SQL Server将整数转换为二进制字符串
我想知道在SQL中是否有一种简单的方法可以将整数转换为二进制表示形式，然后将其存储为varchar。
例如，5将转换为“101”并存储为varchar。
16 个答案:
答案 0 :(得分：17)
可以将以下内容编码为函数。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
答案 1 :(得分：16)
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，可以单独和一起使用 - 位值和二进制字符串（大联合只是一种产生数字的hacky方式，可以在DB中运行：
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
答案 2 :(得分：6)
这是一个通用的基本转换器
http://dpatrickcaldwell.blogspot.com/2009/05/converting-decimal-to-hexadecimal-with.html
你可以做到
select reverse(dbo.ConvertToBase(5, 2)) -- 101
答案 3 :(得分：3)
请参阅此博客文章Converting Integers to Binary Strings，我发布了一段时间。
答案 4 :(得分：2)
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
答案 5 :(得分：1)
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);
答案 6 :(得分：1)
这是accepted answer from Sean的一些变化，因为我发现它限制为仅允许输出中的硬编码数字。在我的日常使用中，我发现只获得最高1位数或者指定我期待的数字更有用。它将自动用0s填充边，使其最多排成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;
答案 7 :(得分：0)
我相信这种方法简化了其他人提出的许多其他想法。它使用逐位算术和带有CTE的SortedDict()技巧来生成二进制数字。
FOR XML答案 8 :(得分：0)
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中测试）。
答案 9 :(得分：0)
在SQL Server上，您可以尝试类似下面的示例：
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
答案 10 :(得分：0)
我知道我在这里比赛有点晚了，但是最近我想出了一个巧妙的解决方案，可以利用理货表（类似于上面的@hkravitz解决方案。）关键区别在于我可以利用我所谓的杠杆虚拟索引，以对结果进行降序排序， 在执行计划中没有排序运算符 。我使用本文末尾包含的dbo.rangeAB完成此操作。
请注意，这将以升序返回数字0到30（对于RowNumber为“ RN”）：
SELECT r.RN
FROM dbo.rangeAB(0,30,1,0) AS r
ORDER BY r.RN;
它这样做没有排序。 RN可以定义为ROW_NUMBER（）OVER（ORDER BY（选择NULL））。同样，按RN排序不需要排序-这就是虚拟索引。
当我尝试降序排序时，我在执行计划中 做 得到排序。
输入有限对立。 RangeAB包括名为 Op 的列-OP RN的有限 Op 正数编号。所谓“有限的对立面”，是指0代表30的对立面，1代表29的对立面，依此类推。与传统的对立数不同（-1代表1的对立面）。有限的对立以降序返回。
SELECT r.RN, r.OP
FROM dbo.rangeAB(0,30,1,0) AS r
ORDER BY r.RN;
返回：
RN OP
----- -------
0 30
1 29
2 28
3 27
....
27 3
28 2
29 1
30 0
我可以使用Op，我可以利用RN的有限对数以降序获取数字，同时仍然利用虚拟索引来避免排序。这两个查询返回相同的内容，但是在比较执行计划时，根据SSMS，删除排序将查询成本降低了50倍。
功能
CREATE FUNCTION dbo.NumberToBinary(@input INT)
RETURNS TABLE WITH SCHEMABINDING AS RETURN
/* Created By Alan Burstein 20191112, Requires RangeAB (code below) */
SELECT BIN = (
SELECT @input/f.Np2%2
FROM dbo.rangeAB(0,30,1,0) AS r
CROSS APPLY (VALUES(POWER(2,r.Op))) AS f(NP2)
WHERE (@input = 0 AND f.Np2 = 1) OR @input >= f.Np2
ORDER BY ROW_NUMBER() OVER (ORDER BY (SELECT NULL))
FOR XML PATH(''));
RangeAB
CREATE FUNCTION dbo.rangeAB
(
@low bigint,
@high bigint,
@gap bigint,
@row1 bit
)
/****************************************************************************************
[Purpose]:
Creates up to 531,441,000,000 sequentia1 integers numbers beginning with @low and ending
with @high. Used to replace iterative methods such as loops, cursors and recursive CTEs
to solve SQL problems. Based on Itzik Ben-Gan's getnums function with some tweeks and
enhancements and added functionality. The logic for getting rn to begin at 0 or 1 is
based comes from Jeff Moden's fnTally function.
