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Editorial Team
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Asked: 2026-06-18T10:24:21+00:00

I need to query in MS Access the difference in value of a column

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I need to query in MS Access the difference in value of a column to 8 and only if it is greater than 8.

So if I have a column of numbers 1-10, I want to query the sum of all the value’s differences from 8. So the result of the query for the below column would be 3. (9-8)+(10-8)

    SELECT Sum(([time1]-8)+([time2]-8)+([time3]-8)+([time4]-8)+([time5]-8)+([time6]-8)+([time7]-8)+([time8]-8)+([time9]-8)+([time10]-8)+([time11]-8)+([time12]-8)+([time13]-8)+([time14]-8)+([time15]-8)+([time16]-8)+([time17]-8)+([time18]-8)+([time19]-8)+([time20]-8)+([time21]-8)+([time22]-8)) AS Total
FROM tblTimeTracking
WHERE (((Month(([Day])))=Month(Now()))) AND ([time1]>8 AND[time2]>8 AND[time3]>8 AND[time4]>8 AND[time5]>8 AND[time6]>8 AND[time7]>8 AND[time8]>8 AND[time9]>8 AND[time10]>8 AND[time11]>8 AND[time12]>8 AND[time13]>8 AND[time14]>8 AND[time15]>8 AND[time16]>8 AND[time17]>8 AND[time18]>8 AND[time19]>8 AND[time20]>8 AND[time21]>8 AND[time22])  ;

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1 Answer

  1. Editorial Team

    How about:

    SELECT Sum([Value]-8) As SumOfVal
FROM table
WHERE [Value]>8

    Edit re complete change in original question.

    It is not clear what you want

     SELECT Sum(([time1]-8)+([time2]-8) ...
 WHERE [time1]>8 And Time2>8 ...

    Time1>8 will exclude nulls, but if that is not what you are doing, you will need to consider:

     Nz([time1],0) + ...

    Edit re comments

    Something like:

     SELECT Sum(times) FROM
 (SELECT IIf(Time1>8,Time1-8,Time1) As times FROM Table
 UNION ALL
 SELECT IIf(Time2>8,Time2-8,Time2)  As times FROM Table) As b

    As b is an alias: Access SQL

    UNION / UNION ALL: View a unified result from multiple queries with a union query

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