I need to list columns from customer table, the date from first order and all data from last one, in a 1:N relationship between customer and order tables. I'm using Oracle 10g.
How the best way to do that?
TABLE CUSTOMER
---------------
id NUMBER
name VARCHAR2(200)
subscribe_date DATE
TABLE ORDER
---------------
id NUMBER
id_order NUMBER
purchase_date DATE
purchase_value NUMBER
Here is one way of doing it, using the row_number function, one join, and on aggregation:
select c.*,
min(o.purchase_date) as FirstPurchaseDate,
min(case when seqnum = 1 then o.id_order end) as Last_IdOrder,
min(case when seqnum = 1 then o.purchase_date end) as Last_PurchaseDate,
min(case when seqnum = 1 then o.purchase_value end) as Last_PurchaseValue
from Customer c join
(select o.*,
row_number() over (partition by o.id order by purchase_date desc) as seqnum
from orders o
) o
on c.customer_id = o.order_id
group by c.customer_id, c.name, c.subscribe_date
It's not obvious how to join the customer table to the orders table (order is a reserved word in Oracle so your table can't be named order). If we assume that the id_order in orders joins to the id in customer
SELECT c.id customer_id,
c.name name,
c.subscribe_date,
o.first_purchase_date,
o.id last_order_id,
o.purchase_date last_order_purchase_date,
o.purchase_value last_order_purchase_value
FROM customer c
JOIN (SELECT o.*,
min(o.purchase_date) over (partition by id_order) first_purchase_date,
rank() over (partition by id_order order by purchase_date desc) rnk
FROM orders o) o ON (c.id = o.id_order)
WHERE rnk = 1
I'm confused by your field names, but I'm going to assume that ORDER.id is the id in the CUSTOMER table.
The earliest order date is easy.
select CUSTOMER.*, min(ORDER.purchase_date)
from CUSTOMER
inner join ORDER on CUSTOMER.id = ORDER.id
group by CUSTOMER.*
To get the last order data, join this to the ORDER table again.
select CUSTOMER.*, min(ORD_FIRST.purchase_date), ORD_LAST.*
from CUSTOMER
inner join ORDER ORD_FIRST on CUSTOMER.id = ORD_FIRST.id
inner join ORDER ORD_LAST on CUSTOMER.id = ORD_LAST.id
group by CUSTOMER.*, ORD_LAST.*
having ORD_LAST.purchase_date = max(ORD_FIRST.purchase_date)
Maybe something like this assuming the ID field in the Order table is actually the Customer ID:
SELECT C.*, O1.*, O2.purchase_Date as FirstPurchaseDate
FROM Customer C
LEFT JOIN
(
SELECT Max(purchase_date) as pdate, id
FROM Orders
GROUP BY id
) MaxPurchaseOrder
ON C.Id = MaxPurchaseOrder.Id
LEFT JOIN Orders O1
ON MaxPurchaseOrder.pdate = O1.purchase_date
AND MaxPurchaseOrder.id = O1.id
LEFT JOIN
(
SELECT Min(purchase_date) as pdate, id
FROM Orders
GROUP BY id
) MinPurchaseOrder
ON C.Id = MinPurchaseOrder.Id
LEFT JOIN Orders O2
ON MinPurchaseOrder.pdate = O2.purchase_date
AND MinPurchaseOrder.id = O2.id
And the sql fiddle.
Related
For example, let's say that I have a table of purchases, and I want to return a list of all purchases that were made on the same day as the 1st purchase, grouping by individual customers. I don't believe that I can use min(purchase_date), and group by customer, since that will return just one row. How would I go about doing this?
Here's an example. I believe this would return only 1 row, whereas I want to return all orders that fall on the initial purchase date.
select c.name, min(o.purchase_date)
from customers c
join orders o on c.id = o.customer_id
group by c.name
Qualify can come in handy
select c.name, o.purchase_date
from customers c
join orders o on c.id = o.customer_id
qualify o.purchase_date = min(o.purchase_date) over (partition by c.name)
select c.name, min(o.purchase_date)
from(
select o.customer_id,o.purchase_date,
row_number()
over(partition by customer_id, date order by id) as firstOrder
from orders o
)t
join customers c on c.id = o.customer_id and o.firstOrder = 1
Using QUALIFY and RANK:
SELECT *
FROM customers c
JOIN orders o
ON c.id = o.customer_id
QUALIFY RANK() OVER(PARTITION BY c.name ORDER BY o.purchase_date) = 1
I am using SQLite3 for this following query.
