I am trying to create a query that does a calculation in a subquery that requires the SUM function and a group by. My query returns the error "Subquery returns more than 1 row". Essentially I am trying to return the amount "Due" for each order. If the order total is greater than the sum of total_collected (for that order_id) from the payments table there will be amount due. Here is the query:
SELECT o.order_id
, o.server
, o.subtotal
, o.discount
, o.tax, o.total
, (SELECT (o.total - SUM(p.total_collected))
from orders o
join payments p
on o.order_id = p.order_id
group by p.order_id) as 'Due'
FROM orders o
join payments p
on o.order_id = p.order_id
WHERE...;
I cannot include 'p.order_id' in the sub select because it should only contain one column. I understand why I am getting the error, I just don't know how to get the sub select to only perform the SUM on a per order_id basis.
Without changing the structure much, I think the subquery is looking at all of the data in the orders/payments tables. I think you need to filter it down to look only at the relevant order_id like so.
(I also added a SUM around the order total because I am pretty sure that would give a different error without it.)
SELECT o.order_id
, o.server
, o.subtotal
, o.discount
, o.tax
, o.total
, (SELECT (SUM(o2.total) - SUM(p.total_collected))
from orders o2
JOIN payments p
ON o2.order_id = p.order_id
WHERE o2.order_id = o.order_id) as 'Due'
FROM orders o
WHERE...;
Although, if you adjust this so that it uses a join instead of a subquery, I think you will get better performance. Something like this:
SELECT o.order_id
, o.server
, o.subtotal
, o.discount
, o.tax
, o.total
, o.total - c.Collected AS 'Due'
FROM orders o
JOIN (
SELECT p2.order_id, SUM(p2.total_collected) AS 'Collected'
FROM payments p2
GROUP BY p2.order_id) AS c
ON o.order_id = c.order_id
WHERE...;
You do not need sub-query:
SELECT
o.order_id,
o.server,
o.subtotal,
o.discount,
o.tax,
o.total,
o.total - ifnull(sum(p.total_collected),0) As Due
FROM orders AS o
LEFT JOIN payments AS p ON o.order_id = p.order_id
WHERE ...
GROUP BY o.order_id
Related
I'm trying to (let's say) gather a report on customers.
In that report I want to include sum of orders and ticket number for each client.
Tables:
Customer(id, name)
Order(id, customer_id, amount)
support_ticket(id, customer_id)
query:
select
c.id as 'Customer',
count(distinct t.id) as "Ticket count",
count(distinct o.id) as "Order count",
sum(o.amount) as 'Order Amount'
from customer as c
inner join `order` as o on c.id = o.customer_id
inner join support_ticket as t on c.id = t.customer_id
group by c.id
Since I join with customer.id on the two tables, I get all the rows "duplicated", since I get all possible combinations, so if the client as multiple tickets, the sum(o.amount) will we multiplied because of "duplicated rows"
sqlFiddle (mysql): http://sqlfiddle.com/#!9/ba39ba/13
sqlFiddle (pg): http://sqlfiddle.com/#!17/bc32e/7
It seems like a simple case but I've been looking at it too much I think, I can't find the proper way to do that report.
What am I doing wrong?
your best bet is to re-write the Aggregation off the Order table as as Derived Table;
EG
select
c.id as 'Customer',
count(distinct t.id) as "Ticket count",
o.amount as 'Order Amount' ,
o.[Order count]
from customer as c
inner join
(SELECT
o.customer_id,
sum(amount) as amount ,
count(distinct o.id) as "Order count"
from [order]
group by o.customer_id)
as o on c.id = o.customer_id
inner join support_ticket as t on c.id = t.customer_id
group by
c.id ,
o.amount ,
o.[Order count]
Note that the Derived Table Columns then are added to the group by clause at the bottom.
Cheers!
Just calculate order values in a sub-query and join it.
SELECT
c.id as 'Customer'
,count(DISTINCT st.id) as 'Ticket Count'
,o.`Order Count`
,o.amount as `Order Amount`
FROM customer c
INNER JOIN support_ticket st
on c.id = st.customer_id
INNER JOIN (
SELECT
customer_id
,SUM(amount) as 'amount'
,count(distinct id) as 'Order Count'
FROM `order`
group by customer_id
) o
on c.id = o.customer_id
GROUP BY c.id;
select c.id as 'Customer'
,t2.count_ticket as "Ticket count"
,t1.count_order as "Order count"
,t1.amount as 'Order Amount'
from customer as c
inner join (select customer_id
,count(id) as count_order
,sum(amount) as amount
from Order group by customer_id) t1
on c.id = t1.customer_id
inner join (select customer_id
,count(id) as count_ticket
from support_ticket group by customer_id) t2
on c.id = t2.customer_id
In cases like yours, when I think the solution of my problem should be fairly simple but I cant wrap my head around it, I tend to use a WITH clause.
