I just started learning Postgres, and I'm trying to make an aggregation table that has the columns:
user_id
booking_sequence
booking_created_time
booking_paid_time
booking_price_amount
total_spent
All columns are provided, except for the booking_sequence column. I need to make a query that shows the first five flights of each user that has at least x purchases and has spent more than a certain amount of money, then sort it by the amount of money spent by the user, and then sort it by the booking sequence column.
I've tried :
select user_id,
row_number() over(partition by user_id order by user_id) as booking_sequence,
booking_created_time as booking_created_date,
booking_price_amount,
sum(booking_price_amount) as total_booking_price_amount
from fact_flight_sales
group by user_id, booking_created_time, booking_price_amount
having count(user_id) > 5
and total_booking_price_amount > 1000
order by total_booking_price_amount;
I got 0 when I added count(user_id) > 5, and total_booking_price_amount is not found when I add the second condition in the HAVING clause.
Edit:
I managed to make the code function correctly, for those who are curious:
select x.user_id, row_number() over(partition by x.user_id)
as booking_sequence, x.booking_created_time::date as booking_created_date, x.booking_price_amount,
sum(y.booking_price_amount) as total_booking_price_amount from
(
select user_id, booking_created_time, booking_price_amount from fact_flight_sales
group by user_id, booking_created_time, booking_price_amount
) as x
join
(
select user_id, booking_price_amount
from fact_flight_sales group by user_id, booking_price_amount
) as y
on x.user_id = y.user_id
group by x.user_id, x.booking_created_time, x.booking_price_amount
having count(x.user_id) >= 1 and sum(y.booking_price_amount) >250000
order by total_booking_price_amount desc, booking_sequence asc;
Big thanks to Laurenz for the help!
About count(user_id) > 5:
HAVING is calculated before window functions are evaluated, So result rows excluded by the HAVING clause will not be used to calculate the window function.
About total_booking_price_amount in HAVING:
You cannot use aliases from the SELECT list in the HAVING clause. You will have to repeat the expression (or use a subquery).
Related
I am writing a query in Redshift to answer the question "Give the average lifetime spend of users who spent more on their first order than their second order." This is based off of an order_items table which has one row for every item ordered (so an order with 3 items would be represented in 3 rows). Here's a snapshot of the first 10 rows:
First 10 rows of order_items:
Here is my solution:
with
cte1_lifetime as (
select oi.user_id, sum(oi.sale_price) as lifetime_spend
from order_items as oi
group by oi.user_id
),
cte2_order as (
select oi.user_id, oi.order_id, sum(oi.sale_price) as order_total, rank() over(partition by oi.user_id order by oi.created_at) as order_rank
from order_items as oi
group by oi.user_id, oi.order_id, oi.created_at
order by oi.user_id, oi.order_id
),
cte3_first_order as (
select user_id, order_id, order_total
from cte2_order
where order_rank=1
order by user_id, order_id
),
cte4_second_order as (
select user_id, order_id, order_total
from cte2_order
where order_rank=2
order by user_id, order_id
)
select avg(cte1.lifetime_spend) as average_lifetime_spend
from cte1_lifetime as cte1
where exists (
select *
from cte3_first_order as cte3, cte4_second_order as cte4
where cte3.user_id=cte4.user_id
and cte1.user_id=cte3.user_id
and cte3.order_total > cte4.order_total)
And here is the answer key:
WITH
table1 AS
(SELECT user_id, order_id,
SUM(sale_price) OVER (PARTITION BY order_id ROWS BETWEEN UNBOUNDED PRECEDING AND UNBOUNDED FOLLOWING) as order_total,
RANK() OVER (PARTITION BY user_id ORDER BY created_at) AS "sequence"
FROM order_items)
,
table2 AS
(SELECT user_id, SUM(sale_price) AS lifetime_spend
FROM order_items
WHERE EXISTS
(SELECT t1.user_id
FROM table1 t1, table1 t2
WHERE t1.user_id = t2.user_id AND t1.sequence = 1 AND t2.sequence = 2 AND t1.order_total>t2.order_total
AND t1.user_id = order_items.user_id)
GROUP BY 1
ORDER BY 1)
SELECT AVG(lifetime_spend)
FROM table2
These answers yield slightly different results on the same data- an average lifetime spend of $215 vs $220. I'd really like to understand why they are different but so far I can't figure it out. Any ideas?