I have a table called "products" that looks like this:
I have a table called "transactions" that looks like this:
I have a table called "segments" that looks like this:
For each active segment, I want to find the category that produces the highest revenue.
I think that I know how to do this in 3 different queries.
create table table1 as
SELECT s.seg_name, p.category, t.item_qty * t.item_price as revenue
from segments s
JOIN
transactions t
on s.cust_id = t.cust_id
JOIN products p
on p.prod_id = t.prod_id
where s.active_flag = 'Y'
order by s.seg_name, p.category
;
create table table2 as
select seg_name, category, sum(revenue) as revenue
from table1
group by seg_name, category;
select seg_name, category, max(revenue) as revenue
from table2
group by seg_name;
How can I do it in 1 query?
here is one way :
select seg_name,category,revenue
from (
select
s.seg_name,
p.category,
sum(t.item_qty * t.item_price) as revenue,
rank() over (partition by seg_name order by sum(t.item_qty * t.item_price) desc) rn
from segments s
join transactions t on s.cust_id = t.cust_id
join products p on p.prod_id = t.prod_id
where s.active_flag = 'Y'
group by seg_name, p.category
) t where rn = 1
I'm asked to find the top user for different countries, however, one of the countries has 2 users with the same amount spent so they should both be the top users, but I can't get the max value for 2 values in this country.
Here is the code:
WITH t1 AS (
SELECT c.customerid,SUM(i.total) tot
FROM invoice i
JOIN customer c ON c.customerid = i.customerid
GROUP BY 1
ORDER BY 2 DESC
),
t2 AS (
SELECT c.customerid as CustomerId ,c.firstname as FirstName,c.lastname as LastName, i.billingcountry as Country,MAX(t1.tot) as TotalSpent
FROM t1
JOIN customer c
ON c.customerid = t1.customerid
JOIN invoice i ON i.customerid = c.customerid
GROUP BY 4
ORDER BY 4
)
SELECT *
FROM t2
BILLINGCOUNTRY is in Invoice, and it has the name of all the countries.
TOTAL is also in invoice and it shows how much is spent for each purchase by Customer (so there are different fees and taxes for each purchase and total shows the final price payed by the user at each time)
Customer has id,name,last name and from its' ID I'm extracting the total of each of his purchases
MAX was used after finding the sum for each Customer and it was GROUPED BY country so that i could find the max for each country, however I can't seem to find the max of the last country that had 2 max values
Use rank() or dense_rank():
SELECT c.*, i.tot
FROM (SELECT i.customerid, i.billingCountry, SUM(i.total) as tot,
RANK() OVER (PARTITION BY i.billingCountry ORDER BY SUM(i.total) DESC) as seqnum
FROM invoice i
GROUP BY 1, 2
) i JOIN
customer c
ON c.customerid = i.customerid
WHERE seqnum = 1;
The subquery finds the amount per customer in each country -- and importantly calculates a ranking for the combination with ties having the same rank. The outer query just brings in the additional customer information that you seem to want.
here is how it worked for me since i was restricted from using many Commands such RIGHT JOIN and RANK() (As what Gordon Linoff suggessted) so i had to create a 3rd case for the anamoly and join it using union. this solution works only on this case, the good solution is the one posted by Gordon Linoff:
WITH t1 AS (
SELECT c.customerid,SUM(i.total) tot
FROM invoice i
JOIN customer c ON c.customerid = i.customerid
GROUP BY 1
ORDER BY 2 DESC
),
t2 AS (
SELECT c.customerid as CustomerId ,c.firstname as FirstName,c.lastname as LastName, i.billingcountry as Country,MAX(t1.tot) as TotalSpent
FROM t1
JOIN customer c
ON c.customerid = t1.customerid
JOIN invoice i ON i.customerid = c.customerid
GROUP BY 4
ORDER BY 4
) ,
t3 AS (
SELECT DISTINCT c.customerid as CustomerId ,c.firstname as FirstName,c.lastname as LastName, i.billingcountry as Country,t1.tot as TotalSpent
FROM t1
JOIN customer c
ON c.customerid = t1.customerid
JOIN invoice i ON i.customerid = c.customerid
WHERE i.billingcountry = 'United Kingdom'
ORDER BY t1.tot DESC
LIMIT 2
)
SELECT *
FROM t2
UNION
SELECT * FROM t3
ORDER BY t2.country
I have two tables
Order with columns:
OrderID,OrderDate,CID,EmployeeID
And OrderItem with columns:
OrderID,ItemID,Quantity,SalePrice
I need to return the CustomerID(CID), number of orders per customer, and each customers total amount for all orders.
