i have tried with below code. but i didnt get ouput as per multiple column.
select Name=stuff((select ','+
name from Input t1 WHERE t1.Name=t2.name for xml path('')),1,9,''),NAME
=stuff((select ','+
name from Input t1 WHERE t1.Name=t2.name for xml path('')),1,1,'')FROM
Input T2
GROUP BY NAME
I have separated the character and numeric parts of the given string in CTE and used a simple group by and sum clause on that.
;With CTE
As
(
Select
Left(Names, 1) As String,
SUBSTRING(Names, 2, Len(Names) - 1) As Number
From SeparateColumns
)
Select
String,
Sum(Cast(Number As Int)) As SumOfDigits
From CTE
Group By String
Order By String;
Assuming you will have single character in your column, the following code works
CREATE TABLE #TEMP_SPLIT
(VALUE VARCHAR(25))
INSERT INTO #TEMP_SPLIT
SELECT 'A10'
UNION
SELECT 'B20'
UNION
SELECT 'A30'
UNION
SELECT 'B40'
UNION
SELECT 'A10'
UNION
SELECT 'C1'
SELECT c, sum(tot)
FROM
(
SELECT SUBSTRING(VALUE,1,1) c ,CONVERT(FLOAT,SUBSTRING(VALUE,2,LEN (VALUE)-1)) Tot
FROM #TEMP_SPLIT
)T
GROUP BY C
DROP TABLE #TEMP_SPLIT
If the names column is always in the specified format, then use LEFT function to extract the character part and RIGHT function to extract the digits part and use these two in the sub-query and use GROUP BY clause and SUM function.
Query
select t.col_a as [string character],
sum(cast(t.col_b as int)) as [sum of digits] from(
select left(names, 1) as col_a,
right(names, len(names) - 1) as col_b
from [your_table_name]
) t
group by t.col_a;
You could use PATINDEX() with SUBSTRING() function if names always in this specified format
select A.CHAR [Strings], SUM(CAST(A.VALUE AS INT)) [Sum] from
(
SELECT SUBSTRING(Name, 1, PATINDEX('%[^A-Z]%', Name)-1) [CHAR], SUBSTRING(Name, PATINDEX('%[0-9]%', Name), len(Name)) [VALUE] FROM <table>
) a GROUP BY A.CHAR
Here's simple solution using subquery and group by clause:
select [StringChar], SUM([Number]) from (
select SUBSTRING(Names, 1, 1) as [StringChar],
convert(int, SUBSTRING(Names, 2, LEN(Names))) as [Number]
from [Input]
) as a group by [StringChar]