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Hi, This was the question ask to one of my friend during an interview. He was ask to perform the addition of the digits. Suppose the number is 985 the output would be 22(9+8+5).




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Here is my solution
declare @i int = 985;
with nums(n) as(select n from (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10))nums(n)) select sum(convert(int,substring(convert(varchar(10),@i),n,1))) from nums where n <= len(convert(varchar(10),@i))
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Thanks that help




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SomewhereSomehow (8/3/2012)
Here is my solution declare @i int = 985;
with nums(n) as(select n from (values (1),(2),(3),(4),(5),(6),(7),(8),(9),(10))nums(n)) select sum(convert(int,substring(convert(varchar(10),@i),n,1))) from nums where n <= len(convert(varchar(10),@i))
This is good. A couple things to note: First, the length of the input variable (@i) is limited to the size of the two varchar declarations. For example, say @i = 99999999999 (that's eleven 9's), you would get an overflow error when converting the expression to varchar...
No big deal; who cares?... Just change the varchars to varchar(20) or varchar(50) and declare @i as bigint. There. Problem solved!!!
Nope. Still have one thing to address and this [b]WILL NOT produce an error[/b]. Instead you will just an incorrect aggregation. For the agregation to be accurate you would have to add additional values to your CTE. Again, say @i = 99999999999 (11 9's) you would return a 90(incorrect) instead of 99(correct). To fix this you would have to add an (11). If @i was 20 characters long you would have to add (11),(12)...(20).
A better way to write this would be:
DECLARE @i BIGINT=99999999999;
with nums(n) as ( SELECT 1 UNION ALL SELECT n+1 FROM nums WHERE n<30 ) select sum(convert(int,substring(convert(varchar(30),@i),n,1))) from nums where n <= len(convert(varchar(30),@i));
Now @i can be 18 characters long (limited to 18 because of the bigint). How about we change @i to varchar(50). Now it works for a number that's 50 chars long.
Instead of:
select n from (values (1),(2),(3),(4)...(50))nums(n) We are using some recursion:
with nums(n) as ( SELECT 1 UNION ALL SELECT n+1 FROM nums WHERE n<LEN(@i) )
... and then we pull it all together:
DECLARE @i varchar(50)='99999999999999999999999999999999999999999999999999';
with nums(n) as ( SELECT 1 UNION ALL SELECT n+1 FROM nums WHERE n<LEN(@i) ) select sum(convert(int,substring(convert(varchar(50),@i),n,1))) from nums where n <= len(convert(varchar(50),@i)); GO
 Alan Burstein
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Here is my solution:
DECLARE @i varchar(50)='122333444455555';
WITH val(x,n) AS ( SELECT LEFT(@i,LEN(@i)),'0' UNION ALL SELECT LEFT(x,LEN(x)1),RIGHT(x,1) FROM val WHERE LEN(x)>0 ) SELECT SUM(CAST(n AS int)) FROM val What's cool is you can replace SELECT SUM(CAST(n AS int)) FROM val with SELECT * FROM val to see how it works.
Result set:
x n   122333444455555 0 12233344445555 5 1223334444555 5 122333444455 5 12233344445 5 1223334444 5 122333444 4 12233344 4 1223334 4 122333 4 12233 3 1223 3 122 3 12 2 1 2 1
 Alan Burstein
Read this article for best practices on asking questions. Need to split a string? Try this (Jeff Moden) Need a patternbased string spitter? Try this (Dwain Camps) My blog




