How to join on _first_ record? - mysql

Question

With my limited knowledge of complex mysql queries I'm trying to retrieve some information from a database.
The story is this; users get an invite to come to our company. On the basis of this one invite, users can get multiple notifications and multiple appointments. I've got three relevant tables:
invites
-------
| id | name | created |
-----------------------------
| 1 | someth1 | 2018-02-03 |
| 2 | someth2 | 2018-02-03 |
| 3 | someth3 | 2018-02-03 |
notifications
-------------
| id | inv_id | message |
--------------------------
| 1 | 101 | hello |
| 2 | 287 | hi |
| 3 | 827 | hey |
appointments
------------
| id | inv_id | start_at |
-----------------------------
| 1 | 101 | 2018-02-03 |
| 2 | 287 | 2018-02-08 |
| 3 | 827 | 2018-02-15 |
I currently have a query, which shows a list of notifications send to users, for all invites done after 1 feb 2018, and which have an appointment no later than '2018-03-10'.
SELECT id, inv_id, message
FROM notifications
WHERE inv_id IN (
SELECT id
FROM invites as invite
WHERE created > '2018-02-01'
AND id IN (
SELECT inv_id
FROM appointments
WHERE invite.id = inv_id
AND start_at < '2018-03-10'
)
)
ORDER BY inv_id ASC;
The result looks something like this:
| id | inv_id | message |
--------------------------
| 1 | 101 | hello |
| 2 | 287 | hi |
| 3 | 827 | hey |
I now want to add the start_at of the first appointment for these notifications
| id | inv_id | message | start_at |
---------------------------------------
| 1 | 101 | hello | 2018-02-03 |
| 2 | 287 | hi | 2018-02-08 |
| 3 | 827 | hey | 2018-02-15 |
But from here I'm kinda lost in how I should do that.
Does anybody know how I can add the start_at of the first appointment which corresponds to the invite for the respective notification? So it should show the start_at of the first appointment for the invite of the notification inv_id?

Answer 1

Try this:
SELECT id, N.inv_id, message,A.start_at
FROM notifications N
JOIN(
SELECT inv_id,MIN(start_at) start_at
FROM appointments
WHERE inv_id IN (
SELECT id
FROM invites as invite
WHERE created > '2018-02-01'
AND id IN (
SELECT inv_id
FROM appointments
WHERE invite.id = inv_id
AND start_at < '2018-03-10'
)
)
GROUP BY inv_id
)A ON N.inv_id = A.inv_id
WHERE inv_id IN (
SELECT id
FROM invites as invite
WHERE created > '2018-02-01'
AND id IN (
SELECT inv_id
FROM appointments
WHERE invite.id = inv_id
AND start_at < '2018-03-10'
)
)
ORDER BY inv_id ASC;

Answer 2

You don't need a lot of subqueries to get the desired result. Just good use of join operations and just one subquery to get the first appointment so here is the query you need:
select n.id, n.inv_id, n.message, a.startat as start_at
from invites i
inner join notifications n
on i.id = n.inv_id
inner join (select inv_id, min(start_at) startat
from appointments
where start_at < '2018-03-10'
group by inv_id) a
on n.inv_id = a.inv_id
where i.created > '2018-02-01';
Notice that for your current sample data your desired result is impossible since in your query you use the notifications.inv_id IN invites.id and there is no equivalence (1,2,3 are different from 101, 287, 827).
Because of that I created a SQLFiddle to show the working query but with those ids "101, 287, 827" as invites. Here it is: http://sqlfiddle.com/#!9/2d5175/2
Also, if you want the notifications even if there is no appointment for it change the join operation between notifications and the subquery from inner join to left join