以它排成8,16或任何你想要的位数。
Create function f_DecimalToBinaryString ( @Dec int, @MaxLength int = null ) Returns varchar(max) as Begin Declare @BinStr varchar(max) = ''; -- Perform the translation from Dec to Bin While @Dec > 0 Begin Set @BinStr = Convert(char(1), @Dec % 2) + @BinStr; Set @Dec = Convert(int, @Dec /2); End; -- Either pad or trim the output to match the number of digits specified. If (@MaxLength is not null) Begin If @MaxLength <= Len(@BinStr) Begin -- Trim down Set @BinStr = SubString(@BinStr, Len(@BinStr) - (@MaxLength - 1), @MaxLength); End Else Begin -- Pad up Set @BinStr = Replicate('0', @MaxLength - Len(@BinStr)) + @BinStr; End; End; Return @BinStr; End;
这个怎么样…
SELECT number_value ,MOD(number_value / 32768, 2) AS BIT15 ,MOD(number_value / 16384, 2) AS BIT14 ,MOD(number_value / 8192, 2) AS BIT13 ,MOD(number_value / 4096, 2) AS BIT12 ,MOD(number_value / 2048, 2) AS BIT11 ,MOD(number_value / 1024, 2) AS BIT10 ,MOD(number_value / 512, 2) AS BIT9 ,MOD(number_value / 256, 2) AS BIT8 ,MOD(number_value / 128, 2) AS BIT7 ,MOD(number_value / 64, 2) AS BIT6 ,MOD(number_value / 32, 2) AS BIT5 ,MOD(number_value / 16, 2) AS BIT4 ,MOD(number_value / 8, 2) AS BIT3 ,MOD(number_value / 4, 2) AS BIT2 ,MOD(number_value / 2, 2) AS BIT1 ,MOD(number_value , 2) AS BIT0 FROM your_table;
我相信这种方法简化了其他人提出的其他想法。 它使用按位算术和CTE的FOR XML技巧来生成二进制数字。
DECLARE @my_int INT = 5 ;WITH CTE_Binary AS ( SELECT 1 AS seq, 1 AS val UNION ALL SELECT seq + 1 AS seq, power(2, seq) FROM CTE_Binary WHERE seq < 8 ) SELECT ( SELECT CAST(CASE WHEN B2.seq IS NOT NULL THEN 1 ELSE 0 END AS CHAR(1)) FROM CTE_Binary B1 LEFT OUTER JOIN CTE_Binary B2 ON B2.seq = B1.seq AND @my_int & B2.val = B2.val ORDER BY B1.seq DESC FOR XML PATH('') ) AS val
我使用了以下ITVF函数将十进制转换为二进制，因为它是内联函数，不需要担心优化器执行的多次读取操作。
CREATE FUNCTION dbo.udf_DecimalToBinary ( @Decimal VARCHAR(32) ) RETURNS TABLE AS RETURN WITH Tally (n) AS ( --32 Rows SELECT TOP 30 ROW_NUMBER() OVER (ORDER BY (SELECT NULL)) -1 FROM (VALUES (0),(0),(0),(0)) a(n) CROSS JOIN (VALUES(0),(0),(0),(0),(0),(0),(0),(0)) b(n) ) , Anchor (n, divisor , Result) as ( SELECT tN , CONVERT(BIGINT, @Decimal) / POWER(2,TN) , CONVERT(BIGINT, @Decimal) / POWER(2,TN) % 2 FROM Tally t WHERE CONVERT(bigint,@Decimal) >= POWER(2,tn) ) SELECT TwoBaseBinary = '' + (SELECT Result FROM Anchor ORDER BY N DESC FOR XML PATH ('') , TYPE).value('.','varchar(200)') /*How to use*/ SELECT TwoBaseBinary FROM dbo.udf_DecimalToBinary ('1234') /*result -> 10011010010*/
with t as (select * from (values (0),(1)) as t(c)), t0 as (table t), t1 as (table t), t2 as (table t), t3 as (table t), t4 as (table t), t5 as (table t), t6 as (table t), t7 as (table t), t8 as (table t), t9 as (table t), ta as (table t), tb as (table t), tc as (table t), td as (table t), te as (table t), tf as (table t) select '' || t0.c || t1.c || t2.c || t3.c || t4.c || t5.c || t6.c || t7.c || t8.c || t9.c || ta.c || tb.c || tc.c || td.c || te.c || tf.c as n from t0,t1,t2,t3,t4,t5,t6,t7,t8,t9,ta,tb,tc,td,te,tf order by n limit 1 offset 5
标准SQL（在PostgreSQL中testing）。
在SQL Server上，你可以尝试类似下面的例子：
DECLARE @Int int = 321 SELECT @Int ,CONCAT (CAST(@Int & power(2,15) AS bit) ,CAST(@Int & power(2,14) AS bit) ,CAST(@Int & power(2,13) AS bit) ,CAST(@Int & power(2,12) AS bit) ,CAST(@Int & power(2,11) AS bit) ,CAST(@Int & power(2,10) AS bit) ,CAST(@Int & power(2,9) AS bit) ,CAST(@Int & power(2,8) AS bit) ,CAST(@Int & power(2,7) AS bit) ,CAST(@Int & power(2,6) AS bit) ,CAST(@Int & power(2,5) AS bit) ,CAST(@Int & power(2,4) AS bit) ,CAST(@Int & power(2,3) AS bit) ,CAST(@Int & power(2,2) AS bit) ,CAST(@Int & power(2,1) AS bit) ,CAST(@Int & power(2,0) AS bit) ) AS BitString ,CAST(@Int & power(2,15) AS bit) AS BIT15 ,CAST(@Int & power(2,14) AS bit) AS BIT14 ,CAST(@Int & power(2,13) AS bit) AS BIT13 ,CAST(@Int & power(2,12) AS bit) AS BIT12 ,CAST(@Int & power(2,11) AS bit) AS BIT11 ,CAST(@Int & power(2,10) AS bit) AS BIT10 ,CAST(@Int & power(2,9) AS bit) AS BIT9 ,CAST(@Int & power(2,8) AS bit) AS BIT8 ,CAST(@Int & power(2,7) AS bit) AS BIT7 ,CAST(@Int & power(2,6) AS bit) AS BIT6 ,CAST(@Int & power(2,5) AS bit) AS BIT5 ,CAST(@Int & power(2,4) AS bit) AS BIT4 ,CAST(@Int & power(2,3) AS bit) AS BIT3 ,CAST(@Int & power(2,2) AS bit) AS BIT2 ,CAST(@Int & power(2,1) AS bit) AS BIT1 ,CAST(@Int & power(2,0) AS bit) AS BIT0
