how can I add 2 digit together in SQL Server 2000 - sql-server

Question

suppose I have a number 62. This is composed of 2 digits
How can add 2 digit together divide by 10 and if result = something like 6.2 just take reminder
declare #Number int,#Result int
set #Number =62
if len(#Number) > 1
set #Result=????=--Add 6 and 2 =8
set #result=#result % 10 --Mod operator
print #result
-- the result should be 2 in this case
what Am I doing wrong?
Thanks a lot

Answer 1

set #tens = floor(#Number / 10);
set #ones = #number - #tens;
set #Result = #tens + #ones;
Or use left and right to access substrings.

Answer 2

I have no way to try it:
DECLARE #Number VARCHAR(2)
SET #Number = '62'
declare #firstNum INT, #secondNum INT
SET #firstNum = CAST(SUBSTRING(#Number, 1, 1) AS INT)
SET #secondNum = CAST(SUBSTRING(#Number, 2, 1) AS INT)
DECLARE #Result int
SET #Result = (#firstNum + #secondNum) % 10

Answer 3

I think #Number % 10is what you are looking for. It returns the last digit of any number. 62 -> 2, 97 -> 7, etc...
Update:
I may have misunderstood the question. Maybe you want 10 % ((#Number / 10) + #Number) instead.
(#Number / 10) + #Number is the sum of the digits of a two-digit number.

Answer 4

If the number is always two digits, then I would use LEFT(#Number,1) and RIGHT(#Number,1) to access each digit. If it's not, comment back and I'll give you some help with the loop you'll need.
More pressing an issue is that you're expecting 2, but the result of 8 mod 10 is 8. If you're looking for 2, the sum is 10 mod 8 (10 % #Result).
In any case, hit me back if this doesn't answer exactly what you wanted.

Answer 5

Try this
declare #myNumber float,#result float
set #myNumber = 62
select
#result = (case when len(#myNumber) = 1 then #myNumber
else (#myNumber /10) + cast(#myNumber as int) % 10 end)
select [Output] = STUFF(
cast(#result as varchar(50))
,1
,charindex('.',cast(#result as varchar(50)))
,'')
Output
2
Hope this helps
And if u just want to add 2 numbers try this(though it has been implemented above)
declare #myNumber int,#result int
set #myNumber = 62
select
Result = (case when len(#myNumber) = 1 then #myNumber
else (#myNumber /10) + #myNumber % 10 end)
Result
8

Answer 6

So I guess you want something like this - parse the string representing your number, adding up the individual digits as integer values. This gives you a total result at the end - then you do whatever you need to do with that. This code works for any length of string (up to 50 characters = 50 digits in your original number):
DECLARE #Number INT
SET #Number = 62
DECLARE #NumString VARCHAR(50)
SET #NumString = CAST(#Number AS VARCHAR(50))
DECLARE #Index INT
SET #Index = 1
DECLARE #Sum INT
SET #Sum = 0
WHILE #Index <= LEN(#NumString)
BEGIN
SET #Sum = #Sum + CAST(SUBSTRING(#NumString, #Index, 1) AS INT)
SET #Index = #Index + 1
END
SELECT #Sum AS 'Sum of all digits'
With the initial value of "62" for #Number, I get a result of 8 - now you can continue on using that value.
If you need this function often, I would probably encapsulate it into a user-defined function so you can call it from everywhere in your code.