Not because its better, but because it helps me to understand my code better by splitting up complexity. First I create a relatively simple temp. Solving the first part of my problem.
WITH temp AS (
SELECT
c.id AS "customer",
COUNT(DISTINCT o.id) AS "order_count",
SUM(o.amount) AS "order_amount"
FROM customer AS c
INNER JOIN "order" AS o on c.id = o.customer_id
GROUP BY c.id
)
Then I simply select the first half of my solution from temp, adding this way all intermediate results, and solve the second part of my initial sql.
SELECT
temp.customer,
COUNT(DISTINCT t.id) as "ticket_count",
temp.order_count,
temp.order_amount
FROM temp
INNER JOIN support_ticket as t on temp.customer = t.customer_id
GROUP BY temp.customer, temp.order_count, temp.order_amount
The principle is the same like in all previous answers, but SELECTS are separated and I can check them fast, and continue on if I'm happy with parts of the solution.
Here's my query
Select email_address, count(order_id) AS order_count, (sum(item_price - discount_amount) * (quantity)) AS order_total,
(avg(item_price - discount_amount) * (quantity)) AS avg_order_total
from customers join orders
using(customer_id)
join order_items
using(order_id)
group by customer_id > 1, email_address
Output of my query
Output Expected
I'm trying to produce the wanted output but I'm not sure how to only display only 3 of them. Tried a where statement but it made it worst.
Another way to display output without the use of limit?
Here's my REVISED query
Select email_address, count(customer_id) AS order_count, sum((item_price - discount_amount) * (quantity)) AS order_total,
round(avg((item_price - discount_amount) * (quantity)),2) AS avg_order_total
from customers join orders
using(customer_id)
join order_items
using(order_id)
group by customer_id
order by count(customer_id) desc limit 3
I'm trying to produce the expected output without using limit but I can't figure how. Above is my query that I used limit on to display the result and it shows the output I wanted. I tried using a where statement, but it doesn't run my query when I did.
Could you try this query?
SELECT c.email_address,
count(o.order_id) AS order_count,
(sum(i.item_price - i.discount_amount) * (i.quantity)) AS order_total,
(avg(sum(i.item_price - i.discount_amount) * (i.quantity))) AS avg_order_total
from customers c
INNER JOIN orders o ON c.customer_id = o.customer_id
INNER JOIN order_items i ON o.order_id = i.order_id
GROUP BY email_address;
A few things to note:
You need to include any field in the SELECT clause that isn't being aggregated into your GROUP BY. So, I've added email_address.
It's far better to specify the actual columns being joined, instead of the USING keyword. This is because it's easier to maintain if any of the columns change.
I've made some assumptions about which table your fields belong to and prefixed them with the table alias (c, o, or i). If this is wrong they can be changed. But it's better to use table aliases as it improves the maintainability of the code, and it's easier to read & write.
There are several guesses in this suggestion:
SELECT
c.email_address
, COUNT(DISTINCT o.order_id) AS order_count
, (SUM(oi.item_price - oi.discount_amount) * (oi.quantity)) AS order_total
, (AVG(oi.item_price - oi.discount_amount) * (oi.quantity)) AS avg_order_total
FROM customers c
JOIN orders o ON c.customer_id = o.customer_id
JOIN order_items oi ON o.order_id = oi.order_id
WHERE c.customer_id > 1
GROUP BY
c.email_address
When joining several tables it is good practice to nominate which table a column comes from. To make this brief each table can be given an alias and a common method for that is to use first letters e.g. for this set of tables: c, o, oi
I may have guessed incorrectly when placing the aliases in some places and I cannot fix that as I don't know your tables.
+EDIT
By the way, I also included DISTINCT into the count of orders, if you don't do that you are counting the number of order items instead.
SELECT
c.email_address
, COUNT(DISTINCT o.order_id) AS order_count
, (SUM(oi.item_price - oi.discount_amount) * (oi.quantity)) AS order_total
, (SUM(oi.item_price - oi.discount_amount) * (oi.quantity))
/ COUNT(DISTINCT o.order_id) AS avg_order_total
FROM customers c
JOIN orders o ON c.customer_id = o.customer_id
JOIN order_items oi ON o.order_id = oi.order_id
WHERE c.customer_id > 1
GROUP BY
c.email_address
I have a question my professor gave me, on making a statement that goes like this
How many customers are “whales” i.e., have spent, in their lifetime, more than $4,000? How many are “shrimps,” having spent less than $20?