The name range because it's similar to clojure's range function. The name "rangeAB" as
used because "range" is a reserved SQL keyword.
[Author]: Alan Burstein
[Compatibility]:
SQL Server 2008+ and Azure SQL Database
[Syntax]:
SELECT r.RN, r.OP, r.N1, r.N2
FROM dbo.rangeAB(@low,@high,@gap,@row1) AS r;
[Parameters]:
@low = a bigint that represents the lowest value for n1.
@high = a bigint that represents the highest value for n1.
@gap = a bigint that represents how much n1 and n2 will increase each row; @gap also
represents the difference between n1 and n2.
@row1 = a bit that represents the first value of rn. When @row = 0 then rn begins
at 0, when @row = 1 then rn will begin at 1.
[Returns]:
Inline Table Valued Function returns:
rn = bigint; a row number that works just like T-SQL ROW_NUMBER() except that it can
start at 0 or 1 which is dictated by @row1.
op = bigint; returns the "opposite number that relates to rn. When rn begins with 0 and
ends with 10 then 10 is the opposite of 0, 9 the opposite of 1, etc. When rn begins
with 1 and ends with 5 then 1 is the opposite of 5, 2 the opposite of 4, etc...
n1 = bigint; a sequential number starting at the value of @low and incrimentingby the
value of @gap until it is less than or equal to the value of @high.
n2 = bigint; a sequential number starting at the value of @low+@gap and incrimenting
by the value of @gap.
[Dependencies]:
N/A
[Developer Notes]:
1. The lowest and highest possible numbers returned are whatever is allowable by a
bigint. The function, however, returns no more than 531,441,000,000 rows (8100^3).
2. @gap does not affect rn, rn will begin at @row1 and increase by 1 until the last row
unless its used in a query where a filter is applied to rn.
3. @gap must be greater than 0 or the function will not return any rows.
4. Keep in mind that when @row1 is 0 then the highest row-number will be the number of
rows returned minus 1
5. If you only need is a sequential set beginning at 0 or 1 then, for best performance
use the RN column. Use N1 and/or N2 when you need to begin your sequence at any
number other than 0 or 1 or if you need a gap between your sequence of numbers.
6. Although @gap is a bigint it must be a positive integer or the function will
not return any rows.
7. The function will not return any rows when one of the following conditions are true:
* any of the input parameters are NULL
* @high is less than @low
* @gap is not greater than 0
To force the function to return all NULLs instead of not returning anything you can
add the following code to the end of the query:
UNION ALL
SELECT NULL, NULL, NULL, NULL
WHERE NOT (@high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0)
This code was excluded as it adds a ~5% performance penalty.
8. There is no performance penalty for sorting by rn ASC; there is a large performance
penalty for sorting in descending order WHEN @row1 = 1; WHEN @row1 = 0
If you need a descending sort the use op in place of rn then sort by rn ASC.
Best Practices:
--===== 1. Using RN (rownumber)
-- (1.1) The best way to get the numbers 1,2,3...@high (e.g. 1 to 5):
SELECT RN FROM dbo.rangeAB(1,5,1,1);
-- (1.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 0 to 5):
SELECT RN FROM dbo.rangeAB(0,5,1,0);
--===== 2. Using OP for descending sorts without a performance penalty
-- (2.1) The best way to get the numbers 5,4,3...@high (e.g. 5 to 1):
SELECT op FROM dbo.rangeAB(1,5,1,1) ORDER BY rn ASC;
-- (2.2) The best way to get the numbers 0,1,2...@high-1 (e.g. 5 to 0):
SELECT op FROM dbo.rangeAB(1,6,1,0) ORDER BY rn ASC;
--===== 3. Using N1
-- (3.1) To begin with numbers other than 0 or 1 use N1 (e.g. -3 to 3):
SELECT N1 FROM dbo.rangeAB(-3,3,1,1);
-- (3.2) ROW_NUMBER() is built in. If you want a ROW_NUMBER() include RN:
SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,1);
-- (3.3) If you wanted a ROW_NUMBER() that started at 0 you would do this:
SELECT RN, N1 FROM dbo.rangeAB(-3,3,1,0);
--===== 4. Using N2 and @gap
-- (4.1) To get 0,10,20,30...100, set @low to 0, @high to 100 and @gap to 10:
SELECT N1 FROM dbo.rangeAB(0,100,10,1);
-- (4.2) Note that N2=N1+@gap; this allows you to create a sequence of ranges.
