Lets consider the following table-
ID Score
1 95
2 100
3 88
4 100
5 73
I am a total SQL noob but how do I return the Scores featuring both IDs 2 and 4?
So it should return 100 since its featured in both ID 2 and 4
This is an example of a "sets-within-sets" query. I recommend aggregation with the having clause, because it is the most flexible approach.
select score
from t
group by score
having sum(id = 2) > 0 and -- has id = 2
sum(id = 4) > 0 -- has id = 4
What this is doing is aggregating by score. Then the first part of the having clause (sum(id = 2)) is counting up how many "2"s there are per score. The second is counting up how many "4"s. Only scores that have at a "2" and "4" are returned.
SELECT score
FROM t
WHERE id in (2, 4)
HAVING COUNT(*) = 2 /* replace this with the number of IDs */
This selects the rows with ID 2 and 4. The HAVING clause then ensures that we found both rows; if either is missing, the count will be less than 2.
This assumes that id is a unique column.
select Score
from tbl a
where a.ID = 2 -- based off Score with ID = 2
--include Score only if it exists with ID 6 also
and exists (
select 1
from tbl b
where b.Score = a.Score and b.ID = 6
)
-- optional? ignore Score that exists with other ids as well
and not exists (
select 1
from tbl c
where c.Score = a.Score and c.ID not in (2, 6)
)
Related
Lets consider the following table-
ID Score
1 95
2 100
3 88
4 100
5 73
I am a total SQL noob but how do I return the Scores featuring both IDs 2 and 4?
So it should return 100 since its featured in both ID 2 and 4
This is an example of a "sets-within-sets" query. I recommend aggregation with the having clause, because it is the most flexible approach.
select score
from t
group by score
having sum(id = 2) > 0 and -- has id = 2
sum(id = 4) > 0 -- has id = 4
What this is doing is aggregating by score. Then the first part of the having clause (sum(id = 2)) is counting up how many "2"s there are per score. The second is counting up how many "4"s. Only scores that have at a "2" and "4" are returned.
SELECT score
FROM t
WHERE id in (2, 4)
HAVING COUNT(*) = 2 /* replace this with the number of IDs */
This selects the rows with ID 2 and 4. The HAVING clause then ensures that we found both rows; if either is missing, the count will be less than 2.
This assumes that id is a unique column.
select Score
from tbl a
where a.ID = 2 -- based off Score with ID = 2
--include Score only if it exists with ID 6 also
and exists (
select 1
from tbl b
where b.Score = a.Score and b.ID = 6
)
-- optional? ignore Score that exists with other ids as well
and not exists (
select 1
from tbl c
where c.Score = a.Score and c.ID not in (2, 6)
)
Lets consider the following table-
ID Score
1 95
2 100
3 88
4 100
5 73
I am a total SQL noob but how do I return the Scores featuring both IDs 2 and 4?
So it should return 100 since its featured in both ID 2 and 4
This is an example of a "sets-within-sets" query. I recommend aggregation with the having clause, because it is the most flexible approach.
select score
from t
group by score
having sum(id = 2) > 0 and -- has id = 2
sum(id = 4) > 0 -- has id = 4
What this is doing is aggregating by score. Then the first part of the having clause (sum(id = 2)) is counting up how many "2"s there are per score. The second is counting up how many "4"s. Only scores that have at a "2" and "4" are returned.
SELECT score
FROM t
WHERE id in (2, 4)
HAVING COUNT(*) = 2 /* replace this with the number of IDs */
This selects the rows with ID 2 and 4. The HAVING clause then ensures that we found both rows; if either is missing, the count will be less than 2.
This assumes that id is a unique column.
select Score
from tbl a
where a.ID = 2 -- based off Score with ID = 2
--include Score only if it exists with ID 6 also
and exists (
select 1
from tbl b
where b.Score = a.Score and b.ID = 6
)
-- optional? ignore Score that exists with other ids as well
and not exists (
select 1
from tbl c
where c.Score = a.Score and c.ID not in (2, 6)
)
Lets consider the following table-
ID Score
1 95
2 100
3 88
4 100
5 73
I am a total SQL noob but how do I return the Scores featuring both IDs 2 and 4?
