There are 2 tables customer (id, name) and orders (id, customer_id, date, price) where customer_id is the foreign key.
How would one get the top 5 individuals with the most purchases in the last 6 months?
needed output format is (customer.id, customer.name, sum(price))
select c.id, c.name, SUM(o.price) as total
from customer c inner join orders o
on c.id = o.customer_id
order by total desc limit 5;
I'm not sure why this query I have doesn't display anything
SUM() is an aggregate function that calculates the sum after a grouping operation. Try using a GROUP BY clause like this:
select c.id, c.name, SUM(o.price) as total
from customer c inner join orders o
on c.id = o.customer_id
group by c.id, c.name
order by total desc limit 5;
DB-fiddle: https://dbfiddle.uk/?rdbms=mysql_5.7&fiddle=7f5e3726d6af5a9144a03d503da5ae37
Related
Finding the city in which the most orders were sent leads to the assignment of the city and the number of orders (the named amount column). I have 2 tables the named Customers and Orders
SELECT Customers.City,count( Orders.OrderID) as amount
FROM voodoo.Customers
inner join voodoo.Orders on Customers.CustomerID=Orders.CustomerID
group by Customers.City
having amount >= all(select count(Orders.OrderID)
from voodoo.Customers
inner join voodoo.Orders on Customers.CustomerID=Orders.CustomerID
group by Customers.City);
tables
You don't need a subquery as you can just order by amount (descending) and limit the result to 1:
SELECT Customers.City, count(Orders.OrderID) as amount
FROM voodoo.Customers INNER JOIN voodoo.Orders
ON Customers.CustomerID=Orders.CustomerID
GROUP BY Orders.OrderID
ORDER BY amount DESC
LIMIT 1;
EDIT: as Thorsten Kettner pointed out, I made a copy & paste error; the correct version would GROUP BY Customers.City.
You are looking for the order count per city, not per order. So, don't group by order, but by city. For the ranking of the cities you can use RANK or DENSE_RANK.
SELECT city, amount
FROM
(
SELECT
c.city,
COUNT(o.orderid) AS amount,
RANK() OVER (ORDER BY COUNT(o.orderid) DESC) AS rnk
FROM voodoo.customers c
INNER JOIN voodoo.orders o ON o.customerid = c.customerid
group by c.city
) counted_and_ranked
WHERE rnk = 1;
I currently am trying to write a query that shows customers with at least 5 orders and customer with no orders. Orders are tracked in their own table and in order to find customers with 0 orders we have to find the customers NOT IN orders. Below is my query I'm trying to use and it returns the same customer 5 times for zero orders.
with t1 as
(select o.customerNumber, c.customerName, count(o.orderNumber) as FiveOrders
from orders o join customers c on (o.customerNumber = c.customerNumber)
group by o.customerNumber having count(o.orderNumber) = 5),
t2 as
(select distinct o.customerNumber, c.customerName, count(o.orderNumber) as NoOrders
from orders o join customers c on (o.customerNumber = c.customerNumber)
group by c.customerNumber not in(select customerNumber from orders))
select distinct t1.customerNumber as FiveOrderNumber, t1.customerName as FiveOrderName,
t2.customerNumber as NoOrderNumber, t2.customerName as NoOrderName
from t1 join t2
order by NoOrderName;
Any and all help is appreciated thanks!
If the errors were only in the second table to, I think it is after using
having with condition NOT IN without any logical comparison, I think you can get wanted results easily like:
select distinct customerNumber, customerName, "0" as NoOrders
from customers
where customerNumber not in (Select customerNumber from orders)
If the group by is important, you can use it like in your code.
Zero or five could be counted together with LEFT JOIN
select c.customerNumber, max(c.customerName) customerName, count(o.orderNumber) as FiveOrdersOrZero
from customers c
left join orders o on o.customerNumber = c.customerNumber
group by c.customerNumber
having count(o.orderNumber) in ( 0, 5 )
order by FiveOrdersOrZero
I have this schema here, and I need to find the name of the customer with the highest total amount for the orders. I have a SQL query here:
SELECT Name
FROM (SELECT Name, SUM(Amount) AS Total
FROM customer JOIN orders ON cust_id = ID
GROUP BY Name) AS Totals
WHERE Total = (SELECT MAX(Total)
FROM (SELECT Name, SUM(Amount) AS Total
FROM customer JOIN orders ON cust_id = ID
GROUP BY Name) AS X);
But this is very inefficient as it creates the same table twice. Is there any more efficient way to get the name?
