I've 2 tables, purchases and customers.
Customers has cid, cname...
Purchases has total_price of purchase, one cid has multiple purchases.
I need to find cname and maximum total price spent by top 3 customers.
I'm doing this
select c.cname,
sum(p.total_price)
from purchases p
inner join
customers c
on p.cid=c.cid
where p.total_price <= (select max(total_price)
from purchases
)
group by p.cid,
c.cname
order by c.cname;
I'm getting the sum total but I can't order it and get the top 3 spending customers.
Where am I going wrong?
All you need is a ROWNUM <= 3 condition and an appropriate ORDER BY:
SELECT
cname
, total_purchase
FROM (SELECT
c.cname
, sum(p.total_price) as total_purchase
FROM customers c
INNER JOIN purchases p on p.cid = c.cid
GROUP BY c.cname
ORDER BY total_purchase DESC) RS
WHERE ROWNUM <= 3;
In Oracle you need to order the result set before limiting it to the top 3 customers, which is why ROWNUM <= 3 is in the outer query.
Related
I am trying to add a new column to my table which will be the average value calculated as the division of two existing columns. Therefore Average value = Total Sales / Number of Orders.
My data looks like this:click to view picture
I don't understand why Example Code A does not work but Example Code B does. Please can someone explain?
Example Code A
%%sql
SELECT
c.country,
count(distinct c.customer_id) customer_num,
count(i.invoice_id) order_num,
ROUND(SUM(i.total),2) total_sales,
order_num / total_sales avg_order_value
FROM customer c
LEFT JOIN invoice i ON c.customer_id = i.customer_id
GROUP BY 1
ORDER BY 4 DESC;
Example Code B
%%sql
WITH
customer_sales AS
(
SELECT
c.country,
count(distinct c.customer_id) customer_num,
count(i.invoice_id) order_num,
ROUND(SUM(i.total),2) total_sales
FROM customer c
LEFT JOIN invoice i ON c.customer_id = i.customer_id
GROUP BY 1
ORDER BY 4 DESC
)
SELECT
country,
customer_num,
order_num,
total_sales,
total_sales / order_num avg_order_value
FROM customer_sales;
Thank you!
Depending on the DBMS some allow you to reference the alias in the calculation (in the same select) and others require you to either bring it outside in an outer query or state your previous aggregation/functions, such as counts or sums.
SELECT
c.country,
count(distinct c.customer_id) customer_num,
count(i.invoice_id) order_num,
ROUND(SUM(i.total),2) total_sales,
count(i.invoice_id) / ROUND(SUM(i.total),2) avg_order_value
FROM customer c
LEFT JOIN invoice i ON c.customer_id = i.customer_id
GROUP BY 1
ORDER BY 4 DESC;
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
Below is a SQL script to see how many customers purchased both products, as well as how many customers purchased either of the two products.
I would like to be able to return a result set with one column for the first product, one column for the second, one column for Count with Both, and one column for Count with Either.
Instead it returns a column for each of the counts and a single NULL column for the ProductID.
SELECT
COUNT(DISTINCT b.CustomerID) AS "Count with Both",
COUNT(DISTINCT c.CustomerID) AS "Count with Either",
b.ProductID
FROM
LocationCode z
LEFT JOIN
(SELECT DISTINCT ProductID, CustomerID
FROM LocationCode a
WHERE a.ProductID IN ('MP040') AND a.ProductID IN ('OG010')) b ON z.CustomerID = b.CustomerID
LEFT JOIN
(SELECT DISTINCT ProductID, CustomerID
FROM LocationCode b
WHERE b.ProductID IN ('MP040', 'OG010')) c ON z.CustomerID = c.CustomerID
GROUP BY
b.ProductID
If I understand correctly, you can use a self-join and aggregation. The following does what you want for all pairs of products:
with cp as (
select customerid, productid,
count(*) over (partition by customerid) as num_customers
from locationcode
group by customerid, productid
)
select cp1.productid, cp2.productid, cp1.cnt as cnt1, cp2.cnt as cnt2,
count(*) as both
from cp cp1 join
cp cp2
on cp1.customerid = cp2.customerid and
cp1.productid < cp2.productid
group by cp1.productid, cp2.productid, cp1.cnt, cp2.cnt ;
You can of course add where filters if you only want results for certain products.
I have two tables:
Buyers:
Orders:
I want to get the name of a buyer with the highest amount of orders (in this case it would be buyer C with id 3).
I wrote the following query:
select top 1 b.Name, sum (o.Amount) as amt from Buyers b
join Orders o on o.BuyerId = b.Id
group by Name
order by amt desc
It gives me Name and amt in resulting row.
How to exclude amt from the result and display only Name of the buyer?
remove sum (o.Amount) as amt from select and add sum (o.Amount) desc in order by
select top 1 b.Name from Buyers b
join Orders o on o.BuyerId = b.Id
group by Name
order by sum (o.Amount) desc
I've a customer table and purchases table,
need to show cname, cid with max(customer_visits) from customer table
and sum of total_purchases by customer in purchases table.
I'm doing something like this
select p.cid, c.cname, sum(p.total_price)
from customers c where exists
(select max(visits_made) from customers having visits_made=max(visits_made)
and cid=p.cid)
inner join purchases p on p.cid=c.cid
group by p.cid,c.cname
and
select p.cid, c.cname, sum(p.total_price)
(select max(visits_made) from customers c where c.cid=p.cid)
from purchases p
inner join customers c on c.cid=p.cid
group by p.cid,c.cname
What's going wrong with these queries?
Found the solution, had to include where clause after inner join :D
I think this is just an aggregation query:
select p.cid, c.cname, sum(p.total_price) as total_price,
max(visits_made) as visits_made
from purchases p inner join
customers c
on c.cid = p.cid
group by p.cid, c.cname;