Select max version number for each group consisting of multiple columns - mysql

Question

I have the following tables:
Apps
TYPE_ID | BUILD_ID | CONFIG_ID | VERSION_ID | (All foreign keys to the respective tables)
1 | 1 | 1 | 1 |
1 | 1 | 1 | 2 |
2 | 2 | 3 | 3 |
2 | 2 | 3 | 4 |
Versions
ID | major | minor | patch
1 | 1 |0 |1
2 | 2 |0 |0
3 | 3 |0 |3
4 | 4 |0 |0
I need to select highest version rows from Apps table for each unique combinations of TYPE_ID, BUILD_ID and CONFIG_ID.
The version number should be calculated by MAX(major * 1000000 + minor * 1000 + patch) in the versions table.
So from the given example of the Apps table the result would be:
TYPE_ID | BUILD_ID | CONFIG_ID | VERSION_ID |
1 | 1 | 1 | 2 |
2 | 2 | 3 | 4 |
Have tried something like this:
SELECT p1.* FROM Apps p1
INNER JOIN (
SELECT max(VERSION_ID) MaxVersion, CONFIG_ID
FROM Apps
GROUP BY CONFIG_ID
) p2
ON p1.CONFIG_ID = p2.CONFIG_ID
AND p1.VERSION_ID = p2.MaxVersion
GROUP BY `TYPE_ID`, `BUILD_ID`, `CONFIG_ID`
But MAX is applied on the VERSION_ID and I need MAX to be applied on major, minor and patch combinations.
MySQL Version 15.1 distribution 5.5.56-MariaDB
Any help would be appreciated.
Cheers!

Answer 1

You can compute the maximum version per type_id, build_id, config_id using the formula described in your question, use it again same formula to locate the version:
SELECT sq.type_id, sq.build_id, sq.config_id, versions.id AS version_id_max
FROM (
SELECT type_id, build_id, config_id, MAX(major * 1000000 + minor * 1000 + patch) AS max_version
FROM apps
INNER JOIN versions ON apps.version_id = versions.id
GROUP BY type_id, build_id, config_id
) sq
INNER JOIN versions ON max_version = major * 1000000 + minor * 1000 + patch
+---------+----------+-----------+----------------+
| type_id | build_id | config_id | version_id_max |
+---------+----------+-----------+----------------+
| 1 | 1 | 1 | 2 |
| 2 | 2 | 3 | 4 |
+---------+----------+-----------+----------------+

Answer 2

Try this :
select type_id, build_id, config_id,
max(1000000*v.major+1000*v.minor+v.patch) as version
from apps a left join versions v on a.version_id=v.id
group by type_id, build_id, config_id

Answer 3

Utilizing Nested Derived subqueries, and a bit of hacky way of identifying VERSION_ID corresponding to MAX VERSION_NO.
We basically first get a derived table determining the VERSION_NO for each row in the Apps table.
Now using that derived table as a source for SELECT, we group by on the TYPE_ID, BUILD_ID and CONFIG_ID, and using a GROUP_CONCAT and string manipulation based trick, we determine the VERSION_ID corresponding to maximum VERSION_NO, for a group.
Try the following:
SELECT nest.TYPE_ID,
nest.BUILD_ID,
nest.CONFIG_ID,
SUBSTRING_INDEX(GROUP_CONCAT(DISTINCT nest.VERSION_ID
ORDER BY nest.VERSION_NO DESC
SEPARATOR ','), ',', 1) AS VERSION_ID
FROM (
SELECT A.TYPE_ID,
A.BUILD_ID,
A.CONFIG_ID,
A.VERSION_ID,
(V.major*1000000 + V.minor*1000 + V.patch) AS VERSION_NO
FROM Apps AS A
INNER JOIN Versions AS V ON V.ID = A.VERSION_ID
) AS nest
GROUP BY nest.TYPE_ID, nest.BUILD_ID, nest.CONFIG_ID
SQL FIDDLE

Answer 4

Try this:
SELECT a1.type_id, a1.build_id, a1.config_id, a1.version_id
FROM apps a1
WHERE NOT EXISTS(
(SELECT 'NEXT'
FROM apps a2
WHERE a2.type_id = a1.type_id
AND a2.build_id = a1.build_id
AND a2.config_id = a1.config_id
AND a2.version_id > a1.version_id))

Answer 5

Try this query, what I do here, I imitate well-known function ROW_NUMBER() OVER (PARTITION BY Type_id, Build_id, Config_id ORDER BY major desc, minor desc, patch desc).
select #type_id_lag := 0, #build_id_lag :=0, #config_id_lag := 0, #rn := 0;
select type_id, build_id, config_id, major, minor, patch from (
select case when #type_id_lag = type_id and
#build_id_lag = build_id and
#config_id_lag = config_id then #rn := #rn + 1 else #rn := 1 rn,
#type_id_lag := type_id type_id,
#build_id_lag := build_id build_id,
#config_id_lag := config_id config_id,
v.major, v.minor, v.patch
from Apps a
left join Versions v on a.version_id = v.id
order by a.type_id, a.build_id, a.config_id,
v.major desc, v.minor desc, v.patch desc
) a where rn = 1;