Now that I got the Select all forums and get latest post too.. how? question answered, I am trying to write a query to select all threads in one particular forum and order them by the date of the latest post (column "updated_at").
This is my structure again:
forums forum_threads forum_posts
---------- ------------- -----------
id id id
parent_forum (NULLABLE) forum_id content
name user_id thread_id
description title user_id
icon views updated_at
created_at created_at
updated_at
last_post_id (NULLABLE)
I tried writing this query, and it works.. but not as expected: It doesn't order the threads by their last post date:
SELECT DISTINCT ON(t.id) t.id, u.username, p.updated_at, t.title
FROM forum_threads t
LEFT JOIN forum_posts p ON p.thread_id = t.id
LEFT JOIN users u ON u.id = p.user_id
WHERE t.forum_id = 3
ORDER BY t.id, p.updated_at DESC;
How can I solve this one?
Assuming you want a single row per thread and not all rows for all posts.
DISTINCT ON is still the most convenient tool. But the leading ORDER BY items have to match the expressions of the DISTINCT ON clause. If you want to order the result some other way, you need to wrap it into a subquery and add another ORDER BY to the outer query:
SELECT *
FROM (
SELECT DISTINCT ON (t.id)
t.id, u.username, p.updated_at, t.title
FROM forum_threads t
LEFT JOIN forum_posts p ON p.thread_id = t.id
LEFT JOIN users u ON u.id = p.user_id
WHERE t.forum_id = 3
ORDER BY t.id, p.updated_at DESC
) sub
ORDER BY updated_at DESC;
If you are looking for a query without subquery for some unknown reason, this should work, too:
SELECT DISTINCT
t.id
, first_value(u.username) OVER w AS username
, first_value(p.updated_at) OVER w AS updated_at
, t.title
FROM forum_threads t
LEFT JOIN forum_posts p ON p.thread_id = t.id
LEFT JOIN users u ON u.id = p.user_id
WHERE t.forum_id = 3
WINDOW w AS (PARTITION BY t.id ORDER BY p.updated_at DESC)
ORDER BY updated_at DESC;
There is quite a bit going on here:
The tables are joined and rows are selected according to JOIN and WHERE clauses.
The two instances of the window function first_value() are run (on the same window definition) to retrieve username and updated_at from the latest post per thread. This results in as many identical rows as there are posts in the thread.
The DISTINCT step is executed after the window functions and reduces each set to a single instance.
ORDER BY is applied last and updated_at references the OUT column (SELECT list), not one of the two IN columns (FROM list) of the same name.
Yet another variant, a subquery with the window function row_number():
SELECT id, username, updated_at, title
FROM (
SELECT t.id
, u.username
, p.updated_at
, t.title
, row_number() OVER (PARTITION BY t.id
ORDER BY p.updated_at DESC) AS rn
FROM forum_threads t
LEFT JOIN forum_posts p ON p.thread_id = t.id
LEFT JOIN users u ON u.id = p.user_id
WHERE t.forum_id = 3
) sub
WHERE rn = 1
ORDER BY updated_at DESC;
Similar case:
Return records distinct on one column but order by another column
You'll have to test which is faster. Depends on a couple of circumstances.
Forget the distinct on:
SELECT t.id, u.username, p.updated_at, t.title
FROM forum_threads t
LEFT JOIN forum_posts p ON p.thread_id = t.id
LEFT JOIN users u ON u.id = p.user_id
WHERE t.forum_id = 3
ORDER BY p.updated_at DESC;
Related
I have a query that selects users with the districts which they visited and visits count.
select users.id, places.district, count(users.id) as counts from users
left join visits on users.id = visits.user_id
inner join places on visits.place_id = places.id
group by users.id, places.district
I need to select only those users who have visited provided district the most. For example, I have a user with id 1 who visited district A one time and district B three times. If I provide district B as parameter, user 1 will be in select. If I want to select users from district A, user 1 will not be in select.
I think that's ranking, then filtering:
select *
from (
select u.id, p.district, count(*) as cnt_visits,
rank() over(partition by u.id order by count(*) desc)
from users u
inner join visits v on u.id = v.user_id
inner join places p on p.id = v.place_id
group by u.id, p.district
) t
where rn = 1 and district = ?
