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non normalized table

Posted on 2014-01-26
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Last Modified: 2014-01-30
I have a non normalized table - I think that's the correct description.  I have a column that is a code that indicates the type of data that is contained in a second column, so let's say "code" and "value" are the column names.    I need the average of all value for each code returned in a single record.  I only need this for, say 3 specific codes, 11, 22, 33.  How can I do this?
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Question by:HLRosenberger
7 Comments
 
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Expert Comment

by:magarity
ID: 39810300
select code, avg(value) from table where code in (11,22,33) group by code
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Expert Comment

by:magarity
ID: 39810308
PS - A 'denormalized' table is one where two or more normalized tables are combined.  For example, if 'code' in your example also has 'code long description' in this same table then that would be denormalized.  If code long description were in a lookup table, that would be normalized,
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Expert Comment

by:PortletPaul
ID: 39810818
>>" returned in a single record. "

Why must it be returned as a single record?

(the solution provided above will produce 3 rows)
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Accepted Solution

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Scott Pletcher earned 2000 total points
ID: 39812202
SELECT
    AVG(CASE WHEN code = '11' THEN value END) AS code_11_avg,
    AVG(CASE WHEN code = '22' THEN value END) AS code_22_avg,
    AVG(CASE WHEN code = '33' THEN value END) AS code_33_avg
FROM dbo.tablename
WHERE
    code IN ( '11', '22', '33' )
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Expert Comment

by:magarity
ID: 39812491
It's not clear either way that it's supposed to be a single record result set. That sentence doesn't parse well. The second query will do if that's the case.
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Expert Comment

by:awking00
ID: 39812740
Perhaps some sample data and your expected output would help clarify. I think one of the two solutions provided will accomplish what you want. Assume the average value of code 11 is 5, and the average value of code 22 is 10, and the average value of code 33 is 15, the first produces
code value
11     5
22    10
33    15
and the second produces
11avgval  22avgval  33avgval
5              10             15
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Author Closing Comment

by:HLRosenberger
ID: 39823205
thanks.
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