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A little dynamic SQL with a string of 100 digits as input:
declare @val varchar(100) = '20876543914578560430730723092317208765439145785604'+ '3073072309231720876543914578560430730723092317208'
declare @cmd varchar(300)
set @cmd = 'select [Sum] = '+ reverse(substring(reverse(replace(replace(replace( replace(replace(replace(replace(replace(replace(replace( convert(varchar(300),@val),'9','9+'),'8','8+'),'7','7+'),'6','6+') ,'5','5+'),'4','4+'),'3','3+'),'2','2+'),'1','1+'),'0','0+')),2,300))
print '@val = '+@val print '@cmd = '+@cmd
exec (@cmd)
Results:
@val = 208765439145785604307307230923172087654391457856043073072309231720876543914578560430730723092317208 @cmd = select [Sum] = 2+0+8+7+6+5+4+3+9+1+4+5+7+8+5+6+0+4+3+0+7+3+0+7+2+3+0+9+2+3+1+7+2+0+8+7+6+5+4+3+9+1+4+5+7+8+5+6+0+4+3+0+7+3+0+7+2+3+0+9+2+3+1+7+2+0+8+7+6+5+4+3+9+1+4+5+7+8+5+6+0+4+3+0+7+3+0+7+2+3+0+9+2+3+1+7+2+0+8 Sum  403
(1 row(s) affected)




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XMLSQLNinja, Using the same logic. Yor solution will not work if there will be 51 digits. Try to understand  my solution was intended to work only with int (and 999 999 999 99 is not int) in the same way, as yours only with 50 digits (btw, why 50, not 49 or 53?). And one note, specifying input as string  not good idea imho, it coul be easily broken if there will be not a digit char in the string. If we talk about numbers, the input should be only one of numeric types  this is good form.
I am really sorry for my poor gramma. And I hope that value of my answers will outweigh the harm for your eyes. Blog: http://somewheresomehow.ru Twitter: @SomewereSomehow




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http://beyondrelational.com/modules/2/blogs/70/posts/14472/sumupdigitsofanumber.aspx
We can just convert each digit into a hash mark. Concatenate them and measure the length. This is pure ANSI/ISO Standard SQL without any XML or other foreign language contamenation.
SET @digits_sum = LEN ( REPLACE(REPLACE(REPLACE(REPLACE(REPLACE( REPLACE(REPLACE(REPLACE(REPLACE(REPLACE (CAST (@in_integer AS VARCHAR(150)) ,'0', '') ,'1', '#') ,'2', '##') ,'3', '###') ,'4', '####') ,'5', '#####') ,'6', '######') ,'7', '#######') ,'8', '########') ,'9', '#########'));
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Another way, up to BIGINTs only
DECLARE @num BIGINT = 985;
WITH Tens(Pos,Val) AS ( SELECT 1, CAST(1 AS BIGINT) UNION ALL SELECT 2, CAST(10 AS BIGINT) UNION ALL SELECT 3, CAST(100 AS BIGINT) UNION ALL SELECT 4, CAST(1000 AS BIGINT) UNION ALL SELECT 5, CAST(10000 AS BIGINT) UNION ALL SELECT 6, CAST(100000 AS BIGINT) UNION ALL SELECT 7, CAST(1000000 AS BIGINT) UNION ALL SELECT 8, CAST(10000000 AS BIGINT) UNION ALL SELECT 9, CAST(100000000 AS BIGINT) UNION ALL SELECT 10,CAST(1000000000 AS BIGINT) UNION ALL SELECT 11,CAST(10000000000 AS BIGINT) UNION ALL SELECT 12,CAST(100000000000 AS BIGINT) UNION ALL SELECT 13,CAST(1000000000000 AS BIGINT) UNION ALL SELECT 14,CAST(10000000000000 AS BIGINT) UNION ALL SELECT 15,CAST(100000000000000 AS BIGINT) UNION ALL SELECT 16,CAST(1000000000000000 AS BIGINT) UNION ALL SELECT 17,CAST(10000000000000000 AS BIGINT) UNION ALL SELECT 18,CAST(100000000000000000 AS BIGINT) UNION ALL SELECT 19,CAST(1000000000000000000 AS BIGINT)) SELECT SUM((@num / Val) % 10) FROM Tens WHERE Val<=@num;
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