Joining a second table and selecting the first entry - mysql

Question

I have been having some troubles getting my head around achieving the below.
I have an 'applications' table and a 'application_logs' table. I am attempting to select all the applications where the 'type' is equal to 'test' and then join the 'application_logs' table and retrieve only the first log entry for the application.
One of the queries I tried and understood most was: (whilst this didn't fail it looked like an endless loop and completed the query.
SELECT applications.id FROM applications JOIN application_logs ON application_logs.application_id =
(
SELECT application_logs.id FROM application_logs
WHERE application_logs.application_id = applications.id
ORDER BY id DESC
LIMIT 1
)
WHERE type = 'test';
There were some other queries (using CROSS APPLY/distinct) I attempted but they didn't make sense to me and didn't look like they were trying to achieve the same thing. I appreciate all the help :)

Answer 1

There are many ways to achieve this in standard SQL. A lateral join (CROSS APPLY) is the first to come to mind, but MySQL doesn't support it. Another would be FETCH FIRST ROWS WITH TIES to get all latest application logs, but MySQL doesn't support it (and its counterpart LIMIT doesn't have a ties clause either). If you are only interested in the application ID alone (as shown in your query) one could even combine this with an INTERSECT operation, but MySQL doesn't support it.
After all you want to find out the maximum log ID per application ID. As of MySQL 8 you can do this on-the-fly:
select *
from applications a
left join
(
select
application_logs.*,
max(id) over (partition by application_id) as max_application_id
from application_logs
) al on al.application_id = a.id and al.application_id = al.max_application_id
where a.type = 'test';
In earlier vesions you would do this in separate steps. One way would be this:
select *
from applications a
left join application_logs al
on al.application_id = alm.application_id
and (al.application_id, al.id) in
(
select application_id, max(id)
from application_logs
group by application_id
)
where a.type = 'test';
Another is your own query, where you only got confused with the IDs:
SELECT *
FROM applications a
JOIN application_logs al ON al.id =
(
SELECT almax.id
FROM application_logs almax
WHERE almax.application_id = a.id
ORDER BY almax.id DESC
LIMIT 1
)
WHERE a.type = 'test';

Answer 2

try this
SELECT
a.*, c.*
FROM
applications a
INNER JOIN (
SELECT
al.application_id AS application_id,
MAX(al.id) AS al_id
FROM
application_logs al
GROUP BY
al.application_id
) c ON a.id = c.application_id
WHERE
a.type = 'test';

Answer 3

To get the first entry from application_logs table with respect to application_id. You need to use Row_Number Over (partition by order by ).
SELECT *
FROM applications A
LEFT JOIN (
SELECT Id AS applications.id
FROM (
SELECT ROW_NUMBER() OVER (Partition By applications.id ORDER BY application_logs.id) as R, application_logs.id, applications.id
FROM application_logs
) AS S
WHERE R = 1
) AS L ON L.applications.id = A.applications.id

Answer 4

Try this -
SELECT a.id, ay.*
FROM applications AS a
INNER JOIN (
SELECT al.application_id, min(al.id) as Min_Id
FROM application_logs AS al
GROUP BY al.application_id
) AS ax ON ax.application_id = a.id
INNER JOIN application_logs AS ay ON ay.id = ax.id
WHERE a.type = 'test';

Answer 5

Your query suggests that "first" means the largest id. (Colloquially, I would expect "first" to mean the smallest id or earliest chronologically.)
I usually recommend a correlated subquery for the filtering. It is worth testing if this is faster than other methods:
select . . .
from applications a join
application_logs al
on al.application_id = a.id
where a.type = 'test' and
al.id = (select max(al2.id)
from application_logs al2
where al2.application_id = al.application_id
);
The optimal indexes for performance are:
application(type, id)
application_logs(application_id, id)