If you want customer with the greatest total mount, then you can just join, order by and limit:
select c.name
from customer c
inner join orders o on o.cust_id = c.id
group by c.id, c.name
order by sum(o.amount) desc
limit 1
Note that this does not handle possible top ties. For this, you need a little more code. Instead of ordering, you would typically filter with a having clause:
select c.name
from customer c
inner join orders o on o.cust_id = c.id
group by c.id, c.name
having sum(o.amount) = (
select sum(o1.amount)
from orders o1
group by cust_id
order by sum(o1.amount) desc
limit 1
)
Finally: if you are running MySQL 8.0, this is simpler done with window function rank():
select name
from (
select c.name, rank() over(order by sum(o.amount) desc) rn
from customer c
inner join orders o on o.cust_id = c.id
group by c.id, c.name
) t
where rn = 1
I have these tables : customers, customer_invoices, customer_invoice_details, each customer has many invoices, and each invoice has many details.
The customer with the ID 574413 has these invoices :
select customer_invoices.customer_id,
customer_invoices.id,
customer_invoices.total_price
from customer_invoices
where customer_invoices.customer_id = 574413;
result :
customer_id invoice_id total_price
574413 662146 700.00
574413 662147 250.00
each invoice here has two details (or invoice lines) :
first invoice 662146:
select customer_invoice_details.id as detail_id,
customer_invoice_details.customer_invoice_id as invoice_id,
customer_invoice_details.total_price as detail_total_price
from customer_invoice_details
where customer_invoice_details.customer_invoice_id = 662146;
result :
detail_id invoice_id detail_total_price
722291 662146 500.00
722292 662146 200.00
second invoice 662147 :
select customer_invoice_details.id as detail_id,
customer_invoice_details.customer_invoice_id as invoice_id,
customer_invoice_details.total_price as detail_total_price
from customer_invoice_details
where customer_invoice_details.customer_invoice_id = 662147;
result :
detail_id invoice_id detail_total_price
722293 662147 100.00
722294 662147 150.00
I have a problem with this query :
select customers.id as customerID,
customers.last_name,
customers.first_name,
SUM(customer_invoices.total_price) as invoice_total,
SUM(customer_invoice_details.total_price) as details_total
from customers
join customer_invoices
on customer_invoices.customer_id = customers.id
join customer_invoice_details
on customer_invoice_details.customer_invoice_id = customer_invoices.id
where customer_id = 574413;
unexpected result :
customerID last_name first_name invoice_total details_total
574413 terry amine 1900.00 950.00
I need to have the SUM of the total_price of the invoices, and the SUM of the total_price of the details for each customer. In this case I'm supposed to get 950 as total_price for both columns (invoice_total& details_total) but it's not the case. what am I doing wrong & how can I get the correct result please. The answers in similar topics don't have the solution for this case.
When you mix normal columns with aggregate functions (for example SUM), you need to use GROUP BY where you list the normal columns from the SELECT.
The reason for the excessive amount in total_price for invoices is that the SUM is also calculated over each detail row as it is part of the join. Use this:
select c.id as customerID,
c.last_name,
c.first_name,
SUM(ci.total_price) as invoice_total,
SUM((select SUM(d.total_price)
from customer_invoice_details d
where d.customer_invoice_id = ci.id)) as 'detail_total_price'
from customers c
join customer_invoices ci on ci.customer_id = c.id
where c.id = 574413
group by c.id, c.last_name, c.first_name
db-fiddle
I used join against sub queries and then did a sum on the sums
SELECT c.id as customerID,
c.last_name,
c.first_name
SUM(i.sum) as invoice_total,
SUM(d.sum) AS details_total
FROM customers c
JOIN (SELECT id, customer_id, SUM(total_price) AS sum
FROM customer_invoices
GROUP BY id, customer_id) AS i ON i.customer_id = c.id
JOIN (SELECT customer_invoice_id as id, SUM(total_price) AS sum
FROM customer_invoice_details
GROUP BY customer_invoice_id) AS d ON d.id = i.id
WHERE c.id = 574413
GROUP BY c.id, c.name
The issue is in the joining logic. The table customers is used as the driving table in the joins. But in the second join, you are using a derivative key column from the first join, to join with the third tables. This is resulting in a Cartesian output doubling the records from the result from the nth-1 join, which is leading to customer_invoices.total_price getting repeated twice, hence the rolled up value of this field is doubled.
