Get the column order of a query - mysql

Question

I have a table with two columns [id, value] both numeric.
In this example:
[ id, value ]
[ 1, 6 ]
[ 2, 4 ]
[ 3, 10 ]
[ 4, 2 ]
[ 5, 7 ]
[ 6, 3 ]
For a given id I'd like to retrieve the top 3 id's (those with highest value), their top position and if the given id is not in the top 3, also get its position, id and value:
Example 1: ask_id = 5 Return:
[ position, id, value ]
[ 1, 3, 10 ]
[ 2, 5, 7 ]
[ 3, 1, 6 ]
Example 2: ask_id = 4. Return:
[ position, id, value ]
[ 1, 3, 10 ]
[ 2, 5, 7 ]
[ 3, 1, 6 ]
[ 6, 4, 2 ]
So the important points are:
How to get for the position column?
How to get the additional row if possible (anyway there's no problem if I need two queries)?

Answer 1

select t2.pos, t1.id, t1.value
from test as t1
inner join
(select id, value, #pos:=if(#pos is null, 0, #pos)+1 as pos
from test order by value desc) as t2
on t1.id=t2.id
where t2.pos<=3 or t2.id={$ask_id}
order by t2.pos;

Answer 2

Basically, the idea is like this:
Rank the rows by value.
Retrieve rows where at least one of the following is true:
position BETWEEN 1 AND 3
id = #given_id
These posts give examples of how you could substitute ranking functions (at least the most fundamental of them, ROW_NUMBER()) in MySQL:
ROW_NUMBER() in MySQL
MSSQL Row_Number() over(order by) in MySql
This method should be used with caution, though, as this article explains.
That said, one possible implementation of the above steps might look like this:
SET #pos = 0;
SELECT
position,
id,
value
FROM (
SELECT
id,
value,
#pos := #pos + 1 AS position
FROM atable
ORDER BY value DESC
) s
WHERE position BETWEEN 1 AND 3
OR id = #given_id
ORDER BY position

Answer 3

Tested in MySQL
to retrieve the top 3 id's (those with highest value) with position in ascending order.
set #num = 0;
SELECT #num := #num + 1 as position_sequence,id,value FROM tablename
ORDER BY value desc
limit 3;

