SELECT distinct on (prices.item_id) *
FROM prices
ORDER BY prices.item_id, prices.updated_at DESC
The above query retrieves the most recent prices, how would I get the total sum of all the current prices?
Is it possible without using a subselect?
This is trivial using a subquery:
select sum(p.price)
from (select distinct on (p.item_id) p.*
from prices p
order by p.item_id, p.updated_at desc
) p
If you don't mind repeated rows, I think the following might work:
select distinct on (p.item_id) sum(prices.price) over ()
from prices p
order by p.item_id, p.updated_at desc
You might be able to add a limit clause to this to get what you want. By the way, I would write this as:
select sum(p.price)
from (select p.*,
row_number() over (partition by p.item_id order by updated_at desc) as seqnum
from prices p
order by p.item_id, p.updated_at desc
) p
where seqnum = 1
ROW_NUMBER() is standard SQL. The DISTINCT ON clause is specific to Postgres.
Related
I am using MySQL database. So, there are two columns I am working on, CustomerId, and OrderDate. I want to find a second-order date (2nd minimum order date) for each customer.
If you are using MySQL 8+, then ROW_NUMBER can be used here:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY CustomerId ORDER BY OrderDate) rn
FROM yourTable
)
SELECT CustomerId, OrderDate
FROM cte
WHERE rn = 2;
I would recommend using dense_rank as it can give you correct result even if there is duplicate order_date as follows:
SELECT * FROM
(SELECT t.*, DENSE_RANK() OVER (PARTITION BY CustomerId ORDER BY OrderDate) dr
FROM yourTable t
) t where dr = 2;
You can use corelated sub-query as follows if your MySQL version do not support analytical functions as follows:
SELECT T.*
FROM YOURTABLE T
WHERE 1 = (
SELECT COUNT(DISTINCT ORDER_DATE)
FROM YOURTABLE TT
WHERE TT.ORDER_DATE > T.ORDER_DATE
)
I would use a subquery like this:
select o.*
from orders o
where o.order_date = (select o2.order_date
from orders o2
where o2.customer_id = o.customer_id
order by o2.order_date
limit 1 offset 1
);
The subquery is a correlated subquery that returns the second date. If you want the second date with other columns, it can be moved to the select.
With an index on (customer_id, order_date), this is likely to be the fastest solution.
This assumes that there is one row per date (or that if there are multiple rows, "second" can be the earliest date). If you want the second distinct date then use select distinct int he subquery -- however select distinct and group by would incur additional overhead.
I have a products table that contains date-of-create, price, rate,....... and multiple columns I need one select statement that returns the 10 newest records, 10 lowest prices and the top rated products as separated result of each other something like
(select top 10 from products Order By Date ASC ) as newest_list
(select top 10 from products Order By price DESC ) as price_list
(select top 10 from products Order By Rate ASC ) as rateList_list
where ( newest_list price_list rateList_list) are different table result
Which better way to approach that result. Thanks.
You basically want union all . . . and a subquery:
select *
from ((select top 10 'newest' as which, p.*
from products p
Order By Date ASC
) union all
(select top 10 'priciest', p.*
from products p
Order By price DESC
) union all
(select top 10 'ratiest', p.*
from products p
Order By Rate ASC
)
) p
If you don't want duplicates, you can use union or window functions:
select p.*
from (select p.*,
row_number() over (order by date desc) as seqnum_d,
row_number() over (order by price desc) as seqnum_p,
row_number() over (order by rate asc) as seqnum_r
from p
) p
where seqnum_d <= 10 or seqnum_p <= 10 or seqnum_r <= 10;
The results are below. I need to get the records (seller and purchaser) with the max count- grouped by purchaser (marked with yellow)
You can use window functions:
with q as (
<your query here>
)
select q.*
from (select q.*,
row_number() over (order by seller desc) as seqnum_s,
row_number() over (order by purchaser desc) as seqnum_p
from q
) q
where seqnum_s = 1 or seqnum_p = 1;
Try this:
SELECT COUNT,seller,purchaser FROM YourTable ORDER BY seller,purchaser DESC
SELECT T2.MaxCount,T2.purchaser,T1.Seller FROM <Yourtable> T1
Inner JOIN
(
Select Max(Count) as MaxCount, purchaser
FROM <Yourtable>
GROUP BY Purchaser
)T2
On T2.Purchaser=T1.Purchaser AND T2.MaxCount=T1.Count
First you select the Seller from which will give you a list of all 5 sellers. Then you write another query where you select only the Purchaser and the Max(count) grouped by Purchaser which will give you the two yellow-marked lines. Join the two queries on fields Purchaser and Max(Count) and add the columns from the joined table to your first query.
I can't think of a faster way but this works pretty fast even with rather large queries. You can further-by order the fields as needed.