Note that you don't actually need table users to get this result. We could simplify the query as:
select *
from (
select v.user_id, p.district, count(*) as cnt_visits,
rank() over(partition by u.id order by count(*) desc)
from visits v
inner join places p on p.id = v.place_id
group by v.user_id, p.district
) t
where rn = 1 and district = ?
This query handles top ties: if a user had the same, maximum number of visits in two different districts, both are taken into account. If you don't need that feature, then we can simplify the subquery with distinct on:
select *
from (
select distinct on (v.user_id) v.user_id, p.district, count(*) as cnt_visits
from visits v
inner join places p on p.id = v.place_id
group by v.user_id, p.district
order by v.user_id, cnt_visits desc
) t
where district = ?
I am working on a small forum component for a site and I am creating a page where I want to display each topic along with its highest rated answer. Here are what the tables look like:
POST USER TOPIC
id id id
date name title
text bio date
views
likes
topic_id
author_id
My query looks like so:
select
u.id, u.name, u.bio,
p.id, p.date, p.text, p.views, p.likes,
t.id, t.title, t.date
from
( select p.id, max(p.likes) as likes, p.topic_id
from post as p group by p.topic_id ) as q
inner join post as p on q.id = p.id
inner join topic as t on t.id = q.topic_id
inner join user as u on u.id = p.author_id
order by date desc;
One of the problems I'm having running this is withing "q". Postgresql wont let me run the "q" query because it wants "p.id" to be in the "group by" clause or in an aggregate function. I tried to use "distinct on (p.id)" but I got the same error message: p.id must appear in the GROUP BY clause or be used in an aggregate function.
Without the p.id attribute, I cannot meaningfully link it to the other tables; is there another way of accomplishing this?
;WITH cte AS (
SELECT
u.id AS UserId
,u.name
,u.bio
,p.id AS PostId
,p.[date] AS PostDate
,p.text
,p.views
,p.Likes
,t.id AS TopidId
,t.title
,t.[date] AS TopicDate
,p.Likes
,ROW_NUMBER() OVER (PARTITION BY t.id ORDER BY p.Likes DESC, p.[date] DESC) AS RowNum
,DENSE_RANK() OVER (PARTITION BY t.id ORDER BY p.Likes DESC) AS RankNum
FROM
topic t
INNER JOIN post p
ON t.id = p.topic_id
INNER JOIN [user] u
ON p.author_id = u.id
)
SELECT *
FROM
cte
WHERE
RowNum = 1
;
switch RowNum to RankNum if you want to see ties for most liked
This is a common need: when grouping, show each group's first/last a ranked by some other criteria b. I don't have a name for it, but this seems to be the canonical question. You can see there are a lot of choices! My favorite solution is probably a lateral join:
SELECT u.id, u.name, u.bio,
p.id, p.date, p.text, p.views, p.likes,
t.id, t.title, t.date
FROM topic t
LEFT OUTER JOIN LATERAL (
SELECT *
FROM post
WHERE post.topic_id = t.id
ORDER BY post.likes DESC
LIMIT 1
) p
ON true
LEFT OUTER JOIN "user" u
ON p.author_id = u.id
;
SELECT
u.id AS uid, u.name, u.bio
, p.id AS pid, p."date" AS pdate, p.text, p.views, p.likes
, t.id AS tid, t.title, t."date" AS tdate
FROM post p
JOIN topic t ON t.id = p.topic_id
JOIN user u ON u.id = p.author_id
WHERE NOT EXISTS ( SELECT *
FROM post nx
WHERE nx.topic_id = p.topic_id
AND nx.likes > p.likes)
ORDER BY p."date" DESC
;
I need to construct a join that will give me the most recent price for each product. I vastly simplified the table structures for the purpose of the example, and each table row counts will be in the millions. My previous stabs at this have not exactly been very effecient.
In PostgreSQL, you could try DISTINCT ON to only get the first row per product id in descending create_date order;
SELECT DISTINCT ON (products.id) products.*, prices.*
FROM products
JOIN prices
ON products.id = prices.product_id
ORDER BY products.id, create_date DESC
(of course, except for illustrative purposes, you should of course select the exact columns you need)
The simplest way to do it is using the row_number function.
