September 26, 2007 at 12:35 pm
How do I take a number like maybe 10.73 and remove the decimal place and making it a number like 1073. In other words I have to remove the decimal.
I need to do it using T-SQL
Thanks for any help
Ted
September 26, 2007 at 1:19 pm
declare @mynum decimal(10,2)
declare @mynum1 int
set @mynum = 10.73
select @mynum1 = @mynum * 100
print @mynum
print @mynum1
or
declare @mynum decimal(10,2)
set @mynum = 10.73
select cast(@mynum * 100 as int)
September 26, 2007 at 1:54 pm
Thanks for your prompt answer.
I was looking for something that would get rid of the decimal point
if the number was 10.73 or 10.733 or 10.7333 or 10.7
The decimal place being in any position
Thanks
Ted
September 26, 2007 at 2:10 pm
what is the datatype of the field containing the decimal point?
If it's varchar you could:
declare @mynum varchar(20)
set @mynum = '10.7333'
select replace(@mynum,'.','')
September 26, 2007 at 3:04 pm
The field is numeric(5,4)
but you have something there.
would it be possible to move it to a char field and then apply the logic you gave me
Again, thank you so much for your generous help
Ted
September 26, 2007 at 3:29 pm
declare @i decimal(10,2)
Set @i = 1078.734
select (left(@i, Charindex('.', @i, 1) - 1 ))
******************
Dinakar Nethi
Life is short. Enjoy it.
******************
September 27, 2007 at 12:02 am
Doesn't matter if it's varchar or not... your method still works...
declare @i decimal(10,2)
Set @i = 1078.734
PRINT REPLACE(@i,'.','')
--Jeff Moden
Change is inevitable... Change for the better is not.
September 27, 2007 at 3:58 pm
Thank you all for your quick responses
I finally made it work with your help
Ted
September 27, 2007 at 7:47 pm
Cool... but it's customary to thank folks by posting the code that you used to make it work 😉
--Jeff Moden
Change is inevitable... Change for the better is not.
September 28, 2007 at 12:19 pm
The problem with using the Replace string function is that 10.341 ends up as 1034100 if the numeric is defined with five places. If you want 10.342 to end up as 10341, 10.73 as 1073, 10.733 as 10733, 10.7333 as 107333 and 10.7 as 107, then the best way is to use numeric processes rather than string.
The only thing is, you have to define your working number to have enough places to the left to contain the entire value. So your example of storing 10.341 into a variable defined as decimal(5,4) is bogus--it only has one place to the left of the decimal point. So if your value is defined as "decimal(x,y)" then you have to declare a working variable as "decimal(x+y,y)" to contain the entire finished value.
declare @original decimal( 10, 5 ),
@Working decimal( 15, 5 ), -- 10 + 5 = 15
@Result decimal( 15, 0 ); -- Doesn't need scale, only precision
Set @original = 1078.734; -- This would be, say, an input parameter
-- First, make a copy into the working variable capable of handling it.
Set @Working = @original;
-- Now set up the loop
Set @Result = Floor( @Working );
While @Result < @Working
begin
Set @Working = @Working * 10;
Set @Result = floor( @Working );
end--while
select @Result as Result, @Working as Working;The loop executes one time through for each significant digit to the right of the decimal point -- in this example, three times. The result is 1078734 instead of 107873400.
As an aside, does anyone know how to get the old "<pre></pre>" formatting back? This code IFCode shortcut sucks. Sure, the code goes into a nice text field 🙂 but everything is double spaced. :angry:
Tomm Carr
--
Version Normal Form -- http://groups.google.com/group/vrdbms
September 29, 2007 at 1:25 pm
DECLARE @MYDECVAR DECIMAL
DECLARE @MYINTVAR INT
SET @MYDECVAR = 100.123
SET @MYINTVAR = CAST(@MYDECVAR AS INT)
SELECT @MYINTVAR
CHEERS 🙂
CHANDRA
September 29, 2007 at 2:15 pm
DECLARE @MYDECVAR DECIMAL
DECLARE @MYINTVAR INT
SET @MYDECVAR = 100.123
SET @MYINTVAR = CAST(@MYDECVAR AS INT)
SELECT @MYINTVAR
This returns 100, not 100123 as it should.
September 29, 2007 at 2:37 pm
This should do the trick... 100% of the time :
DECLARE @val DECIMAL (11,8)
SET @val = 100.10300100
SELECT REPLACE(RTRIM(LTRIM(REPLACE(REPLACE(CONVERT(VARCHAR(30), @val), '.', ''), '0', ' '))), ' ', '0')
SET @val = 0.01
SELECT REPLACE(RTRIM(LTRIM(REPLACE(REPLACE(CONVERT(VARCHAR(30), @val), '.', ''), '0', ' '))), ' ', '0')
SET @val = 1
SELECT REPLACE(RTRIM(LTRIM(REPLACE(REPLACE(CONVERT(VARCHAR(30), @val), '.', ''), '0', ' '))), ' ', '0')
SET @val = 0010.20301
SELECT REPLACE(RTRIM(LTRIM(REPLACE(REPLACE(CONVERT(VARCHAR(30), @val), '.', ''), '0', ' '))), ' ', '0')
September 29, 2007 at 3:56 pm
Why do you think you need the extra overhead of RTRIM/LTRIM when converting Decimal to Varchar?
--Jeff Moden
Change is inevitable... Change for the better is not.
September 29, 2007 at 4:18 pm
For dropping leading and trailing 0s from the string. Maybe I just missed the problem altogether...
Must be having a bad day or something.
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