This is the online database we are using: http://www.w3schools.com/sql/trysql.asp?filename=trysql_select_all
run this query to create another table before helping me out if you can
CREATE TABLE ByCustomerOrders AS
SELECT
o.OrderID
, p.ProductName
, p.ProductID
, Price
, Quantity
, Price * Quantity AS subtotal
, c.CustomerID
, s.SupplierID
FROM OrderDetails AS od
LEFT JOIN Orders AS o
ON od.OrderID = o.OrderID
LEFT JOIN Products AS p
ON p.ProductID = od.ProductID
LEFT JOIN Customers AS c
on c.CustomerID = o.CustomerID
LEFT JOIN Suppliers AS s
ON s.SupplierID = p.SupplierID;
I am trying to combine every customers order together grouping the sum by the customerID and then showing it in the table as a row for each customer ID and total amount they have order from subtotal
SELECT customerID, SUM(subtotal) AS 'total_money_spent' FROM ByCustomerOrders GROUP BY customerID ORDER BY 'total_money_spent' DESC LIMIT 1;
That didn't seem to work as it shows a value of 111. anyone see an issue?
You have a LIMIT 1 at the end of your statement which will only show the first result.
When you run:
SELECT customerID, SUM(subtotal) AS 'total_money_spent' FROM ByCustomerOrders GROUP BY customerID ORDER BY 'total_money_spent' DESC;
It outputs all the totals grouped
I need to do this but with a subquery, not a join. My problem is, how can I use a subquery to display another column? I could grab the info from there, but I'll be missing the order_date column from the orders table. Can I use a subquery to display it?
SELECT CONCAT(c.customer_first_name, ' ' , c.customer_last_name) AS customer_name, MAX(o.order_date) AS recent_order_date
FROM customers AS c
JOIN orders AS o
ON c.customer_id = o.customer_id
GROUP BY customer_name
ORDER BY MAX(o.order_date) DESC
It's not at all clear what resultset you are trying to return, but it looks an awful like the like the ubiquitous "latest row" problem.
The normative pattern for the solution to that problem is to use a JOIN to the inline view. If there's not a unique constraint, you run the possibility of returning more than one matching row.
To get the latest order (the row in the orders table with the maximum order_date for each customer, assuming that the (customer_id, order_date) tuple is unique, you can do something like this:
SELECT o.*
FROM ( SELECT n.customer_id
, MAX(n.order_date) AS latest_order_date
FROM orders n
GROUP BY n.customer_id
) m
JOIN orders o
ON o.customer_id = m.customer_id
AND o.order_date = m.latest_order_date
If you want to also retrieve columns from the customers table based on the customer_id returned from orders, you'd use a JOIN (not a subquery)
SELECT CONCAT(c.customer_first_name,' ',c.customer_last_name) AS customer_name
, c.whatever
, o.order_date AS recent_order_date
, o.whatever
FROM ( SELECT n.customer_id
, MAX(n.order_date) AS latest_order_date
FROM orders n
GROUP BY n.customer_id
) m
JOIN orders o
ON o.customer_id = m.customer_id
AND o.order_date = m.latest_order_date
JOIN customers c
ON c.customer_id = o.customer_id
ORDER BY o.order_date DESC, o.customer_id DESC
As I mentioned before, if a given customer can have two orders with the exact same value for order_date, there's potential to return more than one order for each customer_id.
To rectify that, we can return a unique key from the inline view, and use that in the join predicate to guarantee only a single row returned from orders.
(NOTE: this approach is specific to MySQL, with this syntax, other RDBMS will throw an error that essentially says "the GROUP BY must include all non-aggregates". But MySQL allows it.)
SELECT CONCAT(c.customer_first_name,' ',c.customer_last_name) AS customer_name
, c.whatever
, o.order_date AS recent_order_date
, o.whatever
FROM ( SELECT n.customer_id
, MAX(n.order_date) AS latest_order_date
, n.order_id
FROM orders n
GROUP BY n.customer_id
) m
JOIN orders o
AND o.customer_id = m.customer_id
AND o.order_date = m.latest_order_date
AND o.order_id = n.order_id
JOIN customers c
ON c.customer_id = o.customer_id
ORDER BY o.order_date DESC, o.customer_id DESC
I am not really sure i understand your question, but i think this works... (not tested though...)
SELECT
(
SELECT
CONCAT(c.customer_first_name, ' ' , c.customer_last_name)
FROM
customers c
WHERE
c.customer_id = o.customer_id
LIMIT 1
) AS customer_name,
MAX(o.order_date) AS recent_order_date
FROM
orders o
GROUP BY
customer_name
ORDER BY
MAX(o.order_date) DESC
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;