Logic to check if exact ids (3+ records) are present in a group in SQL Server - sql

Question

I have some sample data like:
INSERT INTO mytable
([FK_ID], [TYPE_ID])
VALUES
(10, 1),
(11, 1), (11, 2),
(12, 1), (12, 2), (12, 3),
(14, 1), (14, 2), (14, 3), (14, 4),
(15, 1), (15, 2), (15, 4)
Now, here I am trying to check if in each group by FK_ID we have exact match of TYPE_ID values for 1, 2 & 3.
So, the expected output is like:
(10, 1) this should fail
As in group FK_ID = 10 we only have one record
(11, 1), (11, 2) this should also fail
As in group FK_ID = 11 we have two records.
(12, 1), (12, 2), (12, 3) this should pass
As in group FK_ID = 12 we have two records.
And all the TYPE_ID are exactly matching 1, 2 & 3 values.
(14, 1), (14, 2), (14, 3), (14, 4) this should also fail
As we have 4 records here.
(15, 1), (15, 2), (15, 4) this should also fail
Even though we have three records, it should fail as the TYPE_ID here (1, 2, 4) are not matching with required match (1, 2, 3).
Here is my attempt:
select * from mytable t1
where exists (select COUNT(t2.TYPE_ID)
from mytable t2 where t2.FK_ID = t1.FK_ID
and t2.TYPE_ID IN (1, 2, 3)
group by t2.FK_ID having COUNT(t2.TYPE_ID) = 3);
This is not working as expected, because it also pass for FK_ID = 14 which has four records.
Demo: SQL Fiddle
Also, how we can make it generic so that if we need to check for 4 or more TYPE_ID values like (1,2,3,4) or (1,2,3,4,5), we can do that easily by updating few values.

Answer 1

The following query will do what you want:
select fk_id
from t
group by fk_id
having sum(case when type_id in (1, 2, 3) then 1 else 0 end) = 3 and
sum(case when type_id not in (1, 2, 3) then 1 else 0 end) = 0;
This assumes that you have no duplicate pairs (although depending on how you want to handle duplicates, it might be as easy as using, from (select distinct * from t) t).
As for "genericness", you need to update the in lists and the 3.
If you want something more generic:
with vals as (
select id
from (values (1), (2), (3)) v(id)
)
select fk_id
from t
group by fk_id
having sum(case when type_id in (select id from vals) then 1 else 0 end) = (select count(*) from vals) and
sum(case when type_id not in (select id from vals) then 1 else 0 end) = 0;

Answer 2

You can use this code:
SELECT y.fk_id FROM
(SELECT x.fk_id, COUNT(x.type_id) AS count, SUM(x.type_id) AS sum
FROM mytable x GROUP BY (x.fk_id)) AS y
WHERE y.count = 3 AND y.sum = 6
For making it generic, you can equal y.count with N and y.sum with N*(N-1)/2, where N is the number you are looking for (1, 2, ..., N).

Answer 3

You can try this query. COUNT and DISTINCT used for eliminate duplicate records.
SELECT
[FK_ID]
FROM
#mytable T
GROUP BY
[FK_ID]
HAVING
COUNT(DISTINCT CASE WHEN [TYPE_ID] IN (1,2,3) THEN [TYPE_ID] END) = 3
AND COUNT(CASE WHEN [TYPE_ID] NOT IN (1,2,3) THEN [TYPE_ID] END) = 0

Answer 4

Try this:
select FK_ID,count(distinct TYPE_ID) from mytable
where TYPE_ID<=3
group by FK_ID
having count(distinct TYPE_ID)=3

Answer 5

You should use CTE with Dynamic pass Value which you have mentioned in Q.
WITH CTE
AS (
SELECT FK_ID,
COUNT(*) CNT
FROM #mytable
GROUP BY FK_ID
HAVING COUNT(*) = 3) <----- Pass Value here What you want to Display Result,
CTE1
AS (
SELECT T.[ID],
T.[FK_ID],
T.[TYPE_ID],
ROW_NUMBER() OVER(PARTITION BY T.[FK_ID] ORDER BY
(
SELECT NULL
)) RN
FROM #mytable T
INNER JOIN CTE C ON C.FK_ID = T.FK_ID),
CTE2
AS (
SELECT C1.FK_ID
FROM CTE1 C1
GROUP BY C1.FK_ID
HAVING SUM(C1.TYPE_ID) = SUM(C1.RN))
SELECT TT1.*
FROM CTE2 C2
INNER JOIN #mytable TT1 ON TT1.FK_ID = C2.FK_ID;
From above SQL Command which will produce Result (I have passed 3) :
ID FK_ID TYPE_ID
4 12 1
5 12 2
6 12 3