2

我有一个带有一个二进制列的表。包含 0x00010100000101010101......00 列数据长度是 35040，我准备这个数据就像

@Jobbyte = COALESCE(@Jobbyte , 0x) + Cast ((Case When Sum(A.bit) >= 1 then 1 else 0 end)as binary(1))

我必须将此二进制数据与另一个二进制数据进行比较并获得匹配的二进制计数。二进制数据都具有相同的数据长度

请看下图，它们是 2 个二进制数据 Binary1 和 Binary2 我想比较两个二进制数据并获得匹配二进制数据的总和。只有一个条件是 binary2 的单个位是 0x01

最后一行表示 0 = 不匹配，1 = 匹配，最后一列是最后一行的总和 (4) 请建议我如何比较，如果你可以发布查询会更好

更新 1

我正在使用 2 个函数并尝试解决，但是执行 10000 条记录需要很长时间，1 条记录的执行时间为 50 毫秒

Create FUNCTION [dbo].[Fn_BinaryToTable]
(   
    @BinaryData VARBINARY(max)
)
RETURNS TABLE 
AS
RETURN 
(
    Select  ((N.Number / 96) - (Case (N.Number % 96) when 0 Then 1 else 0 end))+1 As [FNNoDay],
            (Case (N.Number % 96) when 0 then 96 else (N.Number % 96) end) * 15  AS [FnMinutes],
            SUBSTRING(@BinaryData,(N.Number),1) AS [FNBIT]
    from    Numbers N 
    Where N.Number between 1 and (DATALENGTH(@BinaryData))
)
---------------------------------------------------------
Create FUNCTION [dbo].[fn_GetPercentage]
(
    @JobValue int,  
    @CandidateBinary VARBINARY(max),
    @JobBinary VARBINARY(max)
)
RETURNS Decimal
AS
BEGIN
    DECLARE @RValue Decimal;

    SELECT  @RValue = SUM(cast(JB.FNBIT as int))            
    FROM    dbo.Fn_BinaryToTable(@CandidateBinary)   CB,
            dbo.Fn_BinaryToTable(@JobBinary) JB
    WHERE   CB.FNNoDay = JB.FNNoDay
      AND   CB.FnMinutes = JB.FNMinutes
      AND   JB.FNBIT = CB.FNBIT
      AND   JB.FNBIT = 0x01

    Return ((@RValue * 100)/ @JobValue);
END
--------------------------------------------------------
Declare @Jobbyte varbinary(max);
Declare @JobValue int; 

Select @Jobbyte = JobBinary from Job;
Select @JobValue = count(*) from dbo.Fn_BinaryToTable(@Jobbyte) Where FNBIT = 0x01

----Select @JobValue = Sum(Cast(FNBIT as int)) from dbo.Fn_BinaryToTable(@Jobbyte)
set statistics time on
set statistics io on

select cid,dbo.fn_GetPercentage(@JobValue,cal,@Jobbyte) from eCal

set statistics time oFF
set statistics io oFF

----------------------------------------------------------
  • Numbers 表仅包含一个包含 1 到 99999 值的 int 字段
  • 96 使用 15 分钟间隔一天 (24* (60/15))
4

1 回答 1

0

请看下一个代码。它生成行并比较两个字符串的段：

with q as (
    SELECT
        CONVERT(varchar(max), a ,2) a, -- convert 'a' bin data to varchar
        CONVERT(varchar(max), b ,2) b  -- convert 'b' bin data to varchar
    FROM
        (
            SELECT -- a, b - test bin data
                Cast (1 as binary(1)) + Cast (0 as binary(1)) + Cast (1 as binary(1)) a, -- 01 00 01
                Cast (1 as binary(1)) + Cast (1 as binary(1)) + Cast (1 as binary(1)) b  -- 01 01 01 - 2 matches
        ) k
)

select
    --*, substring(a, x + 1, 2), substring(b, x + 1, 2)
    sum(case when 
        substring(a, x + 1, 2) = substring(b, x + 1, 2) -- is block equals
    then 1 else 0 end) sum_of_all_pair_equals
from
    q
        join
    ( -- generate 99999 rows for getting 2 chars block from staring a, b
        select
            x * 10000 + y * 1000 + z * 100 + k * 10 + l as x
        from
            (select 1 x union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) x,
            (select 1 y union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) y,
            (select 1 z union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) z,
            (select 1 k union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) k,
            (select 1 l union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9 union select 0) l
    ) x
        on (
            x between 0 and len(a) - 2 -- start from 1 to len - 2
            and x % 2 = 0 -- get every even x
        )

这个查询必须更快，但我没有测试它的性能：

DECLARE @a VARCHAR(30); -- a bin string
DECLARE @b VARCHAR(30); -- b bin string
DECLARE @itersLeft INT; -- iterations count
DECLARE @i INT; -- counter
DECLARE @outCnt INT; -- result match counter

SELECT
    @a = CONVERT(varchar(max), a ,2), -- convert 'a' bin data to varchar
    @b = CONVERT(varchar(max), b ,2)  -- convert 'b' bin data to varchar
FROM
    (
        SELECT -- a, b - test bin data
            Cast (1 as binary(1)) + Cast (0 as binary(1)) + Cast (1 as binary(1)) a, -- 01 00 01
            Cast (1 as binary(1)) + Cast (1 as binary(1)) + Cast (1 as binary(1)) b  -- 01 01 01 - 2 matches
    ) k
    ;

SET @i = 0; -- init counter
SET @outCnt = 0; -- init result counter
SELECT @itersLeft = LEN(@a) - 2; -- setting max iterations counter

WHILE @i <= @itersLeft
BEGIN
        if substring(@a, @i + 1, 2) = substring(@b, @i + 1, 2) -- compare data
            SET @outCnt = @outCnt + 1; -- increase counter if equals

        SET @i = @i + 2; -- increase counter by block size
END

SELECT @outCnt result; -- selecting result
于 2013-08-30T09:20:35.137 回答