Is there a way to change "LIMIT 1" and get the same output? I have to get client's name, surname and a quantity of books that has the most books
SELECT stud.skaitytojas.name, stud.skaitytojas.surname,
COUNT (stud.skaitytojas.nr) AS quantity
FROM stud.egzempliorius , stud.skaitytojas
WHERE stud.egzempliorius.client = stud.skaitytojas.nr
GROUP BY stud.skaitytojas.nr
ORDER BY quantity DESC
LIMIT 1
Postgres supports the ANSI standard FETCH FIRST 1 ROW ONLY, so you can do:
SELECT s.name, s.surname, COUNT(s.nr) AS quantity
FROM stud.egzempliorius e JOIN
stud.skaitytojas s
ON e.client = s.nr
GROUP BY s.name, s.surname
ORDER BY quantity DESC
FETCH FIRST 1 ROW ONLY;
Also notice the use of table aliases and proper JOIN syntax. I also prefer to list the columns in the SELECT in the GROUP BY, although that is optional if s.nr is unique.
You can select the row with the highest quantity using row_number()
SELECT * FROM (
SELECT * , row_number() over (order by quantity desc) rn FROM (
SELECT stud.skaitytojas.name, stud.skaitytojas.surname, COUNT (stud.skaitytojas.nr) AS quantity
FROM stud.egzempliorius , stud.skaitytojas
WHERE stud.egzempliorius.client = stud.skaitytojas.nr
GROUP BY stud.skaitytojas.name, stud.skaitytojas.surname
) t
) t where rn = 1
If you want to include ties for the highest quantity, then use rank() instead.
I've been using SQL for a few years, and this type of problem comes up here and there, and I haven't found an answer. But perhaps I've been looking in the wrong places - I'm not really sure what to call it.
For the sake of brevity, let's say I have a table with 3 columns: Customer, Order_Amount, Order_Date. Each customer may have multiple orders, with one row for each order with the amount and date.
My Question: Is there a simple way in SQL to get the DATE of the maximum order per customer?
I can get the amount of the maximum order for each customer (and which customer made it) by doing something like:
SELECT Customer, MAX(Order_Amount) FROM orders GROUP BY Customer;
But I also want to get the date of the max order, which I haven't figured out a way to easily get. I would have thought that this would be a common type of question for a database, and would therefore be easy to do in SQL, but I haven't found an easy way to do it yet. Once I add Order_Date to the list of columns to select, I need to add it to the Group By clause, which I don't think will give me what I want.
Apart from self-join you can do:
SELECT o1.*
FROM orders o1 JOIN orders o2 ON o1.Customer = o2.Customer
GROUP BY o1.Customer, o1.Order_Amount
HAVING o1.Order_Amount = MAX(o2.Order_Amount);
There's a good article reviewing various approaches.
And in Oracle, db2, Sybase, SQL Server 2005+ you would use RANK() OVER.
SELECT * FROM (
SELECT *
RANK() OVER (PARTITION BY Customer ORDER BY Order_Amount DESC) r
FROM orders) o
WHERE r = 1;
Note: If Customer has more than one order with maximum Order_Amount (i.e. ties), using RANK() function would get you all such orders; to get only first one, replace RANK() with ROW_NUMBER().
There's no short-cut... the easiest way is probably to join to a sub-query:
SELECT
*
FROM
orders JOIN
(
SELECT Customer, MAX(Order_Amount) AS Max_Order_Amount
FROM orders
GROUP BY Customer
) maxOrder
ON maxOrder.Customer = orders.Customer
AND maxOrder.Max_Order_Amount = orders.Order_Amount
you will want to join on the same table...
SELECT Customer, order_date, amt
FROM orders o,
( SELECT Customer, MAX(Order_Amount) amt FROM orders GROUP BY Customer ) o2
WHERE o.customer = o2.customer
AND o.order_amount = o2.amt
;
Another approach for the collection:
WITH tempquery AS
(
SELECT
Customer
,Order_Amount
,Order_Date
,row_number() OVER (PARTITION BY Customer ORDER BY Order_Amount DESC) AS rn
FROM
orders
)
SELECT
Customer
,Order_Amount
,Order_Date
FROM
tempquery
WHERE
rn = 1
If your DB Supports CROSS APPLY you can do this as well, but it doesn't handle ties correctly
SELECT [....]
FROM Customer c
CROSS APPLY
(SELECT TOP 1 [...]
FROM Orders o
WHERE c.customerID = o.CustomerID
ORDER BY o.Order_Amount DESC) o
See this data.SE query
You could try something like this:
SELECT Customer, MAX(Order_Amount), Order_Date
FROM orders O
WHERE ORDER_AMOUNT = (SELECT MAX(ORDER_AMOUNT) FROM orders WHERE CUSTOMER = O.CUSTOMER)
GROUP BY CUSTOMER, Order_Date
with t as
(
select CUSTOMER,Order_Date ,Order_Amount,max(Order_Amount) over (partition
by Customer) as
max_amount from orders
)
select * from t where t.Order_Amount=max_amount