Identify only when value matches - sql

Question

I need to return only rows that have the match e.g Value = A, but I only need the rows that have A and with no other values.
T1:
ID Value
1 A
1 B
1 C
2 A
3 A
3 B
4 A
5 B
5 D
5 E
5 F
Desired Output:
2
4
how can I achieve this?
when I try the following, 1&3 are also returned:
select ID from T1 where Value ='A'

Answer 1

With NOT EXISTS:
select t.id
from tablename t
where t.value = 'A'
and not exists (
select 1 from tablename
where id = t.id and value <> 'A'
)
From the sample data you posted there is no need to use:
select distinct t.id
but if you get duplicates then use it.
Another way if there are no null values:
select id
from tablename
group by id
having sum(case when value <> 'A' then 1 else 0 end) = 0
Or if you want the rows where the id has only 1 value = 'A':
select id
from tablename
group by id
having count(*) = 1 and max(value) = 'A'

Answer 2

I think the simplest way is aggregation with having:
select id
from tablename
group by id
having min(value) = max(value) and
min(value) = 'A';
Note that this ignores NULL values so it could return ids with both NULL and A. If you want to avoid that:
select id
from tablename
group by id
having count(value) = count(*) and
min(value) = max(value) and
min(value) = 'A';

Answer 3

Oracle Setup:
CREATE TABLE test_data ( ID, Value ) AS
SELECT 1, 'A' FROM DUAL UNION ALL
SELECT 1, 'B' FROM DUAL UNION ALL
SELECT 1, 'C' FROM DUAL UNION ALL
SELECT 2, 'A' FROM DUAL UNION ALL
SELECT 3, 'A' FROM DUAL UNION ALL
SELECT 3, 'B' FROM DUAL UNION ALL
SELECT 4, 'A' FROM DUAL UNION ALL
SELECT 5, 'B' FROM DUAL UNION ALL
SELECT 5, 'D' FROM DUAL UNION ALL
SELECT 5, 'E' FROM DUAL UNION ALL
SELECT 5, 'F' FROM DUAL
Query:
SELECT ID
FROM test_data
GROUP BY ID
HAVING COUNT( CASE Value WHEN 'A' THEN 1 END ) = 1
AND COUNT( CASE Value WHEN 'A' THEN NULL ELSE 1 END ) = 0
Output:
| ID |
| -: |
| 2 |
| 4 |
db<>fiddle here