So it should return 100 since its featured in both ID 2 and 4
This is an example of a "sets-within-sets" query. I recommend aggregation with the having clause, because it is the most flexible approach.
select score
from t
group by score
having sum(id = 2) > 0 and -- has id = 2
sum(id = 4) > 0 -- has id = 4
What this is doing is aggregating by score. Then the first part of the having clause (sum(id = 2)) is counting up how many "2"s there are per score. The second is counting up how many "4"s. Only scores that have at a "2" and "4" are returned.
SELECT score
FROM t
WHERE id in (2, 4)
HAVING COUNT(*) = 2 /* replace this with the number of IDs */
This selects the rows with ID 2 and 4. The HAVING clause then ensures that we found both rows; if either is missing, the count will be less than 2.
This assumes that id is a unique column.
select Score
from tbl a
where a.ID = 2 -- based off Score with ID = 2
--include Score only if it exists with ID 6 also
and exists (
select 1
from tbl b
where b.Score = a.Score and b.ID = 6
)
-- optional? ignore Score that exists with other ids as well
and not exists (
select 1
from tbl c
where c.Score = a.Score and c.ID not in (2, 6)
)
Lets consider the following table-
ID Score
1 95
2 100
3 88
4 100
5 73
I am a total SQL noob but how do I return the Scores featuring both IDs 2 and 4?
So it should return 100 since its featured in both ID 2 and 4
This is an example of a "sets-within-sets" query. I recommend aggregation with the having clause, because it is the most flexible approach.
select score
from t
group by score
having sum(id = 2) > 0 and -- has id = 2
sum(id = 4) > 0 -- has id = 4
What this is doing is aggregating by score. Then the first part of the having clause (sum(id = 2)) is counting up how many "2"s there are per score. The second is counting up how many "4"s. Only scores that have at a "2" and "4" are returned.
SELECT score
FROM t
WHERE id in (2, 4)
HAVING COUNT(*) = 2 /* replace this with the number of IDs */
This selects the rows with ID 2 and 4. The HAVING clause then ensures that we found both rows; if either is missing, the count will be less than 2.
This assumes that id is a unique column.
select Score
from tbl a
where a.ID = 2 -- based off Score with ID = 2
--include Score only if it exists with ID 6 also
and exists (
select 1
from tbl b
where b.Score = a.Score and b.ID = 6
)
-- optional? ignore Score that exists with other ids as well
and not exists (
select 1
from tbl c
where c.Score = a.Score and c.ID not in (2, 6)
)
Lets consider the following table-
ID Score
1 95
2 100
3 88
4 100
5 73
I am a total SQL noob but how do I return the Scores featuring both IDs 2 and 4?
So it should return 100 since its featured in both ID 2 and 4
This is an example of a "sets-within-sets" query. I recommend aggregation with the having clause, because it is the most flexible approach.
select score
from t
group by score
having sum(id = 2) > 0 and -- has id = 2
sum(id = 4) > 0 -- has id = 4
What this is doing is aggregating by score. Then the first part of the having clause (sum(id = 2)) is counting up how many "2"s there are per score. The second is counting up how many "4"s. Only scores that have at a "2" and "4" are returned.
SELECT score
FROM t
WHERE id in (2, 4)
HAVING COUNT(*) = 2 /* replace this with the number of IDs */
This selects the rows with ID 2 and 4. The HAVING clause then ensures that we found both rows; if either is missing, the count will be less than 2.
This assumes that id is a unique column.
select Score
from tbl a
where a.ID = 2 -- based off Score with ID = 2
--include Score only if it exists with ID 6 also
and exists (
select 1
from tbl b
where b.Score = a.Score and b.ID = 6
)
-- optional? ignore Score that exists with other ids as well
and not exists (
select 1
from tbl c
where c.Score = a.Score and c.ID not in (2, 6)
)