Select distinct rows that contain a given set of data - sql

Question

I have a following table:
bid | data
1 | a
1 | b
1 | c
2 | a
3 | c
3 | a
I want to select all bids that contain given set of data.
For example, all bids that 'contains' data "a" and "b" (result should be bid 1), or ones that contain "a" and "c" (1 and 3).
Only solution I could think of is kind of nasty, so I would appreciate some help/suggestions.
My first try:
select bid from my_table as t1 where
exists (select * from my_table t2 where
t2.bid = t1.bid and
t2.data='a'
)
and
exists (select * from my_table t2 where
t2.bid = t1.bid and
t2.data='b'
)
group by bid;
Thanks.

Answer 1

select t1.bid
from table_1 t1
inner join table_1 t2 on t1.bid = t2.bid
where t1.data = 'a' and t2.data = 'c'
By the way:
all bids that 'contains' data "a" and "b" (result should be bid 1)
--> bid 2 also contains data 'a' and 'b'

Answer 2

While I would not recommend this solution for only two variable lookups it's rate of growth for query cost when matching on more variables increases very slowly as opposed to doing an inner join for each match. As a disclaimer I realize that if pipe is a valid field or there are xml encoded charcters that this break.
select e.bid
from myTable e
cross apply ( select '|'+ i.data + '|'
from myTable i
where e.bid = i.bid
for xml path('')) T(v)
where v like '%|A|%' and v like '%|B|%' --and v like '%|C|%'.....
group by e.bid
as a side not about other options your answer could be simplified into
select bid from my_table as t1 where
exists (select * from my_table t2 where
t2.bid = t1.bid and
t2.data='a'
)
and t1.data = 'c'
group by bid;
This is roughly an equivalent of christian's answer. The optimizer will most likely treat these the same.
select distinct t1.bid
from table_1 t1
inner join table_1 t2 on t1.bid = t2.bid
where t1.data = 'a' and t2.data = 'c'

Answer 3

With a subquery, count the number of right occurences you have in your table.
SELECT DISTINCT m.bid
FROM myTable m
WHERE (
SELECT COUNT(1)
FROM myTable m2
WHERE (m2.data = 'a'
OR m2.data = 'b')
AND m.bid = m2.bid
) = 2

Answer 4

Maybe not the best answer but:
select bid from mytable where data = 'a'
intersect
select bid from mytable where data = 'c'

Answer 5

Uses exists:
declare #t table(bid int, data char)
insert #t values(1,'a'),(1,'b'),(1,'c'),(2,'b'),(2,'a'),(3,'c'),(3,'a')
select distinct t1.bid
from #t t1
where exists(
select 1
from #t t2
where t2.bid = t1.bid and t2.data = 'a'
)
and exists(
select 1
from #t t2
where t2.bid = t1.bid and t2.data = 'b'
)
XML PATH and XQuery version:
select distinct t.bid
from
(
select *
, (
select *
from #t t2
where t2.bid = t1.bid
for xml path, root('root'), type
) [x]
from #t t1
) t
where t.x.exist('root[*/data[text() = "a"] and */data[. = "b"]]') = 1