Selecting COUNT and MAX columns with 2 tables and a bridge table - mysql

so what I am trying to do is having 3 tables (pictures, collections, and bridge) with the following columns:
Collections Table:
| id | name |
------------------
| 1 | coll1 |
| 2 | coll2 |
------------------
Pictures Table: (timestamps are unix timestamps)
| id | name | timestamp |
-------------------------
| 5 | Pic5 | 1 |
| 6 | Pic6 | 19 |
| 7 | Pic7 | 3 |
| 8 | Pic8 | 892 |
| 9 | Pic9 | 4 |
-------------------------
Bridge Table:
| id | collection | picture |
-----------------------------
| 1 | 1 | 5 |
| 2 | 1 | 6 |
| 3 | 1 | 7 |
| 4 | 1 | 8 |
| 5 | 2 | 5 |
| 6 | 2 | 9 |
| 7 | 2 | 7 |
-----------------------------
And the result should look like this:
| collection_name | picture_count | newest_picture |
----------------------------------------------------
| coll1 | 4 | 8 |
| coll2 | 3 | 9 |
----------------------------------------------------
newest_picture should always be the picture with the heighest timestamp in that collection and I also want to sort the result by it. picture_count is obviously the count of picture in that collection.
Can this be done in a single statement with table joins and if yes:
how can I do this the best way?

A simple method uses correlated subqueries:
select c.*,
(select count(*)
from bridge b
where b.collection = c.id
) as pic_count,
(select p.id
from bridge b join
pictures p
on b.picture = b.id
where b.collection = c.id
order by p.timestamp desc
limit 1
) as most_recent_picture
from collections c;
A more common approach would use window functions:
select c.id, c.name, count(bp.collection), bp.most_recent_picture
from collections c left join
(select b.*,
first_value(p.id) over (partition by b.collection order by p.timestamp desc) as most_recent_picture
from bridge b join
pictures p
on b.picture = p.id
) bp
on bp.collection = c.id
group by c.id, c.name, bp.most_recent_picture;

Related

MySQL: Finding the most efficient use of INNER JOIN with subquery

I have a working query using INNER JOIN and a subquery but was wondering if there is a more effient way of writing it.
with prl
as
(
SELECT `number`, creator, notes
FROM ratings
INNER JOIN
projects on ratings.project_id = projects.project_id
WHERE ratings.rating = 5 AND projects.active = 1
)
SELECT prl.`number`, creator, notes
FROM prl
INNER JOIN(
SELECT `number`
HAVING COUNT(creator) > 1
)temp ON prl.`number` = temp.`number`
ORDER BY temp.`number`
projects table
project_id| number | creator | active |
| 1 | 3 | bob | 1 |
| 2 | 4 | mary | 1 |
| 3 | 5 | asi | 1 |
rating table
project_id| notes | rating |
| 1 | note1 | 5 |
| 1 | note2 | 5 |
| 3 | note3 | 5 |
| 1 | note4 | 1 |
| 2 | note5 | 5 |
| 3 | note6 | 2 |
result
| number | creator | notes |
| 3 | bob | note1 |
| 3 | bob | note2 |
It seems like you're using MySQL version that support window function. If so, then try this:
SELECT number, creator, notes
FROM
(SELECT p.number, p.creator, r.notes,
COUNT(creator) OVER (PARTITION BY creator) AS cnt
FROM project p
JOIN rating r ON p.project_id=r.project_id
WHERE r.rating=5
AND p.active = 1) v
WHERE cnt=2;
As far as whether this is more efficient, I'm not really sure because it depends in your table indexes but for a small dataset, I assume this will do well.
Demo fiddle