I want to get the number of consecutive days from the current date using Postgres SQL.
enter image description here
Above is the scenario in which I have highlighted consecutive days count should be like this.
Below is the SQL query which I have created but it's not returning the expected result
with grouped_dates as (
select user_id, created_at::timestamp::date,
(created_at::timestamp::date - (row_number() over (partition by user_id order by created_at::timestamp::date) || ' days')::interval)::date as grouping_date
from watch_history
)
select * , dense_rank() over (partition by grouping_date order by created_at::timestamp::date) as in_streak
from grouped_dates where user_id = 702
order by created_at::timestamp::date
Can anyone please help me to resolve this issue?
If anyhow we can able to apply distinct for created_at field to below query then I will get solutions for my issue.
WITH list AS
(
SELECT user_id,
(created_at::timestamp::date - (row_number() over (partition by user_id order by created_at::timestamp::date) || ' days')::interval)::date as next_day
FROM watch_history
)
SELECT user_id, count(*) AS number_of_consecutive_days
FROM list
WHERE next_day IS NOT NULL
GROUP BY user_id
Does anyone have an idea how to apply distinct to created_at for the above mentioned query ?
To get the "number of consecutive days" for the same user_id :
WITH list AS
(
SELECT user_id
, array_agg(created_at) OVER (PARTITION BY user_id ORDER BY created_at RANGE BETWEEN CURRENT ROW AND '1 day' FOLLOWING) AS consecutive_days
FROM watch_history
)
SELECT user_id, count(DISTINCT d.day) AS number_of_consecutive_days
FROM list
CROSS JOIN LATERAL unnest(consecutive_days) AS d(day)
WHERE array_length(consecutive_days, 1) > 1
GROUP BY user_id
To get the list of "consecutive days" for the same user_id :
WITH list AS
(
SELECT user_id
, array_agg(created_at) OVER (PARTITION BY user_id ORDER BY created_at RANGE BETWEEN CURRENT ROW AND '1 day' FOLLOWING) AS consecutive_days
FROM watch_history
)
SELECT user_id
, array_agg(DISTINCT d.day ORDER BY d.day) AS list_of_consecutive_days
FROM list
CROSS JOIN LATERAL unnest(consecutive_days) AS d(day)
WHERE array_length(consecutive_days, 1) > 1
GROUP BY user_id
full example & result in dbfiddle
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
Have a table with 3 columns: ID, Signature, and Datetime, and it's grouped by Signature Having Count(*) > 9.
select * from (
select s.Signature
from #Sigs s
group by s.Signature
having count(*) > 9
) b
join #Sigs o
on o.Signature = b.Signature
order by o.Signature desc, o.DateTime
I now want to select the 1st and 10th records only, per Signature. What determines rank is the Datetime descending. Thus, I would expect every Signature to have 2 rows.
Thanks,
I would go with a couple of common table expressions.
The first will select all records from the table as well as a count of records per signature, and the second one will select from the first where the record count > 9 and add row_number partitioned by signature - and then just select from that where the row_number is either 1 or 10:
With cte1 AS
(
SELECT ID, Signature, Datetime, COUNT(*) OVER(PARTITION BY Signature) As NumberOfRows
FROM #Sigs
), cte2 AS
(
SELECT ID, Signature, Datetime, ROW_NUMBER() OVER(PARTITION BY Signature ORDER BY DateTime DESC) As Rn
FROM cte1
WHERE NumberOfRows > 9
)
SELECT ID, Signature, Datetime
FROM cte2
WHERE Rn IN (1, 10)
ORDER BY Signature desc
Because I don't know what your data looks like, this might need some adjustment.
