Question

I have a table similar to below in sql server 2005

date_column | field1 | field2 
1 June 2012 | xyz    | 53
1 June 2012 | abc    | 87
2 June 2012 | xyz    | 81
3 June 2012 | xyz    | 54
3 June 2012 | abc    | 53
3 June 2012 | abc    | 54
4 June 2012 | mmn    | 53
4 June 2012 | xyz    | 54
4 June 2012 | mmn    | 54
4 June 2012 | mmn    | 55
3 June 2012 | abc    | 55
3 June 2012 | adf    | 86
3 June 2012 | asd    | 33

I want to find all field1 values which has corresponding field2 values ’53’, ’54’ and ’55’. within the same date so output should be as below :

date_column | field1 | field2
3 June 2012 | abc    | 53
3 June 2012 | abc    | 54
3 June 2012 | abc    | 55
4 June 2012 | mmn    | 53
4 June 2012 | mmn    | 54
4 June 2012 | mmn    | 55

I tried the below sql code with inner join but it doesn’t work

select date_column, field1, field2 from table1
inner join (select date_column, field1, field2 from table1 where field2
in ('54',  '55')) as table2
on table1.date_column = table2.date_column and
table1.field1 = table2.field1
where field1 in ('53', '54', '55')
group by date_column, field1, field2
order by date_column, field1, field2

Answer 1

I believe this should work? You might be overcomplicating this or I am misunderstanding the problem

SELECT field1, date_column FROM table1
WHERE field2 in ('53', '54', '55')
GROUP BY field1, date_column
HAVING COUNT(DISTINCT field2) = 3

Here is the SQLFiddle

You say you only need to find field1, so this should work, but if you need all columns, then you can do this:

SELECT Table1.*
FROM Table1
JOIN
(
SELECT field1, date_column FROM table1
WHERE field2 in ('53', '54', '55')
GROUP BY field1, date_column
HAVING COUNT(DISTINCT field2) = 3
) AS MoreThan1
 ON Table1.field1 = MoreThan1.field1 and Table1.date_column = MoreThan1.date_column

Here is the SQLFiddle

OR, just add a WHERE if you only want the all columns for those in the 53-55:

WHERE Table1.field2 in ('53', '54', '55')

Another Fiddle