SELECT
p.name,
t.amount AS latest_price
FROM (
SELECT
p.*,
row_number() OVER (PARTITION BY product_id ORDER BY create_date DESC) AS rn
FROM
prices p) t
JOIN products p ON p.id = t.product_id
WHERE
rn = 1
While the DISTINCT ON answer worked for my instance, I found there's a faster way for me to get what I need.
SELECT
DISTINCT ON(u.id) u.id,
(CAST(data AS JSON) ->> 'Finished') AS Finished,
ee.post_value
FROM
users_user u
JOIN events_event ee on u.id = ee.actor_id
WHERE
u.id > 20000
ORDER BY
u.id DESC,
ee.time DESC;
takes ~25s on my DB, while
SELECT
u.id,
(CAST(data AS JSON) ->> 'Finished') AS Finished,
e.post_value
FROM
users_user u
JOIN events_event e on u.id = e.actor_id
LEFT JOIN events_event ee on ee.actor_id = e.actor_id
AND ee.time > e.time
WHERE
u.id > 20000
AND ee.id IS NULL
ORDER BY
u.id DESC;
takes ~15s.
Calling all sql enthusiasts!
Quick info: using PostgreSQL.
I have a query that return the maximum number of likes for a user per category. What I want now, is to show the top 3 users with the most likes per category.
A helpful resource was using this example to solve the problem:
select type, variety, price
from fruits
where (
select count(*) from fruits as f
where f.type = fruits.type and f.price <= fruits.price
) <= 2;
I understand this, but my query is using joins and I am also a beginner, so I was not able to use this information effectively.
Down to business, this is my query for returning the MAX likes for a user per category.
SELECT category, username, MAX(post_likes) FROM (
SELECT c.name category, u.username username, SUM(p.like_count) post_likes, COUNT(*) post_num
FROM categories c
JOIN topics t ON c.id = t.category_id
JOIN posts p ON t.id = p.topic_id
JOIN users u ON u.id = p.user_id
GROUP BY c.name, u.username) AS leaders
WHERE post_likes > 0
GROUP BY category, username
HAVING MAX(post_likes) >= (SELECT SUM(p.like_count)
FROM categories c
JOIN topics t ON c.id = t.category_id
JOIN posts p ON t.id = p.topic_id
JOIN users u ON u.id = p.user_id WHERE c.name = leaders.category
GROUP BY u.username order by sum desc limit 1)
ORDER BY MAX(post_likes) DESC;
Any and all help would be greatly appreciated. I am having a difficult time wrapping my head around this problem. Thank!
If you want the most likes per category, use window functions:
SELECT cu.*
FROM (SELECT c.name as category, u.username as username,
SUM(p.like_count) as post_likes, COUNT(*) as post_num,
ROW_NUMBER() OVER (PARTITION BY c.name ORDER BY COUNT(*) DESC) as seqnum
FROM categories c JOIN
topics t
ON c.id = t.category_id JOIN
posts p
ON t.id = p.topic_id JOIN
users u
ON u.id = p.user_id
GROUP BY c.name, u.username
) cu
WHERE seqnum <= 3;
This always returns three rows per category, even if there are ties. If you want to do something else, then consider DENSE_RANK() or RANK() instead of ROW_NUMBER().
Also, use as for column aliases in the FROM clause. Although optional, one day you will leave out a comma and be grateful that you are in the habit of using as.
I need to find out the user who has posted the most number of comments. There are two tables 1)users(Id, DisplayName) 2)comments(Id, UserId, test) . I have used the following query
Select DisplayName from users INNER JOIN (Select UserId, max(comment_count) as `max_comments from (Select UserId, count(Id) as comment_count from comments group by UserId) as` T1) as T2 ON users.Id=T2.UserId
However, this returns to me the Display Name of the user with Id = 1 rather than what I want. How do I work around this ?
SELECT TOP 1
U.DisplayName,
COUNT(C.ID) AS CommentCount
FROM
Users AS U
INNER JOIN Comments AS C ON U.ID = C.UserID
GROUP BY
U.DisplayName
ORDER BY
COUNT(C.ID) DESC