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 Posted Thursday, August 02, 2012 11:42 PM
 SSC Veteran Group: General Forum Members Last Login: Today @ 12:04 AM Points: 204, Visits: 473
 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).
Post #1339633
 Posted Friday, August 03, 2012 12:22 AM
 SSC Journeyman Group: General Forum Members Last Login: Thursday, November 28, 2013 10:02 AM Points: 75, Visits: 426
 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))` 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.ruTwitter: @SomewereSomehow
Post #1339641
 Posted Friday, August 03, 2012 7:24 AM
 SSC Veteran Group: General Forum Members Last Login: Today @ 12:04 AM Points: 204, Visits: 473
 Thanks that help
Post #1339813
 Posted Friday, August 03, 2012 2:43 PM
 SSC-Addicted Group: General Forum Members Last Login: Yesterday @ 4:02 PM Points: 444, Visits: 2,028
 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
Post #1340116
 Posted Friday, August 03, 2012 2:59 PM
 SSC-Addicted Group: General Forum Members Last Login: Yesterday @ 4:02 PM Points: 444, Visits: 2,028
 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 012233344445555 51223334444555 5122333444455 512233344445 51223334444 5122333444 412233344 41223334 4122333 412233 31223 3122 312 21 2 1` -- Alan BursteinRead this article for best practices on asking questions.Need to split a string? Try this (Jeff Moden)Need a pattern-based string spitter? Try this (Dwain Camps)My blog
Post #1340118
 Posted Friday, August 03, 2012 3:32 PM
 Hall of Fame Group: General Forum Members Last Login: 2 days ago @ 1:39 PM Points: 3,021, Visits: 10,987
 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 = '+@valprint '@cmd = '+@cmdexec (@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)`
Post #1340131
 Posted Saturday, August 04, 2012 12:10 AM
 SSC Journeyman Group: General Forum Members Last Login: Thursday, November 28, 2013 10:02 AM Points: 75, Visits: 426
 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.ruTwitter: @SomewereSomehow
Post #1340165
 Posted Saturday, August 04, 2012 4:48 PM
 SSCommitted Group: General Forum Members Last Login: Tuesday, January 15, 2013 11:11 AM Points: 1,945, Visits: 2,782
 http://beyondrelational.com/modules/2/blogs/70/posts/14472/sum-up-digits-of-a-number.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', '#########')); Books in Celko Series for Morgan-Kaufmann PublishingAnalytics and OLAP in SQL Data and Databases: Concepts in Practice Data, Measurements and Standards in SQLSQL for SmartiesSQL Programming Style SQL Puzzles and Answers Thinking in SetsTrees and Hierarchies in SQL
Post #1340226
 Posted Sunday, August 05, 2012 1:54 AM
 SSC Journeyman Group: General Forum Members Last Login: Thursday, November 28, 2013 10:02 AM Points: 75, Visits: 426
 CELKO,Good idea, interesting approach! Smth tell's me that it would be also the fastest way of doing this! 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.ruTwitter: @SomewereSomehow
Post #1340246
 Posted Sunday, August 05, 2012 3:34 AM
 SSCommitted Group: General Forum Members Last Login: Friday, November 29, 2013 7:22 AM Points: 1,694, Visits: 19,539
 Another way, up to BIGINTs only`DECLARE @num BIGINT = 985;WITH Tens(Pos,Val) AS (SELECT 1, CAST(1 AS BIGINT) UNION ALLSELECT 2, CAST(10 AS BIGINT) UNION ALLSELECT 3, CAST(100 AS BIGINT) UNION ALLSELECT 4, CAST(1000 AS BIGINT) UNION ALLSELECT 5, CAST(10000 AS BIGINT) UNION ALLSELECT 6, CAST(100000 AS BIGINT) UNION ALLSELECT 7, CAST(1000000 AS BIGINT) UNION ALLSELECT 8, CAST(10000000 AS BIGINT) UNION ALLSELECT 9, CAST(100000000 AS BIGINT) UNION ALLSELECT 10,CAST(1000000000 AS BIGINT) UNION ALLSELECT 11,CAST(10000000000 AS BIGINT) UNION ALLSELECT 12,CAST(100000000000 AS BIGINT) UNION ALLSELECT 13,CAST(1000000000000 AS BIGINT) UNION ALLSELECT 14,CAST(10000000000000 AS BIGINT) UNION ALLSELECT 15,CAST(100000000000000 AS BIGINT) UNION ALLSELECT 16,CAST(1000000000000000 AS BIGINT) UNION ALLSELECT 17,CAST(10000000000000000 AS BIGINT) UNION ALLSELECT 18,CAST(100000000000000000 AS BIGINT) UNION ALLSELECT 19,CAST(1000000000000000000 AS BIGINT))SELECT SUM((@num / Val) % 10)FROM TensWHERE Val<=@num;` ____________________________________________________How to get the best help on a forumhttp://www.sqlservercentral.com/articles/Best+Practices/61537Never approach a goat from the front, a horse from the rear, or a fool from any direction.
Post #1340255

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