In POSTGRESQL 13, I have a table of ids,dates, prices.
I simply want to have the latest date where the price is greater than 0 per id.
One row per id.
So the optimal output is :
id | the_date | price
1 2013-08-09 0.45
2 2013-08-11 0.34
I have an SQL fiddle at this link :
https://dbfiddle.uk/?rdbms=postgres_13&fiddle=a89bbbc922601be5465ad764fd035161
I have tried an INNER JOIN with the MAX date unsuccessfully.
SELECT DISTINCT ON (id)
id, the_date, price
FROM inventory
WHERE price>0
ORDER BY id ASC, the_date DESC
You can do something like this:
select i.id, i.the_date, i.price
from inventory as i, (
select id, max(the_date) as max_date
from inventory
where price > 0
group by id
) as c where c.id = i.id and i.the_date = c.max_date
Demo in dbfiddle.uk
This might work for you.
SELECT inventory.id, the_date, price
FROM inventory
join (select id,max(the_date) md from inventory where price>0 group by id ) d
on inventory.id=d.id and the_date=d.md
If you want a row for id's with not price you'd use left join.
Related
SELECT partner_id
FROM trip_delivery_sales ts
WHERE ts.route_id='152'
GROUP BY ts.partner_id
From the query we can get the partners id.Using that partner id we want check in trip delicery sales lines table and want to find each customer last two sale product quantity sum. If last two sale have product qty as 2 & 5 want result as partner_id | count as Mn2333 - 7
here fore example i take partner id as 34806. But i want to check all partner_id obtained from last query
SELECT product_qty
FROM trip_delivery_sales_lines td
WHERE td.partner_id='34806'
AND td.route_id='152'
AND td.product_id='432'
ORDER BY td.order_date DESC
LIMIT 2
You can run this query
SELECT td.partner_id,sum(product_qty)
FROM trip_delivery_sales_lines td,
(SELECT partner_id FROM trip_delivery_sales ts WHERE ts.route_id='152') as ts
WHERE td.partner_id=ts.partner_id
AND td.product_id='432'
GROUP BY td.partner_id
ORDER BY td.order_date DESC
LIMIT 2
Or this one
with ts as (SELECT distinct partner_id FROM trip_delivery_sales WHERE route_id='152')
SELECT td.partner_id,sum(product_qty)
FROM trip_delivery_sales_lines td,ts
WHERE td.partner_id=ts.partner_id
AND td.product_id='432'
GROUP BY td.partner_id
ORDER BY td.order_date DESC
LIMIT 2
You might be looking for
SELECT DISTINCT ts.partner_id, ARRAY(
SELECT product_qty
FROM trip_delivery_sales_lines td
WHERE td.partner_id=ts.partner_id
AND td.product_id='432'
ORDER BY td.order_date DESC
LIMIT 2
) AS product_qty_arr
FROM trip_delivery_sales ts
WHERE ts.route_id='152'
or just
SELECT
partner_id,
array_agg(product_qty ORDER BY order_date DESC) as product_qty_arr
FROM (
SELECT
td.partner_id,
td.product_qty,
td.order_date,
row_number() OVER (PARTITION BY td.partner_id ORDER BY td.order_date DESC)
FROM trip_delivery_sales_lines td
JOIN trip_delivery_sales ts USING (partner_id)
WHERE ts.route_id='152'
AND td.product_id='432'
) AS enumerated
WHERE row_number <= 2
GROUP BY partner_id
See also PostgreSQL: top n entries per item in same table or Optimize GROUP BY query to retrieve latest row per user
I have a table with people and another with visits. I want to count all visits but if the person signed up with 'emp' or 'oth' on ref_signup then remove the first visit. Example:
This are my tables:
PEOPLE:
id | ref_signup
---------------------
20 | emp
30 | oth
23 | fri
VISITS
id | date
-------------------------
20 | 10-01-2019
20 | 10-05-2019
23 | 10-09-2019
23 | 10-10-2019
30 | 09-10-2019
30 | 10-07-2019
On this example the visit count should be 4 because persons with id's 20 and 30 have their ref_signup as emp or oth, so it should exclude their first visit, but count from the second and forward.
This is what I have as a query:
SELECT COUNT(*) as visit_count FROM visits
LEFT JOIN people ON people.id = visits.people_id
WHERE visits.group_id = 1
Would using a case on the count help on this case as I just want to remove one visit not all of the visits from the person.
Subtract from COUNT(*) the distinct number of person.ids with person.ref_signup IN ('emp', 'oth'):
SELECT
COUNT(*) -
COUNT(DISTINCT CASE WHEN p.ref_signup IN ('emp', 'oth') THEN p.id END) as visit_count
FROM visits v LEFT JOIN people p
ON p.id = v.id
See the demo.
Result:
| visit_count |
| ----------- |
| 4 |
Note: this code and demo fiddle use the column names of your sample data.
