I have a denormalized table with the columns:
buyer_id
order_id
item_id
item_price
item_category
I would like to return something that returns 1 row per buyer_id
buyer_id, sum(item_price), item_category
-- but ONLY for the category with the highest rank of sales along that specific buyer_id.
I can't get row_number() or partition to work because I need to order by the sum of item_price relative to item_category relative to buyer. Am I overlooking anything obvious?
You need a few layers of fudging here:
SELECT buyer_id, item_sum, item_category
FROM (
SELECT buyer_id,
rank() OVER (PARTITION BY buyer_id ORDER BY item_sum DESC) AS rnk,
item_sum, item_category
FROM (
SELECT buyer_id, sum(item_price) AS item_sum, item_category
FROM my_table
GROUP BY 1, 3) AS sub2) AS sub
WHERE rnk = 1;
In sub2 you calculate the sum of 'item_price' for each 'item_category' for each 'buyer_id'. In sub you rank these with a window function by 'buyer_id', ordering by 'item_sum' in descending order (so the highest 'item_sum' comes first). In the main query you select those rows where rnk = 1.
Related
Im trying to find the distributor with the highest and lowest quantity for each country
in two columns distributor with minimum quantity and maximum quantity
I have been able to get the information from other posts but it is in a column however I want it on a row per country
See http://sqlfiddle.com/#!17/448f6/2
Desired result
"country" "min_qty_name" "max_qty_name"
1. "Madagascar" "Leonard Cardenas" "Gwendolyn Mccarty"
2. "Malaysia" "Arsenio Knowles" "Yael Carter"
3. "Palau" "Brittany Burris" "Clark Weaver"
4. "Tanzania" "Levi Douglas" "Levi Douglas"
You can use subqueries:
select distinct country,
(select distributor_name
from product
where country = p.country
order by quantity limit 1) as min_qty_name,
(select distributor_name
from product
where country = p.country
order by quantity desc limit 1) as max_qty_name
from product p;
Fiddle
You can do it with cte too (result here)
WITH max_table AS
(
SELECT ROW_NUMBER() OVER (partition by country order by country,quantity DESC) AS rank,
country, quantity,distributor_name
FROM
product
),
min_table AS
(
SELECT ROW_NUMBER() OVER (partition by country order by country,quantity) AS rank,
country, quantity,distributor_name
FROM
product
)
SELECT m1.country,m2.distributor_name,m1.distributor_name
from max_table m1, min_table m2
where m1.country = m2.country
and m1.rank = 1 and m2.rank = 1
You can do this with a single sort and pass through the data as follows:
with min_max as (
select distinct country,
first_value(distributor_name) over w as min_qty_name,
last_value(distributor_name) over w as max_qty_name
from product
window w as (partition by country
order by quantity
rows between unbounded preceding
and unbounded following)
)
select *
from min_max
order by min_max;
Updated Fiddle
Hi guys I have two tables dbo.Sales (customer_id, order_date, product_id) and dbo.Menu (Product_id, product_name, price). The question is
What was the first item from the menu purchased by each customer?
My solution is
select A.customer_id,m.product_id, m.product_name
from dbo.menu m
cross apply
(select top 1 * from dbo.sales s
where s.product_id=m.product_id
group by s.customer_id,s.order_date, s.product_id
order by s.order_date) A
customer_id product_id product_name
A 1 sushi
A 2 curry
C 3 ramen
Missing customer is B. Instead of B it gives me the second first order by A.
I need for each customer
Murat
You could use a ROW_NUMBER() window function to get the earliest product_id per customer and then join to the Menu table to get your product details.
Edit: Updated ORDER to ASC.
;with cte
as (
select customer_id, product_id, row_number() over (partition by customer_id order by order_date acs) RN
from dbo.Sales)
select c.customer_id, c.product_id, m.product_name
from cte c
join dbo.menu m on c.product_id=m.product_id
where RN = 1
SELECT distinct s.customer_id,
FIRST_VALUE(m.product_name) OVER (partition by s.customer_id order by order_date )
as FirstItem_Customer
FROM [dbo].[sales] S
join [dbo].[menu] M on M.product_id=s.product_id
I have two tables. One is Transactions and the other is Tickets. In Tickets I have the Ticket_Number,the name of the Category(Theater,Cinema,Concert), the Price of the Ticket. In Transactions I also have the Ticket_Number. What i want to do is to Get a SUM of money for each Category, and then with that data I want to Select the Category with the most money.
I already managed to get the SUM for each category but I am stuck here
SELECT category, SUM (Tickets.Price) AS Price
FROM Tickets,Transactions
WHERE Tickets.ticket_num=Transactions.ticket_num
GROUP BY Category
ORDER BY Price DESC;
I know i can add LIMIT 1 but I know it's not correct because 2 or more values can be the same
Using ROW_NUMBER to generate a sequence based on the sum of the price. Then, restrict to only the matching aggregated row with the highest total price.
WITH cte AS (
SELECT category, SUM(t1.Price) AS Price,
ROW_NUMBER() OVER (ORDER BY SUM(t1.Price) DESC) rn
FROM Tickets t1
INNER JOIN Transactions t2
ON t1.ticket_num = t2.ticket_num
GROUP BY Category
)
SELECT category, Price
FROM cte
WHERE rn = 1
ORDER BY Price DESC;
Note that if you want to capture all categories tied for the highest price, should a tie occur, then replace ROW_NUMBER in the above CTE with RANK, keeping everything else the same.
What you are looking for is a window function DENSE_RANK() which will handle ties properly.
RANK() will also work for your case, but if you would like to extend it to get TOP N places with ties (where N > 1), dense rank is the way to go.
SELECT Category, Price
FROM (
SELECT
Category,
SUM(ti.Price) AS Price,
DENSE_RANK() OVER (ORDER BY SUM(ti.Price) DESC) AS rnk
FROM Tickets ti
INNER JOIN Transactions tr ON
ti.ticket_num = tr.ticket_num
GROUP BY Category
) t
WHERE rnk = 1
I've also replaced the old style and not recommended joining of tables as comma separated list in FROM clause to a proper INNER JOIN clause and assigned aliases to tables.
You can use rank() to rank the sums of the prices, more expensive first.
SELECT category,
price
FROM (SELECT category,
sum(tickets.price) price,
rank() OVER (ORDER BY sum(tickets.price) DESC) r
FROM tickets
INNER JOIN transactions
ON transactions.ticket_num = tickets.ticket_num
GROUP BY category) x
WHERE r = 1;
I also took the liberty to rewrite your join from the ancient comma style to a modern, clearer version.
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