how to get this sql query - sql

Question

hello everyone I have a problem for this query, I need to get the rows where the id has the same numbers:
id2 | num
----+------
28 | 6
28 | 104
28 | 106
50 | 6
50 | 104
expected result:
id2 | num
----+-----
28 | 6
28 | 104
50 | 6
50 | 104
result doesn't include 28 106 because there's no 50 106 .
case 2:
id2 | num
----+-----
29 | 1
30 | 1
31 | 1
expected result:
id2 | num
----+-----
29 | 1
30 | 1
31 | 1
retrieves all because all the ids have num equal to 1
these numbers are random the condition is that if there are more than two ids they must have the same numbers in column 2

Answer 1

One way to do this is to count the occurrences of each num value, and compare it with the number of DISTINCT id2 values. If they are the same, then that num value occurs for every id2 value. You can then SELECT rows from the table which match those num values:
SELECT *
FROM data
WHERE num IN (SELECT num
FROM data
GROUP BY num
HAVING COUNT(*) = (SELECT COUNT(DISTINCT id2) FROM data))
ORDER BY id2, num
Output (for first dataset):
id2 num
28 6
28 104
50 6
50 104
Output (for second dataset):
id2 num
29 1
30 1
31 1
Demo on SQLFiddle

Answer 2

Another way is to use an OVER clause to count the occurrences of each value in num and compare to count of distinct values in id2 - which can be laterally joined:
CREATE TABLE mytable(
id2 VARCHAR(11)
,num INTEGER
);
INSERT INTO mytable(id2,num) VALUES ('28',6);
INSERT INTO mytable(id2,num) VALUES ('28',104);
INSERT INTO mytable(id2,num) VALUES ('28',106);
INSERT INTO mytable(id2,num) VALUES ('50',6);
INSERT INTO mytable(id2,num) VALUES ('50',104);
select id2, num
from (
select
id2, num
, count(*) over(partition by num) c_num
, ca.c_id2
from mytable
left join lateral (select count(distinct id2) c_id2 from mytable) ca on true
) d
where c_num = c_id2
;
id2 | num
:-- | --:
28 | 6
50 | 6
28 | 104
50 | 104
CREATE TABLE mytable(
id2 VARCHAR(11)
,num INTEGER
);
INSERT INTO mytable(id2,num) VALUES ('29',1);
INSERT INTO mytable(id2,num) VALUES ('30',1);
INSERT INTO mytable(id2,num) VALUES ('31',1);
select id2, num
from (
select
id2, num
, count(*) over(partition by num) c_num
, ca.c_id2
from mytable
left join lateral (select count(distinct id2) c_id2 from mytable) ca on true
) d
where c_num = c_id2
;
id2 | num
:-- | --:
29 | 1
30 | 1
31 | 1
db<>fiddle here

Answer 3

Basically, you want to count the number of distinct id2 values in the data and the number of distinct id2 values on each num. If only Postgres supported count(distinct) as a window function, you could do:
select id2, num
from (select t.*,
count(distinct t.id2) over (partition by t.num) as cnt_id2_on_num,
count(distinct t.id2) over () as cnt_id2
from t
) t
where cnt_id2_on_num = cnt_id2;
There is a simple work-around, which is the sum of dense_rank()s:
select id2, num
from (select t.*,
(dense_rank() over (partition by t.num order by t.id2) +
dense_rank() over (partition by t.num order by t.id2 desc)
) as cnt_id2_on_num,
(dense_rank() over (order by t.id2) +
dense_rank() over (order by t.id2 desc)
) as cnt_id2
from mytable t
) d
where cnt_id2_on_num = cnt_id2;
If you know there are no duplicates, you can write this as:
select id2, num
from (select t.*,
count(*) (partition by t.num) as cnt_id2_on_num,
(dense_rank() over (order by t.id2) +
dense_rank() over (order by t.id2 desc)
) as cnt_id2
from mytable t
) d
where cnt_id2_on_num = cnt_id2;