Premise, select the count of visits from each person, along with a synthetic column that contains a 1 if the referral was from emp or oth, a 0 otherwise. Select the sum of the count minus the sum of that column.
SELECT SUM(count) - SUM(ignore_first) FROM (SELECT COUNT(*) as count, CASE WHEN ref_signup in ('emp', 'oth') THEN 1 ELSE 0 END as ignore_first as visit_count FROM visits
LEFT JOIN people ON people.id = visits.people_id
WHERE visits.group_id = 1 GROUP BY id) a
where's "people_id" in your example ?
SELECT COUNT(*) as visit_count
FROM visits v
JOIN people p ON p.id = v.people_id
WHERE p.ref_signup IN ('emp','oth');
then remove the first visit.
You cannot select count and delete the first visit at same time.
DELETE FROM visits
WHERE id IN (
SELECT id
FROM visits v
JOIN people p ON p.id = v.people_id
WHERE p.ref_signup IN ('emp','oth')
ORDER BY v.id
LIMIT 1
);
edit: typos
First, I create the tables
create table people (id int primary key, ref_signup varchar(3));
insert into people (id, ref_signup) values (20, 'emp'), (30, 'oth'), (23, 'fri');
create table visits (people_id int not null, visit_date date not null);
insert into visits (people_id, visit_date) values (20, '10-01-2019'), (20, '10-05-2019'), (23, '10-09-2019'), (23, '10-10-2019'), (30, '09-10-2019'), (30, '10-07-2019');
You can use the row_number() window function to mark which visit is "visit number one":
select
*,
row_number() over (partition by people_id order by visit_date) as visit_num
from people
join visits
on people.id = visits.people_id
Once you have that, you can do another query on those results, and use the filter clause to count up the correct rows that match the condition where visit_num > 1 or ref_signup = 'fri':
-- wrap the first query in a WITH clause
with joined_visits as (
select
*,
row_number() over (partition by people_id order by visit_date) as visit_num
from people
join visits
on people.id = visits.people_id
)
select count(1) filter (where visit_num > 1 or ref_signup = 'fri')
from joined_visits;
-- First get the corrected counts for all users
WITH grouped_visits AS (
SELECT
COUNT(visits.*) -
CASE WHEN people.ref_signup IN ('emp', 'oth') THEN 1 ELSE 0 END
AS visit_count
FROM visits
INNER JOIN people ON (people.id = visits.id)
GROUP BY people.id, people.ref_signup
)
-- Then sum them
SELECT SUM(visit_count)
FROM grouped_visits;
This should give you the result you're looking for.
On a side note, I can't help but think clever use of a window function could do this in a single shot without the CTE.
EDIT: No, it can't since window functions run after needed WHERE and GROUP BY and HAVING clauses.
This question has probably been asked in different formats, but I could not find the answer.
I have table orders
date, quantity_ordered, unit_cost_cents , product_model_number, title
I would like to:
SELECT
model_number,
title,
SUM(unit_cost_cents / 100.00 * quantity_ordered) as total
FROM orders
GROUP BY model_number
HAVING SUM(quantity_submitted) > 0
ORDER BY total DESC
But it requires grouping by the title as well.
My problem being is that my title changes over time. I'd like to preserve the titles and simply display/select the most recent title without grouping by title which would make the numbers different.
You can use a subquery to fetch the latest title:
SELECT
model_number,
(select max(title) from orders where date = (
select max(date) from orders where model_number = o.model_number)
) title,
SUM(unit_cost_cents / 100.00 * quantity_ordered) as total
FROM orders o
GROUP BY model_number
HAVING SUM(quantity_submitted) > 0
ORDER BY total DESC
I used select max(title) instead of select title to make sure that the subquery will not return more than 1 rows (just in case).
SELECT
o.model_number
, om.title
, SUM(o.unit_cost_cents / 100.00 * o.quantity_ordered) as total
FROM orders o
JOIN (SELECT model_number, title
,row_number() OVER (PARTITION BY model_number ORDER BY zdate DESC) AS rn
FROM orders) om
ON om.model_number=o.model_number AND om.rn=1
GROUP BY 1,2
HAVING SUM(o.quantity_submitted) > 0
ORDER BY 3 DESC
;
I need a specific SQL query to select last 10 conversations for user inbox.
Inbox shows only conversations(threads) with every user - it selects the last message from the conversation and shows it in inbox.
Edited.
Expecting result: to extract latest message from each of 10 latest conversations. Facebook shows latest conversations in the same way
And one more question. How to make a pagination to show next 10 latest messages from previous latest conversations in the next page?
Private messages in the database looks like:
| id | user_id | recipient_id | text
| 1 | 2 | 3 | Hi John!
| 2 | 3 | 2 | Hi Tom!
| 3 | 2 | 3 | How are you?
| 4 | 3 | 2 | Thanks, good! You?