mysql table ordering incorrect with group by and order by

table 1: forum_threads
+-----+------+-------+
| id | title| status|
+-----+------+-------+
| 1 | a | 1 |
| 2 | b | 1 |
| 3 | c | 1 |
| 4 | d | 1 |
| 5 | e | 1 |
| 6 | f | 1 |
+-----+------+-------+
table 2: forum_comments
+-----+----------+--------------------+
| id | thread_id| comment |
+-----+----------+--------------------+
| 1 | 4 | hai |
| 2 | 4 | hello |
| 3 | 2 | welcome |
| 4 | 2 | whats your name |
| 5 | 6 | how are you |
| 6 | 5 | how old are you |
| 7 | 5 | good |
+-----+----------+--------------------+
wanted output
+-----------+----------+-----------------+
| thread_id | title | comment_count |
+-----------+----------+-----------------+
| 5 | e | 2 |
| 6 | f | 1 |
| 2 | b | 2 |
| 4 | d | 2 |
+-----------+----------+-----------------+
my Query
SELECT forum_threads.*,forum_comments.*,count(forum_comments.id) as comment_count
FROM forum_comments
LEFT JOIN forum_threads ON forum_comments.thread_id = forum_threads.id
GROUP BY forum_threads.id
ORDER BY forum_comments.id desc
Here I am trying to get the titles by the latest comment.
when I give ORDER BY forum_comments.id this returns the wrong order.
I need to order by the latest comments in the forum_comments table.
this query returns the wrong order please help me to find out the correct order.
how could I solve this easily?
This query should give you the expected result:
select t2.thread_id, t1.title, t2.comment_count from forum_threads as t1,
(SELECT id, thread_id, count(comment) as comment_count from forum_comments group by thread_id) as t2
where t1.id = t2.thread_id order by t2.id desc;
Instead of using forum_threads.* and forum_comments.* can you give specific column names and try.
If that doesn't work you should try explicitly assigning primary and foreign keys.

How can I determine which user is the top one in the specific tag?

I have a question and answer website like stackoverflow. Here is the structure of some tables:
-- {superfluous} means some other columns which are not related to this question
// q&a
+----+-----------------+--------------------------+------+-----------+-----------+
| id | title | body | type | related | author_id |
+----+-----------------+--------------------------+------+-----------+-----------+
| 1 | How can I ... | I'm trying to make ... | q | NULL | 3 |
| 2 | | You can do that by ... | a | 1 | 1 |
| 3 | Why should I .. | I'm wonder, why ... | q | NULL | 1 |
| 4 | | First of all you ... | a | 1 | 2 |
| 5 | | Because that thing ... | a | 3 | 2 |
+----+-----------------+--------------------------+------+-----------+-----------+
// users
+----+--------+-----------------+
| id | name | {superfluous} |
+----+--------+-----------------+
| 1 | Jack | |
| 2 | Peter | |
| 3 | John | |
+----+--------+-----------------+
// votes
+----+----------+-----------+-------+-----------------+
| id | user_id | post_id | value | {superfluous} |
+----+----------+-----------+-------+-----------------+
| 1 | 3 | 4 | 1 | |
| 2 | 1 | 1 | -1 | |
| 3 | 2 | 1 | 1 | |
| 4 | 3 | 2 | -1 | |
| 5 | 1 | 4 | 1 | |
| 6 | 3 | 5 | -1 | |
+----+--------+-------------+-------+-----------------+
// tags
+----+------------+-----------------+
| id | name | {superfluous} |
+----+------------+-----------------+
| 1 | PHP | |
| 2 | SQL | |
| 3 | MySQL | |
| 4 | HTML | |
| 5 | CSS | |
| 6 | C# | |
+----+------------+-----------------+
// q&aTag
+-------+--------+
| q&aid | tag_id |
+-------+--------+
| 1 | 1 |
| 1 | 4 |
| 3 | 5 |
| 3 | 4 |
| 4 | 6 |
+-------+--------+
Now I need to find top users in a specific tag. For example, I need to find Peter as top user in PHP tag. Because his answer for question1 (which has PHP tag) has earned 2 upvotes. Is doing that possible?
Try this:
select q1.title, u.id, u.name, sum(v.value) total from `q&a` q1
left join `q&atag` qt ON q1.id = qt.`q&aid`
inner join tags t ON qt.tag_id = t.id
left join `q&a` q2 ON q2.related = q1.id
left join users u ON q2.author_id = u.id
left join votes v ON v.post_id = q2.id
where t.name = 'PHP'
group by q1.id, u.id
and here is a simple divided solution:
Let us divide it into sub queries:
get the id of the tag you will search for: select id from tags where name = 'PHP'
get the questions with this tag: select 'q&aid' from 'q&aTag' where tag_id = 1.
get the ids of answers for that question: select id, author_id fromq&awhere related in (2.)
get the final query: select user_id, sum(value) from votes where post_id in (3.) group by user_id
Now combining them all give the result:
select user_id, sum(`value`) total from votes
where post_id in (
select id from `q&a` where related in (
select `q&aid` from `q&aTag` where tag_id IN (
select id from tags where name = 'PHP'
)
)
)
group by user_id
you can add this at the end if you want only one record:
order by total desc limit 1

