i want to list customer id with highest sum of price of his orders. Please see graph oí orders down.
SQL DEMO
SELECT *
FROM (
SELECT "customerid", SUM("price")
FROM Orders
GROUP BY "customerid"
ORDER BY SUM("price") DESC
) T
WHERE ROWNUM <= 1
You need to group by customer_id to get all prices of each customers.
Then sum these prices filter it with max( sum(price))or get first row by descending order of sum(price).
--for Oracle
select * from (Select c.name,c.id,sum(o.price) from Customer c
inner join order o on o.customer_id=c.id
group by c.name,c.id
order by sum(o.price)desc
)where rownum =1
--For sql server and mysql
Select top 1 c.name,c.id,sum(o.price) from Customer c
inner join order o on o.customer_id=c.id
group by c.name,c.id
order by sum(o.price)desc
Related
Hello I am trying to group multiple customer orders into buckets in SQL, the output should look something like it does below. Do I have to use a case statement to group them?
Table1 looks like:
CustomerID
Order_date
1
somedate
2
somedate
3
somedate
2
somedate
Edit: # of customers meaning if CustomerID 2 had 2 orders he/she would be of the in the bucket of #of orders of 2.
Output should be something like this?
# of Customers
# of Orders
2
1
1
2
My code so far is:
select count(*) CustomerID
FROM Table1
GROUP BY CustomerID;
Use a double aggregation:
SELECT COUNT(*) AS num_customers, cnt AS num_orders
FROM
(
SELECT CustomerID, COUNT(*) AS cnt
FROM Table1
GROUP BY CustomerID
) t
GROUP BY cnt;
The inner subquery finds the number of orders for each customer. The outer query then aggregates by number of orders and finds out the number of customers having each number of orders.
If you want to sort your tables and your users depending on the number of orders they made, this query should work:
SELECT CustomerID, COUNT(CustomerID) as NbOrder
FROM Table1
GROUP BY(NbOrder)
I believe what you want to do is get the count of orders by customer, first, via aggregation. Then get the count of customers by order count from that query.
SELECT count(*) as count_of_customers, count_of_orders
FROM
(
SELECT customerid, count(*) as count_of_orders
FROM your_table
GROUP BY customerid
) sub
GROUP BY count_of_orders
ORDER BY count_of_orders
Let's say there is a table of customers, each customer in this table has an id which is PK (integer).
What is the most efficient way to write a query that returns the 3rd highest id?
And I in here presume it's 3rd, but my intention is for N highest.
I wrote two queries and I wonder which one is better?
select top 1 Customer_Id
from customers c
where c.Customer_Id not in(
select top 2 cc.Customer_Id
from customers cc
order by cc.Customer_Id desc
)
order by c.Customer_Id desc
------------------------------
select top 1 temp.Customer_Id
from (
select top 3 c.Customer_Id
from customers c
order by c.Customer_Id desc) as temp
order by temp.Customer_Id
If someone here has a better approach, I will be glad to hear.
Just use an OFFSET, like I mentioned:
DECLARE #N int = 3; --Demonstrates that the OFFSET can be parametrised
SELECT CustomerID
FROM dbo.Customer C
ORDER BY CustomerID DESC
OFFSET #N-1 ROWS FETCH NEXT 1 ROWS ONLY;
Something like this using the Row_Number function to list them in order, giving you the ability to select the nth number:
DECLARE #TargetRowNum INT = 2;
WITH CustomerList
AS (
SELECT c.Customer_Id
, CustomerRowNum = ROW_NUMBER()OVER(PARTITION BY c.Customer_Id ORDER BY c.Customer_Id DESC)
FROM customers c
)
SELECT *
FROM CustomerList
WHERE CustomerList.CustomerRowNum = #TargetRowNum;
According to #Larnu, I hope this could be a solution:
select temp2.Customer_Id
from(
select temp.Customer_Id, ROW_NUMBER() over(partition by grp order by temp.Customer_Id desc) as rrank
from (
select c.Customer_Id, 1 as grp
from customers c
) as temp
) as temp2
where temp2.rrank = 3
I have query like this.It is working properly on oracle 12.
select * from customers where customerId IN (select custId from Orders where
orderStatus = 'S' and orderCity = 'Amsterdam' and ORDER BY custId DESC FETCH FIRST 10 ROWS ONLY)
But I am using oracle 11.That query is not working on oracle 11.Therefore I changed my query like that
select * from customers where customerId IN (select custId from Orders where
orderStatus = 'S' and orderCity = 'Amsterdam' and ORDER BY custId DESC rownum <= 10)
It gives missing right paranthesis
How can I solve this problem.Do you have any idea?Actually I use a variable instead on 10 number.
Syntax is a bit different - you need an extra subquery so it would need to be more like ...
where customerId IN (select * from (select custId from Orders where orderStatus = 'S' and orderCity = 'Amsterdam' and ORDER BY custId DESC) where rownum <=10)
You wouldn't need the extra subquery if you didn't have the order by clause
For ref see https://docs.oracle.com/cd/B19306_01/server.102/b14200/pseudocolumns009.htm#:~:text=For%20each%20row%20returned%20by,has%202%2C%20and%20so%20on.
I would phrase this with exists and row_number():
select c.*
from customers
where exists (
select 1
from (
select custId, row_number() over(order by custid desc) rn
from orders
where orderStatus = 'S' and orderCity = 'Amsterdam'
) o
where o.rn <= 10 and o.cusid = c.customerid
)
row_number() ranks orders by descending customer id in the subquery. Then, we filter on the first 10 rows, and use exists to filter the corresponding rows in the outer query.
Assume I have these 3 tables :
The first 2 tables define customers of different types ,i.e second table has other columns which are not included in table 1 i just left them the same to save complexity.
The third table defines orders for both types of customers . Each customer has more than one orders
I want to select the last order for every customer, i.e the order with order_id 4 for customer 1 which was created on 23/12/2016 and the order with order_id 5 for customer 2 which was created on 26/12/2016
I tried something like this :
select *
from customertype1
left join order on order.customer_id = customertype1.customer_id
order by order_id desc;
But this gives me multiple records for every customer, as I have stated above I want only the last order for every customertype1.
If you want the last order for each customer, then you only need the orders table:
select o.*
from (select o.*,
row_number() over (partition by customer_id order by datecreated desc) as seqnum
from orders o
) o
where seqnum = 1;
If you want to include all customers, then you need to combine the two tables. Assuming they are mutually exclusive:
with c as (
select customer_id from customers1 union all
select customer_id from customers2
)
select o.*
from c left join
(select o.*,
row_number() over (partition by customer_id order by datecreated desc) as seqnum
from orders o
) o
on c.customer_id = o.customer_id and seqnum = 1;
A note about your data structure: You should have one table for all customers. You can then define a foreign key constraint between orders and customers. For the additional columns, you can have additional tables for the different types of customers.
Use ROW_NUMBER() and PARTITION BY.
ROW_NUMBER(): it will give sequence no to your each row
PARTITION BY: it will group your data by given column
When you use ROW_NUMBER() and PARTITION BY both together then first partition by group your records and then row_number give then sequence no by each group, so for each group you have start sequence from 1
Help Link: Example of ROW_NUMBER() and PARTITION BY
This is the general idea. You can work out the details.
with customers as
(select customer_id, customer_name
from table1
union
select customer_id, customer_name
from table2)
, lastOrder as
(select customer_id, max(order_id) maxOrderId
from orders
group by customer_id)
select *
from lastOrder join customers on lastOrder.Customer_id = customers.customer_id
join orders on order_id = maxOrderId
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.