-- For example, to get (0,10),(10,20),(20,30).... (90,100):
SELECT N1, N2 FROM dbo.rangeAB(0,90,10,1);
-- (4.3) Remember that a rownumber is included and it can begin at 0 or 1:
SELECT RN, N1, N2 FROM dbo.rangeAB(0,90,10,1);
[Examples]:
--===== 1. Generating Sample data (using rangeAB to create "dummy rows")
-- The query below will generate 10,000 ids and random numbers between 50,000 and 500,000
SELECT
someId = r.rn,
someNumer = ABS(CHECKSUM(NEWID())%450000)+50001
FROM rangeAB(1,10000,1,1) r;
--===== 2. Create a series of dates; rn is 0 to include the first date in the series
DECLARE @startdate DATE = '20180101', @enddate DATE = '20180131';
SELECT r.rn, calDate = DATEADD(dd, r.rn, @startdate)
FROM dbo.rangeAB(1, DATEDIFF(dd,@startdate,@enddate),1,0) r;
GO
--===== 3. Splitting (tokenizing) a string with fixed sized items
-- given a delimited string of identifiers that are always 7 characters long
DECLARE @string VARCHAR(1000) = 'A601225,B435223,G008081,R678567';
SELECT
itemNumber = r.rn, -- item's ordinal position
itemIndex = r.n1, -- item's position in the string (it's CHARINDEX value)
item = SUBSTRING(@string, r.n1, 7) -- item (token)
FROM dbo.rangeAB(1, LEN(@string), 8,1) r;
GO
--===== 4. Splitting (tokenizing) a string with random delimiters
DECLARE @string VARCHAR(1000) = 'ABC123,999F,XX,9994443335';
SELECT
itemNumber = ROW_NUMBER() OVER (ORDER BY r.rn), -- item's ordinal position
itemIndex = r.n1+1, -- item's position in the string (it's CHARINDEX value)
item = SUBSTRING
(
@string,
r.n1+1,
ISNULL(NULLIF(CHARINDEX(',',@string,r.n1+1),0)-r.n1-1, 8000)
) -- item (token)
FROM dbo.rangeAB(0,DATALENGTH(@string),1,1) r
WHERE SUBSTRING(@string,r.n1,1) = ',' OR r.n1 = 0;
-- logic borrowed from: http://www.sqlservercentral.com/articles/Tally+Table/72993/
--===== 5. Grouping by a weekly intervals
-- 5.1. how to create a series of start/end dates between @startDate & @endDate
DECLARE @startDate DATE = '1/1/2015', @endDate DATE = '2/1/2015';
SELECT
WeekNbr = r.RN,
WeekStart = DATEADD(DAY,r.N1,@StartDate),
WeekEnd = DATEADD(DAY,r.N2-1,@StartDate)
FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r;
GO
-- 5.2. LEFT JOIN to the weekly interval table
BEGIN
DECLARE @startDate datetime = '1/1/2015', @endDate datetime = '2/1/2015';
-- sample data
DECLARE @loans TABLE (loID INT, lockDate DATE);
INSERT @loans SELECT r.rn, DATEADD(dd, ABS(CHECKSUM(NEWID())%32), @startDate)
FROM dbo.rangeAB(1,50,1,1) r;
-- solution
SELECT
WeekNbr = r.RN,
WeekStart = dt.WeekStart,
WeekEnd = dt.WeekEnd,
total = COUNT(l.lockDate)
FROM dbo.rangeAB(0,datediff(DAY,@StartDate,@EndDate),7,1) r
CROSS APPLY (VALUES (
CAST(DATEADD(DAY,r.N1,@StartDate) AS DATE),
CAST(DATEADD(DAY,r.N2-1,@StartDate) AS DATE))) dt(WeekStart,WeekEnd)
LEFT JOIN @loans l ON l.lockDate BETWEEN dt.WeekStart AND dt.WeekEnd
GROUP BY r.RN, dt.WeekStart, dt.WeekEnd ;
END;
--===== 6. Identify the first vowel and last vowel in a along with their positions
