Question

我有以下功能：

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO    
ALTER FUNCTION [dbo].[IP4toBIGINT](
    @ip4 varchar(15)
) 
RETURNS bigint
WITH SCHEMABINDING
AS
BEGIN
    -- oc3 oc2 oc1 oc0
    -- 255.255.255.255
    -- Declared as BIGINTs to avoid overflows when multiplying later on     DECLARE @oct0 bigint, @oct1 bigint, @oct2 bigint, @oct3 bigint;
    DECLARE @Result bigint;

    SET @oct3 = CAST(PARSENAME(@ip4, 4) as tinyint);
    SET @oct2 = CAST(PARSENAME(@ip4, 3) as tinyint);
    SET @oct1 = CAST(PARSENAME(@ip4, 2) as tinyint);
    SET @oct0 = CAST(PARSENAME(@ip4, 1) as tinyint);

    -- Combine all values, multiply by 2^8, 2^16, 2^24 to bitshift.
    SET @Result = @oct3 * 16777216 + @oct2 * 65536 + @oct1 * 256 + @oct0;
    RETURN @Result;

END

但是...

SELECT 
     OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'IsDeterministic') as IsDeterministic 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'IsPrecise') as IsPrecise 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'IsSystemVerified') as IsSystemVerified 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'SystemDataAccess') as SystemDataAccess 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'UserDataAccess') as UserDataAccess

返回（结果转置）：

IsDeterministic 0

IsPrecise 1

IsSystemVerified 1

SystemDataAccess 0

UserDataAccess 0

我尝试多次删除并重新创建该函数，以确保它不是一些缓存问题。 CAST在这里应该是确定性的，因为我将它用于字符串 - >整数。

我完全难过，有什么想法吗？

Answer 1

PARSENAME总的来说是不确定的。是的，您在确定性的上下文中使用它，但我猜测服务器不知道。尝试替换PARSENAME并查看它是否发生变化。

Answer 2

这是造成问题的PARSENAME。用硬编码字符串替换它会导致确定性。不知道为什么...解析名称应该只是一个奇特的分裂函数。

检查出来：

SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO    
ALTER FUNCTION [dbo].[IP4toBIGINT](
    @ip4 varchar(15)
) 
RETURNS bigint
WITH SCHEMABINDING
AS
BEGIN
    -- oc3 oc2 oc1 oc0
    -- 255.255.255.255
    -- Declared as BIGINTs to avoid overflows when multiplying later on         
    DECLARE @oct0 bigint, @oct1 bigint, @oct2 bigint, @oct3 bigint;
    DECLARE @Result bigint;

    SET @oct3 = CAST('1' as tinyint);
    SET @oct2 = CAST('2' as tinyint);
    SET @oct1 = CAST('3' as tinyint);
    SET @oct0 = CAST('4' as tinyint);

    -- Combine all values, multiply by 2^8, 2^16, 2^24 to bitshift.
    SET @Result = @oct3 * 16777216 + @oct2 * 65536 + @oct1 * 256 + @oct0

    RETURN @Result
END
GO

SELECT 
     OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'IsDeterministic') as IsDeterministic 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'IsPrecise') as IsPrecise 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'IsSystemVerified') as IsSystemVerified 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'SystemDataAccess') as SystemDataAccess 
    ,OBJECTPROPERTYEX(OBJECT_ID('dbo.IP4toBIGINT'), 'UserDataAccess') as UserDataAccess

结果：

IsDeterministic IsPrecise IsSystemVerified  SystemDataAccess UserDataAccess
1               1         1                 0                0

Answer 3

是的，问题是确实使用了PARSENAME。 MSDN explicitly says这是确定性的。也许这是因为SQL假设您将阅读数据库架构？这表明非决定论，但我只是在推测。

SQL 2005为什么说这个UDF是非确定性的？

3 个答案: