arithmatic function

Question

arithmatic function

bruce_reid

SSCrazy

Points: 2419
More actions
November 25, 2003 at 12:42 pm

#160454

i have two numbers 999 and 888.But I want to muliply them as 900 and 800 .
Like that i want to do for other numbers
345*234->300*200
567*345->500*300.
Is there any function to achieve this.
Thanks

Viewing 8 posts - 1 through 7 (of 7 total)

You must be logged in to reply to this topic. Login to reply

cmore SSCrazy Points: 2452 More actions · Answer 1



declare @MyNumbers table (a int not null, b int not null)

insert into @MyNumbers select 345,234

insert into @MyNumbers select 567,345



select round(a,-2,0)*round(b,-2,0) from @MyNumbers

Jonathan SSC-Insane Points: 20427 More actions · Answer 2

Sorry; duplicate post. "Delete reply" still nonfunctional.

Edited by - Jonathan on 11/25/2003 1:44:33 PM

--Jonathan

Jonathan SSC-Insane Points: 20427 More actions · Answer 3

As you're wanting to truncate rather than round, I think cmore means:



SELECT ROUND(1,-2,1)*ROUND(b,-2,1)

If the vales are an integer data type, you could also just do something like this:



SELECT (a/100)*(b/100)*10000

--Jonathan

cmore SSCrazy Points: 2452 More actions · Answer 4

quote:
As you're wanting to truncate rather than round, I think cmore means:

SELECT ROUND(1,-2,1)*ROUND(b,-2,1)
If the vales are an integer data type, you could also just do something like this:

SELECT (a/100)*(b/100)*10000
--Jonathan

Oops. Yes thanks for catching that Jonathan.

Frank Kalis SSC Guru Points: 111183 More actions · Answer 5

Just curious, why?

Frank

http://www.insidesql.de

http://www.familienzirkus.de

--
Frank Kalis
Microsoft SQL Server MVP
Webmaster: http://www.insidesql.org/blogs
My blog: http://www.insidesql.org/blogs/frankkalis/[/url]

Janssen SSC Veteran Points: 233 More actions · Answer 6

for two integers @i & @j-2

the following should work

( looks like a slight overkill I know! )

select POWER(10,FLOOR(LOG10(@i)))*(FLOOR(@i)/POWER(10,FLOOR(LOG10(@i))))

*

POWER(10,FLOOR(LOG10(@j)))*(FLOOR(@j)/POWER(10,FLOOR(LOG10(@j))))

The advantage is that you don't have to worry about the precision of the integers - it'll work for 9 , 99 , 999, 9999 etc.

Regards

DavidT SSCommitted Points: 1727 More actions · Answer 7

Or......

declare @N int

declare @m int

set @N = 86548

set @m = 21458

select

cast(left(@N,1) + replicate(0,len(@N)-1) as int)

*

cast(left(@M,1) + replicate(0,len(@M)-1) as int)