At a high level I feel that the purpose of rolling up the prices is already achieved in SUM(customer_invoice_details.total_price).
But if you have a specific project requirement that SUM(customer_invoices.total_price) should also be obtained and must match with SUM(customer_invoice_details.total_price), then you can do this:
In a separate query, Join customer_invoice_details and customer_invoices. Roll up the pricing fields, and have a result such that you have only one record for one customer ID.
Then use this as a sub-query and join it with the customers table.
You are aggregating along multiple dimensions. This is challenging. I would suggest doing the aggregation along each dimension independently:
select c.id as customerID, c.last_name, c.first_name,
ci.invoice_total,
cid.details_total
from customers c join
(select ci.sum(ci.total_price) as invoice_total
from customer_invoices ci
group by ci.customer_id
) ci
on ci.customer_id = c.id join
(select ci.sum(cid.total_price) as details_total
from customer_invoices ci join
customer_invoice_details cid
on cid.customer_invoice_id = ci.id
group by ci.customer_id
) cid
on cid.customer_id = c.id
where c.id = 574413;
A faster version (for one customer) uses correlated subqueries:
select c.id as customerID, c.last_name, c.first_name,
(select ci.customer_id, sum(ci.total_price) as invoice_total
from customer_invoices ci
where ci.customer_id = c.id
) as invoice_total,
(select ci.customer_id, sum(cid.total_price) as details_total
from customer_invoices ci join
customer_invoice_details cid
on cid.customer_invoice_id = ci.id
where ci.customer_id = c.id
) as details_total
from customers c
where c.id = 574413;
I am working on an mySQL assignment for school and I am stuck on a question. I am still new to mySQL. COUNT(o.customer_id) is not working the way I want. I want it to count the number of orders but it is counting all items. i.e. Customer 1 has 2 orders but it is returning 3 because one order has two items. I have three tables one with customers, another with orders than another with each item on each order. Ive posed my query below. Any help would be great.
SELECT email_address, COUNT(o.order_id) AS num_of_orders,
SUM(((item_price - discount_amount) * quantity)) AS total
FROM customers c JOIN orders o
ON c.customer_id = o.customer_id
JOIN order_items ot
ON o.order_id = ot.order_id
GROUP BY o.customer_id
HAVING num_of_orders > 1
ORDER BY total DESC;
As simple as use Distinct reserved word:
SELECT email_address, COUNT(distinct o.order_id) AS num_of_orders
Looks like you want to count the DISTINCT number of orders. Add a DISTINCT into the COUNT. Although MySQL allows you to use the SELECT expression in the HAVING clause, it's not good practice to do so.
SELECT email_address, COUNT(DISTINCT o.order_id) AS num_of_orders,
SUM(((item_price - discount_amount) * quantity)) AS total
FROM customers c JOIN orders o
ON c.customer_id = o.customer_id
JOIN order_items ot
ON o.order_id = ot.order_id
GROUP BY o.customer_id
HAVING COUNT(DISTINCT o.order_id) > 1
ORDER BY total DESC;
Just take out the join to items. All it is doing is duplicating rows when there are multiple items.
SELECT email_address, COUNT(o.order_id) AS num_of_orders,
SUM(((item_price - discount_amount) * quantity)) AS total
FROM customers c JOIN orders o
ON c.customer_id = o.customer_id
GROUP BY o.customer_id
HAVING COUNT(o.order_id) > 1
ORDER BY total DESC;