Question

I assume this is not deterministic simply because DB_NAME() is not deterministic? If DB_NAME() is not deterministic, why is it not deterministic?

ALTER FUNCTION [TheSchema].[udf_IS_PRODUCTION] ()
RETURNS bit
    WITH SCHEMABINDING
AS 
    BEGIN
        RETURN CASE WHEN DB_NAME() = 'PRODUCTION' THEN CONVERT(bit, 1) ELSE CONVERT(bit, 0) END
    END

Update: This version works, is deterministic, allows the same code to be used in any database and removes the hardcoding of the database name (which also allows me to remove another automatic system health exception about database name coding)

ALTER FUNCTION [TheSchema].[udf_IS_PRODUCTION] ()
RETURNS bit
    WITH SCHEMABINDING
AS 
    BEGIN
        RETURN (SELECT IS_PRODUCTION FROM TheSchema.IS_PRODUCTION)
    END

FYI This is the code snippet in my system health self-reporting system which I use to monitor potential problems.

    SELECT  'Non-deterministic Scalar UDF' AS Problem
           ,QUOTENAME(ROUTINE_SCHEMA) + '.' + QUOTENAME(ROUTINE_NAME) AS ROUTINE_NAME
    FROM    INFORMATION_SCHEMA.ROUTINES WITH (NOLOCK)
    WHERE   IS_DETERMINISTIC = 'NO'
            AND ROUTINE_TYPE = 'FUNCTION'
            AND DATA_TYPE <> 'TABLE'
    ORDER BY ROUTINE_SCHEMA
           ,ROUTINE_NAME

Answer 1

Sure, I can think of one way to make it deterministic. Deploy this function on your production database:

ALTER FUNCTION [TheSchema].[udf_IS_PRODUCTION] ()
RETURNS bit
    WITH SCHEMABINDING
AS 
BEGIN
    RETURN CONVERT(bit, 1)
END

And deploy this one to your test database:

ALTER FUNCTION [TheSchema].[udf_IS_PRODUCTION] ()
RETURNS bit
    WITH SCHEMABINDING
AS 
BEGIN
    RETURN CONVERT(bit, 0)
END

This may seem silly but IMO the database name should not be “hard coded” any more so than the return value of some UDF.

Better yet, just put this information in a configuration table somewhere.