I want to select rows from a table where ALL the values for a column are the same. For example, these two tables:
T1 T2 Num Let Num Val 1 A 1 R 2 A 2 R 3 C 3 S 4 C 3 S 5 D 3 - 6 A 4 S 7 A 4 S 8 A 5 R
I want a query that will return all Num where T1.Num = T2.Num and ALL values for T2.Val are 'S'. The query should return:
Num ----- 4
The reason 3 is not returned is because at least one T2.Val is NOT an 'S'. Thanks if you can answer this.
Seems pretty straightforward. Try this:
select T2.Num from T1 inner join T2 on T1.Num = T2.Num group by T2.Num having count(distinct T2.Val) = 1 and min(T2.Val) = 'S'
Since each row of T1 can join to more than one row of T2, a GROUP BY on T2.Num is required. The HAVING clause ensures that any T2.num is returned only when there's exactly one distinct T2.Val value in its group, and this value is 'S'. The MIN function is used, but it could also have been MAX, since there would be only one distinct value.
Note that a WHERE clause to restrict T2 rows to those containing 'S' would be wrong, unless accompanied by a NOT EXISTS correlated subquery to check for any other rows with any other value besides 'S' -- but this would be needlessly complex, so I won't show it.
UPDATE: Reader B. Humphreys emailed me with a concern about the solution above:
My understanding was that "Count(FieldName)" returned a count of NON-NULL values for the captioned field while "Count(*)" returned a count of all records (NULL or not).If that's the case, will this actually work properly or will it return TRUE even though the field "T2.Val" contains NULLs in addition to one other non-NULL value?
An excellent observation. One way around this is to replace any NULLS with an empty string, and if there are only NULLs, then the MIN() condition would not be 'S':
select ... having count(distinct coalesce(T2.Val,'')) = 1 and min(T2.Val) = 'S'
Mr. Humphreys suggested:
select ... having count(T2.Val) = count(*) and min(T2.Val) = 'S' and max(T2.Val) = 'S'
This works perfectly, even in Microsoft Access (which does not support COUNT DISTINCT, at least not in the version I'm using). Thanks, B. Humphreys!
This was first published in December 2003