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.
Related
From this table, I'm trying to determine the nation (s) that have the highest number of teams (a nation X has a team if it has at least one athlete from that country X).
driver(id,name, team, country)
This solution restores all countries in descending order. Would it be possible to ensure that only the one (s) with the most team (s) return and not all of them? I think you should use the 'max' command but I'm not sure.
SELECT (country) ,count(distinct team)
FROM driver
GROUP BY country
order by count(distinct team) DESC;
I would use your query as a CTE and then select from it like this -
WITH t AS
(
SELECT country, count(distinct team) cnt
FROM driver
GROUP BY country
)
SELECT country, cnt FROM t
WHERE cnt = (SELECT max(cnt) FROM t);
You can combine this with a window function:
with counts as (
SELECT country,
count(distinct team) as num_teams,
dense_rank() over (order by count(distinct team) desc) as rnk
FROM driver
GROUP BY country
)
select country, num_teams
from counts
where rnk = 1;
If you are using Postgres 14, you can use fetch first with the option with ties:
SELECT country,
count(distinct team) as num_teams
FROM driver
GROUP BY country
order by count(distinct team) desc
fetch first 1 rows with ties
If two countries have the same highest number of drivers, this would return both. Without the with ties option (which was introduced in Postgres 14) only one of them would be returned.
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
I am trying to calculate a percentile using the percentile_cont() function in PostgreSQL using common table expressions. The goal is find the top 1% of accounts regards to their balances (called amount here). My logic is to find the 99th percentile which will return those whose account balances are greater than 99% of their peers (and thus finding the 1 percenters)
Here is my query
--ranking subquery works fine
with ranking as(
select a.lname,sum(c.amount) as networth from customer a
inner join
account b on a.customerid=b.customerid
inner join
transaction c on b.accountid=c.accountid
group by a.lname order by sum(c.amount)
)
select lname, networth, percentile_cont(0.99) within group
order by networth over (partition by lname) from ranking ;
I keeping getting the following error.
ERROR: syntax error at or near "order"
LINE 2: ...ame, networth, percentile_cont(0.99) within group order by n..
I am thinking that perhaps I forgot a closing brace etc. but I can't seem to figure out where. I know it could be something with the order keyword but I am not sure what to do. Can you please help me to fix this error?
This tripped me up, too.
It turns out percentile_cont is not supported in postgres 9.3, only in 9.4+.
https://www.postgresql.org/docs/9.4/static/release-9-4.html
So you have to use something like this:
with ordered_purchases as (
select
price,
row_number() over (order by price) as row_id,
(select count(1) from purchases) as ct
from purchases
)
select avg(price) as median
from ordered_purchases
where row_id between ct/2.0 and ct/2.0 + 1
That query care of https://www.periscopedata.com/blog/medians-in-sql (section: "Median on Postgres")
You are missing the brackets in the within group (order by x) part.
Try this:
with ranking
as (
select a.lname,
sum(c.amount) as networth
from customer a
inner join account b on a.customerid = b.customerid
inner join transaction c on b.accountid = c.accountid
group by a.lname
order by networth
)
select lname,
networth,
percentile_cont(0.99) within group (
order by networth
) over (partition by lname)
from ranking;
I want to point out that you don't need a subquery for this:
select c.lname, sum(t.amount) as networth,
percentile_cont(0.99) within group (order by sum(t.amount)) over (partition by lname)
from customer c inner join
account a
on c.customerid = a.customerid inner join
transaction t
on a.accountid = t.accountid
group by c.lname
order by networth;
Also, when using table aliases (which should be always), table abbreviations are much easier to follow than arbitrary letters.
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.
I want to select rows for all employess without repeating the data in one column.
For example I have two rows where salary (before raise) is displayed, how can I display only the largest figure without duplication.
You can use Row_Number function
Here is a sample code
select * from (
select *,
row_number() over (partition by empid, name, department order by salary desc) as rn
from employee
) employee where rn = 1
You can find Row_Number() with Partition By clause sample at http://www.kodyaz.com
If I'm understanding the question correctly, then a simple MAX function and GROUP BY would work.
SELECT EmployeeId, OtherColumns, MAX(Salary)
FROM tblEmployees
GROUP BY EmployeeId, OtherColumns