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 Posted Saturday, July 24, 2004 5:20 PM
 SSC-Enthusiastic Group: General Forum Members Last Login: Thursday, April 03, 2014 6:02 AM Points: 138, Visits: 259
 Comments posted to this topic are about the content posted at http://www.sqlservercentral.com/column
Post #128055
 Posted Tuesday, September 07, 2004 7:46 AM
 Forum Newbie Group: General Forum Members Last Login: Wednesday, August 05, 2009 10:58 AM Points: 1, Visits: 2
 You might be able to reduce a few instructions by using BETWEEN instead of x >= y AND x <= z.  So,   ...WHEN x BETWEEN y AND z.There seems to be an oportunity to simply write a BASE 10 to BASE 36 conversion routine.  I don't know if there is a more straightforward conversion than what you've preposed here but it might be worth a thought.Enjoy
Post #135630
 Posted Tuesday, September 07, 2004 10:35 AM
 SSC Veteran Group: General Forum Members Last Login: Thursday, October 27, 2005 7:23 AM Points: 266, Visits: 1
 The email header for your article states: "User defined functions were added in SQL Server 7 and enhanced in SQL Server 2000...".I was not aware of any capabiltiy for user defined functions in SQL Server 7. Did I miss something?
Post #135672
 Posted Wednesday, September 08, 2004 12:20 PM
 Grasshopper Group: General Forum Members Last Login: Thursday, April 10, 2014 1:49 PM Points: 15, Visits: 50
 <>Why not use an identity column ? There are plenty of number and letters available ! Leave the identity column as anumber until it is needed, and then convert it. If you only use numbers and upper-case letters you have 36 (base 36 !), so your range is from 1 to (36^3 + 36^2 + 36) (about 48K). Using lower-case letters too gets you to 240K+ ! Surely that would be sufficient !Mike
Post #135914
 Posted Tuesday, September 14, 2004 5:58 AM
 SSC-Enthusiastic Group: General Forum Members Last Login: Thursday, April 03, 2014 6:02 AM Points: 138, Visits: 259
 I can't use identity column. The column should be char(3) (or int with 3 digits). Database is growing dynamically. So, identity column has only 999 combinations. I need some where close to 40,000.
Post #136857
 Posted Tuesday, September 14, 2004 6:01 AM
 SSC-Enthusiastic Group: General Forum Members Last Login: Thursday, April 03, 2014 6:02 AM Points: 138, Visits: 259
 I didn't wrote the header. Sorry. You are right that there is no user defined function in SQL Server 7.
Post #136859
 Posted Tuesday, September 14, 2004 6:03 AM
 SSC-Enthusiastic Group: General Forum Members Last Login: Thursday, April 03, 2014 6:02 AM Points: 138, Visits: 259
 You may be right. I will try it. But the article was not written to show the best code, but to show the idea of how to use user defined function for the case.
Post #136861
 Posted Wednesday, September 29, 2004 9:36 AM
 Valued Member Group: General Forum Members Last Login: Monday, February 28, 2005 3:04 AM Points: 50, Visits: 1
 The column REALLY sensless. It's just an identity of a CHAR(3) type... THIS is much easier. declare @INT intset @INT = 32768 --(smallint)declare @VB varbinary(3)set @VB = cast( @INT as varbinary(3))select @VB,CHAR(ASCII(SUBSTRING(@VB, 1, 1)) + 33) +CHAR(ASCII(SUBSTRING(@VB, 2, 1)) + 33) +CHAR(ASCII(SUBSTRING(@VB, 3, 1)) + 33)
Post #139292
 Posted Wednesday, September 29, 2004 4:39 PM
 SSC-Enthusiastic Group: General Forum Members Last Login: Thursday, April 03, 2014 6:02 AM Points: 138, Visits: 259
 As I pointed, it will be more than 999 number of records in a table.Column data must be assigned automatically by the database. No one wants to rebuid the applications. Anyway any mechanism must create a value and place it as default for the column.User defined function is set the value by default if NULL is inserted.
Post #139370
 Posted Thursday, February 24, 2005 3:00 AM
 Forum Newbie Group: General Forum Members Last Login: Monday, April 07, 2014 6:45 AM Points: 5, Visits: 91
 This reminds me of a function I had to write recently, to convert alphanumeric keys to numeric keys.  The company had char(8) keys as '3AJPUL68' and desired to map the alphanumeric space to a numeric space, in order to start using numeric keys.  I've written a function that takes a string and makes up a numeric hash for it.The resolution parameter @p controls how many bits of information are kept, at each position of the string.  Entering 0 will suppress the position, 1 will do (ascii & 2^1 => giving either 0 or 1), entering 2 will do (ascii & 2^2 => resulting in either 0, 1, 2, or 3), etc, up to 6, being the highest resolution, which would make the ascii code correspond to a number between 0 and 63.  As the function iterates through the string positions, from right to left, it shifts the intermediate left results before adding the next hash value.By setting the resolution parameter @p to '00016666' you'd always get integers in return (because the sum of all bits is less than 32), that's what we did, because the leftmost characters were almost always for us non-significant.  You may need to change the return value to bigint if you need a higher resolution for the left-most positions.`CREATE  function fn_char8hash ( @s char(8) , @p char(8) )returns intasbeginif @s is null return @s``set @s = replicate(' ', 8 - len(@s)) + @sdeclare @c char(1)declare @a smallint``declare @e tinyintdeclare @l tinyintset @l = 0``declare @r intset @r = 0``declare @i tinyintset @i = 8while @i > 0begin set @e = cast(substring(@p, @i, 1) as tinyint) set @c = substring(@s, @i, 1)`` set @c = replace(upper(@c), '_', ' ') set @a = (power(2, @e) - 1) & (ascii(@c) -   case when @c < 'A' then 32 else   case when @e < 6 then 64 else 32   end end)-- print @a if @a > 0 begin  set @r = @a * power(2, @l) + @r  set @l = @l + @e end-- print - @r-- print @l`` set @i = @i - 1end``return @r``end`
Post #163689

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