where exists - all - group by? - tsql

Question

I use SQL Server 2008 R2.
I have a weird problem as following. I have a table as shown in
I need to write such a query like:
SELECT DISTINCT Field1
FROM MYTABLE
WHERE Field2 IN (96,102)
in this query, WHERE Field2 IN (96,102) gives me 96 or 102 or both!
More over, I would like to return rows that contains 96 and 102 at the same time!
Is there any suggestion? please write result oriented...

Answer 1

I have made a sqlfiddle for this..
create table a (id int, val int)
go
insert into a select 1, 22
insert into a select 1, 122
insert into a select 2, 22
insert into a select 3, 122
insert into a select 4, 22
insert into a select 4, 122
then select like this
select count(distinct id), id
from a
where val in (22, 122)
group by id
having count(id) > 1
EDIT: count(distinct id) will only show distinct counts..

Answer 2

EDIT:
Here's a sqlfiddle example (thanks to Mark Kremers):
http://sqlfiddle.com/#!3/df201/1
create table mytable (field1 int, field2 int)
go
insert into mytable values (199201, 84)
insert into mytable values (199201, 96)
insert into mytable values (199201, 102)
insert into mytable values (199201, 103)
insert into mytable values (581424, 96)
insert into mytable values (581424, 84)
insert into mytable values (581424, 106)
insert into mytable values (581424, 122)
insert into mytable values (687368, 79)
insert into mytable values (687368, 96)
insert into mytable values (687368, 102)
insert into mytable values (687368, 104)
insert into mytable values (687368, 106)
Here's the query:
select distinct a.field1 from
( select field1 from mytable where field2=96) a
inner join
( select field1 from mytable where field2=102) b
on a.field1 = b.field1
And here are the results:
FIELD1
199201
687368
Finally, here's a simplified version of the query (thans to pst):
select distinct a.field1 from mytable a
inner join mytable b
on a.field1 = b.field1
where a.field2=96 and b.field2=102

Answer 3

Use a self-join? Not the most tidy, but I think it works well for 2 values
SELECT *
FROM T R1
JOIN T R2 -- join table with itself
ON R1.F1 = R2.F1 -- where the first field is the same
WHERE R1.F2 = 96 AND R2.F2 = 102 -- and each has one of the required values
(T = Table, Rx = Relation Alias, Fx = Field)
If there can be an arbitrary number of fields, this can be solved as
CREATE TABLE #T (id int, val int)
GO
INSERT INTO #T (id, val)
VALUES
(1, 22), (1, 22), -- no, only 22 (but 2 records)
(2, 22), (2, 122), -- yes, both values (only)
(3, 122), -- no, only 122
(4, 22), (4,122), -- yes, both values ..
(4, 444), (4, null), -- and extra values
(5, 555) -- no, neither value
GO
-- Using DISTINCT over filtered results first, as
-- SQL Server 2008 does not support HAVING COUNT(DISTINCT F1, F2)
SELECT id
FROM (SELECT DISTINCT id, val
FROM #T
WHERE val IN (22, 122)) AS R1
GROUP BY id
HAVING COUNT(id) >= 2 -- or 3 or ..
GO
-- Or a similar variation, as can COUNT(DISTINCT ..)
-- in the SELECT of a GROUP BY
SELECT id
FROM (SELECT id, COUNT(DISTINCT val) as ct
FROM #T
WHERE val IN (22, 122)
GROUP BY id) AS R1
WHERE ct >= 2 -- or 3 or ..
GO
For larger IN (..) sizes, say above 20 values, it may be advisable to use a separate table or table-value and a JOIN for performance reasons.

Answer 4

Try from your original query:
SELECT DISTINCT Field1
FROM MYTABLE
WHERE rtrim(ltrim(cast(Field2 as varchar))) IN ('96','102')