So far I have two separate queries. One gives me the count of customer orders....
SELECT CID, Count(Order.OrderID) AS TotalOrders
FROM [Order]
Where CID = CID
GROUP BY CID
Order BY Count(Order.OrderID) DESC;
And the other gives me the total sales. I'm having trouble combining them...
SELECT CID, Sum(OrderItem.Quantity*OrderItem.SalePrice) AS TotalDollarAmount
FROM OrderItem, [Order]
WHERE OrderItem.OrderID = [Order].OrderID
GROUP BY CID
I'm doing this in Access 2010.
You would use COUNT(DISTINCT ...) in other SQL engines:
SELECT CID,
Count(DISTINCT O.OrderID) AS TotalOrders,
Sum(OI.Quantity*OI.SalePrice) AS TotalDollarAmount
FROM [Order] O
INNER JOIN [OrderItem] OI
ON O.OrderID = OI.OrderID
GROUP BY CID
Order BY Count(DISTINCT O.OrderID) DESC
Which Access unfortunately does not support. Instead you can first get the Order dollar amounts and then join them before figuring the order counts:
SELECT CID,
COUNT(Orders.OrderID) AS TotalOrders,
SUM(OrderAmounts.DollarAmount) AS TotalDollarAmount
FROM [Orders]
INNER JOIN (SELECT OrderID, Sum(Quantity*SalePrice) AS DollarAmount
FROM OrderItems GROUP BY OrderID) AS OrderAmounts
ON Orders.OrderID = OrderAmounts.OrderID
GROUP BY CID
ORDER BY Count(Orders.OrderID) DESC
If you need to include Customers that have orders with no items (unusual but possible), change INNER JOIN to LEFT OUTER JOIN.
Create a query which uses your 2 existing queries as subqueriers, and join the 2 subqueries on CID. Define your ORDER BY in the parent query instead of in a subquery.
SELECT
sub1.CID,
sub1.TotalOrders,
sub2.TotalDollarAmount
FROM
(
SELECT
CID,
Count(Order.OrderID) AS TotalOrders
FROM [Order]
GROUP BY CID
) AS sub1
INNER JOIN
(
SELECT
CID,
Sum(OrderItem.Quantity*OrderItem.SalePrice)
AS TotalDollarAmount
FROM OrderItem INNER JOIN [Order]
ON OrderItem.OrderID = [Order].OrderID
GROUP BY CID
) AS sub2
ON sub1.CID = sub2.CID
ORDER BY sub1.TotalOrders DESC;
If I turn this sub-query which selects sales persons and their highest price paid for any item they sell:
select *,
(select top 1 highestProductPrice
from orders o
where o.salespersonid = s.id
order by highestProductPrice desc ) as highestProductPrice
from salespersons s
in to this join in order to improve efficiency:
select *, highestProductPrice
from salespersons s join (
select salespersonid, highestProductPrice, row_number(
partition by salespersonid
order by salespersonid, highestProductPrice) as rank
from orders ) o on s.id = o.salespersonid
It still touches every order record (it enumerates the entire table before filtering by salespersonid it seems.) However you cannot do this:
select *, highestProductPrice
from salespersons s join (
select salespersonid, highestProductPrice, row_number(
partition by salespersonid
order by salespersonid, highestProductPrice) as rank
from orders
where orders.salepersonid = s.id) o on s.id = o.salespersonid
The where clause in the join causes a `multi-part identifier "s.id" could not be bound.
Is there any way to join the top 1 out of each order group with a join but without touching each record in orders?
Try
SELECT
S.*,
T.HighestProductPrice
FROM
SalesPersons S
CROSS APPLY
(
SELECT TOP 1 O.HighestProductPrice
FROM Orders O
WHERE O.SalesPersonid = S.Id
ORDER BY O.SalesPersonid, O.HighestProductPrice DESC
) T
would
select s.*, max(highestProductPrice)
from salespersons s
join orders o on o.salespersonid = s.id
group by s.*
or
select s.*, highestProductPrice
from salespersons s join (select salepersonid,
max(highestProductPrice) as highestProductPrice
from orders o) as o on o.salespersonid = s.id
work?