# need help to solve a problem

• ### Question

• My problem is the following sittuation:

I have this table in ACCESS database and i want help to get an SQL expression to solve the issue i explain:

this is the table i have, with 2 columns:

CODE               count

0101                  5

0101                  8

0101                  7

0305                  8

0305                  6

0305                  1

So, what i want to do is to obtain the total of the count for each CODE

Example: for the CODE 0101 i have 5+8+7 = 20

So, i want to have an SQL expression that give me a table with the totals to each code

maybe its not complicated, but i am an inniciate with SQL, can you please help me with that?

Sunday, April 15, 2012 1:19 AM

• ```select code, sum([Count]) as [Total Count]
from myTable GROUP BY [Code]```

For every expert, there is an equal and opposite expert. - Becker's Law

My blog

• Proposed as answer by Sunday, April 15, 2012 8:58 AM
• Marked as answer by Sunday, April 15, 2012 9:27 PM
Sunday, April 15, 2012 4:18 AM
• Hi,

Use Group by function http://msdn.microsoft.com/en-us/library/ms177673.aspx

```SELECT CODE, SUM([count])
FROM tablename
GROUP BY [CODE]```

I hope this is helpful.

Elmozamil Elamir Hamid

### MyBlog

• Marked as answer by Sunday, April 15, 2012 9:27 PM
Sunday, April 15, 2012 6:00 AM

### All replies

• ```select code, sum([Count]) as [Total Count]
from myTable GROUP BY [Code]```

For every expert, there is an equal and opposite expert. - Becker's Law

My blog

• Proposed as answer by Sunday, April 15, 2012 8:58 AM
• Marked as answer by Sunday, April 15, 2012 9:27 PM
Sunday, April 15, 2012 4:18 AM
• Hi,

Use Group by function http://msdn.microsoft.com/en-us/library/ms177673.aspx

```SELECT CODE, SUM([count])
FROM tablename
GROUP BY [CODE]```

I hope this is helpful.

Elmozamil Elamir Hamid

### MyBlog

• Marked as answer by Sunday, April 15, 2012 9:27 PM
Sunday, April 15, 2012 6:00 AM
• as explained above you will need to use the GROUP BY statement and a SUM() function in your SELECT clause

John

http://knowledgy.org

Sunday, April 15, 2012 5:56 PM
• Thanks for your help

It works

joca_santos

Sunday, April 15, 2012 9:27 PM