Answer 4

I've not (yet) tested the selected answer in MySQL on the interesting cases where there are ties in the top three places, but I have tested this code in Informix on those cases, and it produces the answer I think should be produced.
Assuming that the table is called leader_board:
CREATE TABLE leader_board(id INTEGER NOT NULL PRIMARY KEY, value INTEGER NOT NULL);
INSERT INTO leader_board(id, value) VALUES(1, 6);
INSERT INTO leader_board(id, value) VALUES(2, 4);
INSERT INTO leader_board(id, value) VALUES(3, 10);
INSERT INTO leader_board(id, value) VALUES(4, 2);
INSERT INTO leader_board(id, value) VALUES(5, 7);
INSERT INTO leader_board(id, value) VALUES(6, 3);
This query works on the data shown, assuming that the special ID is 4:
SELECT b.position - c.tied + 1 AS standing, a.id, a.value
FROM leader_board AS a
JOIN (SELECT COUNT(*) AS position, d.id
FROM leader_board AS d
JOIN leader_board AS e ON (d.value <= e.value)
GROUP BY d.id
) AS b
ON a.id = b.id
JOIN (SELECT COUNT(*) AS tied, f.id
FROM leader_board AS f
JOIN leader_board AS g ON (f.value = g.value)
GROUP BY f.id
) AS c
ON a.id = c.id
WHERE (a.id = 4 OR (b.position - c.tied + 1) <= 3) -- Special ID = 4; Top N = 3
ORDER BY position, a.id;
Output on original data:
standing id value
1 3 10
2 5 7
3 1 6
6 4 2
Explanation
The two sub-queries are closely related, but they produce different answers. At one time, I used two temporary tables to hold those results. In particular, the first sub-query (AS b) produces a position, but when there are ties, the position is the lowest rather than the highest of the tied positions. That is, given:
ID Value
1 10
2 7
3 7
4 7
The outputs will be:
Position ID
1 1
4 2
4 3
4 4
However, we would like to count them as:
Position ID
1 1
2 2
2 3
2 4
So, the corrected position is the original position minus the number of tied values (3 for ID ∈ { 2, 3, 4 }, 1 for ID 1) plus 1. The second sub-query returns the number of tied values for each ID. There might be a neater way to do that calculation, but I'm not sure what it is at the moment.
Special cases
However, the code should demonstrate that it handles the cases where:
There are 2 or more ID values with the same top value.
There are 2 or more ID values with the same second highest top score (but the top one is unique).
There are 2 or more ID values with the same third highest top score (but the top two are unique).
To save rewriting the query each time, I converted it into an Informix-style stored procedure which take both the Special ID and the Top N (defaulting to 3) values that should be displayed and made them into parameters of the procedure. (Yes, the notation in the RETURNING clause is weird.)
CREATE PROCEDURE leader_board_standings(extra_id INTEGER, top_n INTEGER DEFAULT 3)
RETURNING INTEGER AS standing, INTEGER AS id, INTEGER AS value;
DEFINE standing, id, value INTEGER;
FOREACH SELECT b.position - c.tied + 1 AS standing, a.id, a.value
INTO standing, id, value
FROM leader_board AS a
JOIN (SELECT COUNT(*) AS position, d.id
FROM leader_board AS d
JOIN leader_board AS e ON (d.value <= e.value)
GROUP BY d.id
) AS b
ON a.id = b.id
JOIN (SELECT COUNT(*) AS tied, f.id
FROM leader_board AS f
JOIN leader_board AS g ON (f.value = g.value)
GROUP BY f.id
) AS c
ON a.id = c.id
WHERE (a.id = extra_id OR (b.position - c.tied + 1) <= top_n)
ORDER BY position, a.id
RETURN standing, id, value WITH RESUME;
END FOREACH;
END PROCEDURE;
This can be invoked to produce the same result as before:
EXECUTE PROCEDURE leader_board_standings(4);
To illustrate the various cases outlined above, add and remove extra rows:
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
6 4 2
INSERT INTO leader_board(id, value) VALUES(10, 10);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
1 10 10
3 5 7
7 4 2
INSERT INTO leader_board(id, value) VALUES(11, 10);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
1 10 10
1 11 10
8 4 2
INSERT INTO leader_board(id, value) VALUES(12, 10);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
1 10 10
1 11 10
1 12 10
9 4 2
DELETE FROM leader_board WHERE id IN (10, 11, 12);
EXECUTE PROCEDURE leader_board_standings(6, 4); -- Special ID 6; Top 4
1 3 10
2 5 7
3 1 6
4 2 4
5 6 3
INSERT INTO leader_board(id, value) VALUES(7, 7);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
2 7 7
7 4 2
INSERT INTO leader_board(id, value) VALUES(13, 7);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
2 7 7
2 13 7
8 4 2
INSERT INTO leader_board(id, value) VALUES(14, 7);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
2 7 7
2 13 7
2 14 7
9 4 2
DELETE FROM leader_board WHERE id IN(7, 13, 14);
INSERT INTO leader_board(id, value) VALUES(8, 6);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
3 8 6
7 4 2
INSERT INTO leader_board(id, value) VALUES(9, 6);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
3 8 6
3 9 6
8 4 2
INSERT INTO leader_board(id, value) VALUES(15, 6);
EXECUTE PROCEDURE leader_board_standings(4);
1 3 10
2 5 7
3 1 6
3 8 6
3 9 6
3 15 6
9 4 2
EXECUTE PROCEDURE leader_board_standings(3); -- Special ID 3 appears in top 3
1 3 10
2 5 7
3 1 6
That all looks correct to me.