As per my understanding, you need to get the latest message of the conversation on per-user basis (of the last 10 latest conversations)
Update: I have modified the query to get the latest_conversation_message_id for every user conversation
The below query gets the details for user_id = 2, you can modify, users.id = 2 to get it for any other user
SQLFiddle, hope this solves your purpose
SELECT
user_id,
users.name,
users2.name as sent_from_or_sent_to,
subquery.text as latest_message_of_conversation
FROM
users
JOIN
(
SELECT
text,
row_number() OVER ( PARTITION BY user_id + recipient_id ORDER BY id DESC) AS row_num,
user_id,
recipient_id,
id
FROM
private_messages
GROUP BY
id,
recipient_id,
user_id,
text
) AS subquery ON ( ( subquery.user_id = users.id OR subquery.recipient_id = users.id) AND row_num = 1 )
JOIN users as users2 ON ( users2.id = CASE WHEN users.id = subquery.user_id THEN subquery.recipient_id ELSE subquery.user_id END )
WHERE
users.id = 2
ORDER BY
subquery.id DESC
LIMIT 10
Info: The query gets the latest message of every conversation with any other user, If user_id 2, sends a message to user_id 3, that too is displayed, as it indicates the start of a conversation. The latest message of every conversation with any other user is displayed
To solve groupwise-max in pg you can use DISTINCT ON. Like this:
SELECT
DISTINCT ON(pm.user_id)
pm.user_id,
pm.text
FROM
private_messages AS pm
WHERE pm.recipient_id= <my user id>
ORDER BY pm.user_id, pm.id DESC;
http://sqlfiddle.com/#!12/4021d/19
To get the latest X however we will have to use it in a subselect:
SELECT
q.user_id,
q.id,
q.text
FROM
(
SELECT
DISTINCT ON(pm.user_id)
pm.user_id,
pm.id,
pm.text
FROM
private_messages AS pm
WHERE pm.recipient_id=2
ORDER BY pm.user_id, pm.id DESC
) AS q
ORDER BY q.id DESC
LIMIT 10;
http://sqlfiddle.com/#!12/4021d/28
To get both sent and recieved threads:
SELECT
q.user_id,
q.recipient_id,
q.id,
q.text
FROM
(
SELECT
DISTINCT ON(pm.user_id,pm.recipient_id)
pm.user_id,
pm.recipient_id,
pm.id,
pm.text
FROM
private_messages AS pm
WHERE pm.recipient_id=2 OR pm.user_id=2
ORDER BY pm.user_id,pm.recipient_id, pm.id DESC
) AS q
ORDER BY q.id DESC
LIMIT 10;
http://sqlfiddle.com/#!12/4021d/42
Paste it after your WHERE clause
ORDER BY "ColumnName" [ASC, DESC]
UNION Description at W3Schools it combines the result of this 2 statements.
SELECT "ColumnName" FROM "TableName"
UNION
SELECT "ColumnName" FROM "TableName"
For large data sets I think you might like to try running the two statements and then consolidating the results, as an index scan on (user_id and id) or (recipient_id and id) ought to be very efficient at getting the 10 most recent conversations of each type.
with sent_messages as (
SELECT *
FROM private_messages
WHERE user_id = my_user_id
ORDER BY id desc
LIMIT 10),
received_messages as ( SELECT *
FROM private_messages
WHERE recipient_id = my_user_id
ORDER BY id desc
LIMIT 10),
all_messages as (
select *
from sent_messages
union all
select *
from received_messages)
select *
from all_messages
order by id desc
limit 10
Edit: Actually another query worth trying might be:
select *
from private_messages
where id in (
select id
from (
SELECT id
FROM private_messages
WHERE user_id = my_user_id
ORDER BY id desc
LIMIT 10
union all
SELECT id
FROM private_messages
WHERE recipient_id = my_user_id
ORDER BY id desc
LIMIT 10) all_ids
order by id desc
limit 10) last_ten_ids
order by id desc
This might be better in 9.2+, where the indexes alone could be used to get the id's, or in cases where the most recent number to retrieve is very large. Still a bit unclear on that though. If in doubt I'd go for the former version.
I have a table with the following data
Bldg Suit SQFT Date
1 1 1,000 9/24/2012
1 1 1,500 12/31/2011
1 2 800 8/31/2012
1 2 500 10/1/2005
I want to write a query that will sum the max date for each suit record, so the desired result would be 1,800, and must be in one cell/row. This will ultimately be part of subquery, I am just not getting what I expect with the queries I have writtren so far.
Thanks in advance.
You can use the following (See SQL Fiddle with Demo):
select sum(t1.sqft) Total
from yourtable t1
inner join
(
select max(dt) mxdt, suit, bldg
from yourtable
group by suit, bldg
) t2
on t1.dt = t2.mxdt
and t1.bldg = t2.bldg
and t1.suit = t2.suit
; With Data As
(
Select Bldg, Suit, SQFT, Row_Number() Over (Partition By Bldg, Suit Order By Date DESC) As RowID
From YourTableNameHere
)
Select Bldg, Sum(SQFT) As TotalSQFT
From Data
Where RowId = 1
Group By Bldg