join with count on a joined table with group clause in mysql

I have 3 tables:
applications (has many votes)
votes (belongs to applications and questions)
questions (has many votes)
I need to get number of votes per application per question.
So, my attempt was:
SELECT applications.id, COUNT(votes.id), votes.question_id
FROM applications
LEFT OUTER JOIN votes ON (votes.application_id = application.id)
GROUP BY votes.question_id
However, it displays data only for a single application, so I assume my query is malformed:
+----+-----------------+-------------+
| id | COUNT(votes.id) | question_id |
+----+-----------------+-------------+
| 1 | 1185 | 1 |
| 1 | 1170 | 2 |
| 1 | 1209 | 3 |
| 1 | 1230 | 4 |
| 1 | 1213 | 5 |
+----+-----------------+-------------+
What I need:
+----+-----------------+-------------+
| id | COUNT(votes.id) | question_id |
+----+-----------------+-------------+
| 1 | 1185 | 1 |
| 1 | 1170 | 2 |
| 1 | 1209 | 3 |
| 1 | 1230 | 4 |
| 1 | 1213 | 5 |
| 2 | null | 1 |
| 2 | 50 | 2 |
| 2 | 333 | 3 |
| 2 | 1230 | 4 |
| 2 | 1213 | 5 |
| 3 | null | 1 |
| 3 | 50 | 2 |
| 3 | 333 | 3 |
| 3 | null | 4 |
| 3 | 5555 | 5 |
+----+-----------------+-------------+
The group by clause was missing applications.id.
SELECT applications.id, COUNT(votes.id), votes.question_id
FROM applications
LEFT OUTER JOIN votes ON votes.application_id = application.id
group by applications.id, votes.question_id
You should be grouping by the applications.id as well as the questions.id:
SELECT a.id, COUNT(votes.id), votes.question_id
FROM applications a LEFT OUTER JOIN
votes v
ON v.application_id = a.id
GROUP BY a.id, v.question_id;
However, this will not produce exactly what you want. You seem to want all the questions for the applications, regardless of whether or not there are any votes. If so, this is probably what you want:
SELECT a.id, q.question_id, COUNT(v.application_id)
FROM applications a CROSS JOIN
(SELECT DISTINCT question_id FROM votes) q LEFT JOIN
votes v
ON v.application_id = a.id and v.question_id = q.question_id
GROPU BY a.id, q.question_id;

Left join select using Propel ORM

I have 3 table
major table:
+----+------------+
| id | major |
+----+------------+
| 1 | Computer |
| 2 | Architect |
| 3 | Designer |
+----+------------+
classroom table:
+----+----------+-------+
| id | major_id | name |
+----+----------+-------+
| 1 | 1 | A |
| 2 | 1 | B |
| 3 | 1 | C |
| 4 | 2 | A |
| 5 | 2 | B |
| 6 | 3 | A |
+----+----------+-------+
and finally, student_classroom table
+----+------------+--------------+----------+
| id | student | classroom_id | status |
+----+------------+--------------+----------+
| 1 | John | 1 | Inactive |
| 2 | Defou | 2 | Active |
| 3 | John | 2 | Active |
| 4 | Alexa | 1 | Active |
| 5 | Nina | 1 | Active |
+----+------------+--------------+----------+
how can I use propel to build query below
select
a.id,
a.major,
b.number_of_student,
c.number_of_classroom
from major a
left join (
select
major.major_id,
count(student_classroom.id) as number_of_student
from major
left join classroom on classroom.major_id = major.id
left join student_classroom on student_classroom.classroom_id = classroom.id
where student_classroom.`status` = 'Active'
group by major_id
) b on b.major_id = a.major_id
left join (
select
major.major_id,
count(classroom.id) as number_of_classroom
from major
left join classroom on classroom.major_id = major.id
group by major_id
) c on c.major_id = a.major_id
Because I want the final result would be something like this, I spend hours trying to figure it out without success.
+----+------------+-------------------+---------------------+
| id | major | number_of_student | number_of_classroom |
+----+------------+-------------------+---------------------+
| 1 | Computer | 4 | 3 |
| 2 | Architect | 0 | 2 |
| 3 | Designer | 0 | 1 |
+----+------------+-------------------+---------------------+
Try this
select
m.id,
m.major,
count(distinct s.id) as number_of_student ,
count(distinct c.id) as number_of_classroom
from major m
left join classroom c on
(m.id = c.major_id)
left join student_classroom s
on (s.classroom_id = c.id and c.major_id = m.id and s.status = 'active')
group by m.id
order by m.id