The simplest way here, since you already know your sort order (DateTime DESC) and partitioning (Signature), is probably to assign row numbers and then select the rows you want.
SELECT *
FROM
(
select o.Signature
,o.DateTime
,ROW_NUMBER() OVER (PARTITION BY o.Signature ORDER BY o.DateTime DESC) [Row]
from (
select s.Signature
from #Sigs s
group by s.Signature
having count(*) > 9
) b
join #Sigs o
on o.Signature = b.Signature
order by o.Signature desc, o.DateTime
)
WHERE [Row] IN (1,10)
Let's say I have an orders table with customer_id, order_total, and order_date columns. I'd like to build a report that shows all customers who haven't placed an order in the last 30 days, with a column for the total amount their last order was.
This gets all of the customers who should be on the report:
select customer, max(order_date), (select order_total from orders o2 where o2.customer = orders.customer order by order_date desc limit 1)
from orders
group by 1
having max(order_date) < NOW() - '30 days'::interval
Is there a better way to do this that doesn't require a subquery but instead uses a window function or other more efficient method in order to access the total amount from the most recent order? The techniques from How to select id with max date group by category in PostgreSQL? are related, but the extra having restriction seems to stop me from using something like DISTINCT ON.
demo:db<>fiddle
Solution with row_number window function (https://www.postgresql.org/docs/current/static/tutorial-window.html)
SELECT
customer, order_date, order_total
FROM (
SELECT
*,
first_value(order_date) OVER w as last_order,
first_value(order_total) OVER w as last_total,
row_number() OVER w as row_count
FROM orders
WINDOW w AS (PARTITION BY customer ORDER BY order_date DESC)
) s
WHERE row_count = 1 AND order_date < CURRENT_DATE - 30
Solution with DISTINCT ON (https://www.postgresql.org/docs/9.5/static/sql-select.html#SQL-DISTINCT):
SELECT
customer, order_date, order_total
FROM (
SELECT DISTINCT ON (customer)
*,
first_value(order_date) OVER w as last_order,
first_value(order_total) OVER w as last_total
FROM orders
WINDOW w AS (PARTITION BY customer ORDER BY order_date DESC)
ORDER BY customer, order_date DESC
) s
WHERE order_date < CURRENT_DATE - 30
Explanation:
In both solutions I am working with the first_value window function. The window function's frame is defined by customers. The rows within the customers' groups are ordered descending by date which gives the latest row first (last_value is not working as expected every time). So it is possible to get the last order_date and the last order_total of this order.
The difference between both solutions is the filtering. I showed both versions because sometimes one of them is significantly faster
The window function style is creating a row count within the frames. Every first row can be filtered later. This is done by adding a row_number window function. The benefit of this solution comes out when you are trying to filter the first two or three data sets. You simply have to change the filter from WHERE row_count = 1 to WHERE row_count = 2
But if you want only one single row per group you just need to ensure that the expected row per group is ordered to be the first row in the group. Then the DISTINCT ON function can delete all following rows. DISTINCT ON (customer) gives the first (ordered) row per customer group.
Try to join table on itself
select o1.customer, max(order_date),
from orders o1
join orders o2 on o1.id=o2.id
group by o1.customer
having max(o1.order_date) < NOW() - '30 days'::interval
Subqueries in select is a bad idea, because DB will execute a query for each row
If you use postgres you can also try to use CTE
https://www.postgresql.org/docs/9.6/static/queries-with.html
WITH t as (
select id, order_total from orders o2 where o2.customer = orders.customer
order by order_date desc limit 1
) select o1.customer, max(order_date),
from orders o1
join t t.id=o2.id
group by o1.customer
having max(order_date) < NOW() - '30 days'::interval