DECLARE @string VARCHAR(200) = 'This string has vowels';
SELECT TOP(1) position = r.rn, letter = SUBSTRING(@string,r.rn,1)
FROM dbo.rangeAB(1,LEN(@string),1,1) r
WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
ORDER BY r.rn;
-- To avoid a sort in the execution plan we'll use op instead of rn
SELECT TOP(1) position = r.op, letter = SUBSTRING(@string,r.op,1)
FROM dbo.rangeAB(1,LEN(@string),1,1) r
WHERE SUBSTRING(@string,r.rn,1) LIKE '%[aeiou]%'
ORDER BY r.rn;
---------------------------------------------------------------------------------------
[Revision History]:
Rev 00 - 20140518 - Initial Development - Alan Burstein
Rev 01 - 20151029 - Added 65 rows to make L1=465; 465^3=100.5M. Updated comment section
- Alan Burstein
Rev 02 - 20180613 - Complete re-design including opposite number column (op)
Rev 03 - 20180920 - Added additional CROSS JOIN to L2 for 530B rows max - Alan Burstein
****************************************************************************************/
RETURNS TABLE WITH SCHEMABINDING AS RETURN
WITH L1(N) AS
(
SELECT 1
FROM (VALUES
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),(0),
(0),(0)) T(N) -- 90 values
),
L2(N) AS (SELECT 1 FROM L1 a CROSS JOIN L1 b CROSS JOIN L1 c),
iTally AS (SELECT rn = ROW_NUMBER() OVER (ORDER BY (SELECT 1)) FROM L2 a CROSS JOIN L2 b)
SELECT
r.RN,
r.OP,
r.N1,
r.N2
FROM
(
SELECT
RN = 0,
OP = (@high-@low)/@gap,
N1 = @low,
N2 = @gap+@low
WHERE @row1 = 0
UNION ALL -- ISNULL required in the TOP statement below for error handling purposes
SELECT TOP (ABS((ISNULL(@high,0)-ISNULL(@low,0))/ISNULL(@gap,0)+ISNULL(@row1,1)))
RN = i.rn,
OP = (@high-@low)/@gap+(2*@row1)-i.rn,
N1 = (i.rn-@row1)*@gap+@low,
N2 = (i.rn-(@row1-1))*@gap+@low
FROM iTally AS i
ORDER BY i.rn
) AS r
WHERE @high&@low&@gap&@row1 IS NOT NULL AND @high >= @low AND @gap > 0;
GO
答案 11 :(得分：0)
您可以使用递归CTE表来执行此操作。在此示例代码中，将其设置为16位，但是您可以通过更改16->选择任意长度。 另外，您要转换的数据是表DecimalTable
WITH DecimalTable AS (SELECT 10 decimal_num UNION SELECT 20),
DtoB AS (SELECT decimal_num
,1 n
,CAST(CAST(decimal_num%2 AS bit) AS VARCHAR(16)) binary_num
FROM DecimalTable
UNION ALL
SELECT decimal_num
,n*2 n
,CAST(CONCAT(CAST(decimal_num&n as bit), binary_num)
AS VARCHAR(16)) binary_num
FROM DtoB
WHERE n<POWER(2,16))
SELECT decimal_num, binary_num
FROM DtoB
答案 12 :(得分：0)
此函数是一个通用转换器，允许将整数转换为任何基本编号系统的字符串描述，如二进制、八进制、十六进制等。
-- specify a string and numbering system Base value, for example 16 for hexadecimal
CREATE FUNCTION udf_IntToBaseXStr(@baseVal BIGINT,
@baseX BIGINT)
returns VARCHAR(63)
AS
BEGIN
--bigint : -2^63 (-9,223,372,036,854,775,808) to 2^63-1 (9,223,372,036,854,775,807)
-- or 63 ones (1111111,11111111,11111111,11111111,11111111,11111111,11111111,11111111) in binary
DECLARE @val BIGINT -- value from all
DECLARE @cv BIGINT -- value from a single char
DECLARE @baseStr VARCHAR(63)
SET @baseStr = '';
-- assumes a numbering method of 0123456789ABCDEF.....
