previous item in a list ordered on two columns - mysql

Question

Let's say we have a table with columns DAY and NUMERO.
There can be numerous rows with the same value of DAY but NUMERO is always unique.
What could be the most efficient way to get the item immediately preceding an existing couple (DAY, NUMERO) in a list ordered by DAY, NUMERO ?
I precise that I need this for mysql and I don't want to add a specific index (that's the reason why I don't simply use a linear function of DAY and NUMERO).
Here's an ordered test case :
DAY | NUMERO
1 | 11
1 | 12
1 | 15
4 | 7
4 | 9
4 | 14
5 | 8
6 | 10
6 | 19
My request must do this :
(1,11) => nothing
(1,15) => (1,12)
(4,7) => (1,15)
(4,9) => (4,7)
(4,14) => (4,9)
EDIT :
my current best solution is to have two successive queries :
select * from item where day=? and numero<? order by day desc, numero desc limit 1;
select * from item where day<? order by day desc, numero desc limit 1;
If the first query gives a result, I don't have to run the second one.
A similar solution would be to use a union but mysql doesn't seem to authorize unions with more than one column.
Both solutions look too heavy for a problem seeming so simple...

Answer 1

UPDATE
Here is a solution that actually gives correct results!
SELECT T1.day,
T1.numero,
COALESCE(MAX(T2.[day]), MAX(T3.[day])) AS prev_day,
COALESCE(MAX(T2.numero), MAX(T3.numero)) AS prev_numero
FROM #table AS T1
LEFT JOIN #table AS T2 ON
(
T2.[day] = T1.[day] AND T2.numero < T1.numero
)
LEFT JOIN #table AS T3 ON
(
T3.[day] < T1.[day]
AND NOT EXISTS(
SELECT *
FROM #table AS T4
WHERE T4.[day] > T3.[day] AND T4.[day] < T1.[day]
)
)
-- Add where clause like so to get specific values
-- WHERE T1.day = 4 AND T1.numero = 7
GROUP BY T1.day, T1.numero
ORDER BY T1.day, T1.numero
Results:
day numero prev_day prev_numero
----------- ----------- ----------- -----------
1 11 NULL NULL
1 12 1 11
1 15 1 12
4 7 1 15
4 9 4 7
4 11 4 9
5 8 4 11
6 10 5 8
6 19 6 10

Answer 2

If you are only retrieving the previous pair based on a single pair as opposed to trying to do the entire table, as others have alluded to, you could try the following simple query -
:day = 4
:numero = 9
SELECT day, numero
FROM table
WHERE (day = :day AND numero < :numero)
OR (day < :day)
ORDER BY day DESC, numero DESC
LIMIT 1

Answer 3

This is an vanilla query, avoiding CTE's, aggregates or window functions.
-- (a Before b) := (a.day < b.day
-- OR (a.day = b.day AND a.numero < b.numero))
SELECT d1.day AS DAY
, d1.numero AS numero
, d0.day AS day0
, d0.numero AS numero0
FROM tmp.lutser d1
LEFT JOIN tmp.lutser d0
ON (d0.day < d1.day OR (d0.day = d1.day AND d0.numero < d1.numero ))
WHERE NOT EXISTS (SELECT *
FROM tmp.lutser d
WHERE (dx.day < d1.day OR (dx.day = d1.day AND dx.numero < d1.numero ))
AND (dx.day > d0.day OR (dx.day = d0.day AND dx.numero > d0.numero ))
)
ORDER BY day,numero
;
For reference, this is the query using window functions:
SELECT day
, numero
, lag(day) OVER (w1)
, lag(numero) OVER (w1)
FROM tmp.lutser
WINDOW w1 AS (ORDER BY day, numero)
;
Result:
day | numero | day0 | numero0
-----+--------+------+---------
1 | 11 | |
1 | 12 | 1 | 11
1 | 15 | 1 | 12
4 | 7 | 1 | 15
4 | 9 | 4 | 7
4 | 11 | 4 | 9
5 | 8 | 4 | 11
6 | 10 | 5 | 8
6 | 19 | 6 | 10
(9 rows)