SET @val = @baseVal
WHILE ( @val > 0 )
BEGIN
SET @cv = @val % @basex -- calculate the right most char's value
SET @baseStr = -- add it to (any existing) string
CASE
WHEN @cv < 10 THEN Char(Ascii('0') + @cv)
ELSE Char(Ascii('A') + ( @cv - 10 ))
END
+ @baseStr
SET @val = ( @val - @cv ) / @basex
END
RETURN @baseStr
END
GO
如果你需要保证一个最小长度，下一个函数会包装上面的函数，在前面加上一些零，强制返回的字符串达到你想要的最小长度。它不会截断到指定的长度。
-- specify a string and numbering system Base value, for example, 16 for hexadecimal
-- prepends LEADING ZEROS to force length of returned string to be AT LEAST minLength chars
CREATE FUNCTION udf_IntToBaseXStr_MinLength(@baseVal BIGINT,
@baseX BIGINT,
@minLength INT)
returns VARCHAR(63)
AS
BEGIN
DECLARE @baseStr VARCHAR(63)
SET @baseStr = dbo.udf_IntToBaseXStr(@baseVal, @baseX)
IF Len(@baseStr) < @minLength
SET @baseStr = Replicate('0', @minLength - Len(@baseStr))
+ @baseStr
RETURN @baseStr
END
GO
udf_IntToBaseXStr 用法：
;with CTE as
(
SELECT BaseX = 2, AKA = 'binary'
UNION SELECT 8, 'octal'
UNION SELECT 10, 'decimal'
UNION SELECT 15, 'pentadecimal'
UNION SELECT 16, 'hexadecimal'
)
SELECT BaseX, AKA, Result = dbo.udf_IntToBaseXStr(328239523, BaseX) FROM CTE
udf_IntToBaseXStr 结果：
| BaseX | 又名 | 结果 |
|---|---|---|
| 2 | 二进制 | 10011100100001000100110100011 |
| 8 | 八进制 | 2344104643 |
| 10 | 十进制 | 328239523 |
| 15 | 十进制 | 1DC3B24D |
| 16 | 十六进制 | 139089A3 |
udf_IntToBaseXStr_MinLength 用法：
;with CTE as
(
SELECT BaseX = 2, AKA = 'binary'
UNION SELECT 8, 'octal'
UNION SELECT 10, 'decimal'
UNION SELECT 15, 'pentadecimal'
UNION SELECT 16, 'hexadecimal'
)
SELECT BaseX, AKA, Result = dbo.udf_IntToBaseXStr_MinLength(328239523, BaseX, 24) FROM CTE
udf_IntToBaseXStr_MinLength 结果：
| BaseX | 又名 | 结果 |
|---|---|---|
| 2 | 二进制 | 10011100100001000100110100011 |
| 8 | 八进制 | 000000000000002344104643 |
| 10 | 十进制 | 000000000000000328239523 |
| 15 | 十进制 | 00000000000000001DC3B24D |
| 16 | 十六进制 | 0000000000000000139089A3 |
答案 13 :(得分：0)
想要轻松吗？做一些按位数学运算来映射出每个二进制数字。
CREATE FUNCTION dbo.BinaryRep (@val INT)
RETURNS VARCHAR(32)
WITH EXECUTE AS CALLER
AS
BEGIN
DECLARE @ret VARCHAR(32)
DECLARE @cnt INT = 30; -- 30 to 0 inclusive in loop
-- handle negative (we're using signed magnitude because that's simple)
SET @ret = IIF(@val < 0, '1', '0');
SET @val = ABS(@val); -- totally cheating here.
-- bitwise masking madness, one digit at a time.
WHILE @cnt > -1
BEGIN
SET @ret = CONCAT(@ret, IIF(@val & POWER(2, @cnt) = 0, 0, 1));
SET @cnt = @cnt - 1;
END;
RETURN @ret;
END
唯一的转折点就是what Constantin notes：你觉得你的底片怎么样？
这个版本便宜并使用 signed magnitude ，其中您只有第一个位作为 1 用于否定，没有其他更改。 -123 和 123 的区别仅在于它们的高位。
select dbo.BinaryRep(123) as plus, dbo.BinaryRep(-123) as minus
plus minus
-------------------------------- --------------------------------
00000000000000000000000001111011 10000000000000000000000001111011
注意 SQL Server INT 支持 2-31 到 231 所以我们需要循环 31 次（30 到 0，包括），而不是32.
答案 14 :(得分：0)
为什么不简单地...
#define FILL4 -1, -1, -1, -1
#define FILL16 FILL4, FILL4, FILL4, FILL4
#define FILL64 FILL16, FILL16, FILL16, FILL16
#define FILL256 FILL64, FILL64, FILL64, FILL64
static const int my_array[1024] = {FILL256, FILL256, FILL256, FILL256};
答案 15 :(得分：-1)
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;
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