Customer Placing the Largest Number of OrdersQuery the customer_number from the orders table for the customer who has placed the largest number of orders.It is guaranteed that exactly one customer will have placed more orders than any other customer.The orders table is defined as follows:ColumnTypeorder_number (PK)intcustomer_numberintorder_datedaterequired_datedateshipped_datedatestatuschar(15)commentchar(200)Sample Inputorder_numbercustomer_numberorder_daterequired_dateshipped_date112017-04-092017-04-132017-04-12222017-04-152017-04-202017-04-18332017-04-162017-04-252017-04-20442017-04-182017-04-282017-04-25Sample Outputcustomer_number3ExplanationThe customer with number '3' has two orders, which is greater than either customer '1' or '2' because each of them only has one order.So the result is customer_number '3'. Follow up: What if more than one customer have the largest number of orders, can you find all the customer_number in this case?Optionsselect customer_numberfrom ( select customer_number, count(*) as cnt group by customer_number) as eorder by e.cnt desclimit 1;select customer_numberfrom ( select customer_number, count(*) as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_number select customer_number, count(*) as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_numberfrom ( select customer_number, count(*) as cnt from orders group by customer_number) as elimit 1;
Question
Customer Placing the Largest Number of OrdersQuery the customer_number from the orders table for the customer who has placed the largest number of orders.It is guaranteed that exactly one customer will have placed more orders than any other customer.The orders table is defined as follows:ColumnTypeorder_number (PK)intcustomer_numberintorder_datedaterequired_datedateshipped_datedatestatuschar(15)commentchar(200)Sample Inputorder_numbercustomer_numberorder_daterequired_dateshipped_date112017-04-092017-04-132017-04-12222017-04-152017-04-202017-04-18332017-04-162017-04-252017-04-20442017-04-182017-04-282017-04-25Sample Outputcustomer_number3ExplanationThe customer with number '3' has two orders, which is greater than either customer '1' or '2' because each of them only has one order.So the result is customer_number '3'. Follow up: What if more than one customer have the largest number of orders, can you find all the customer_number in this case?Optionsselect customer_numberfrom ( select customer_number, count() as cnt group by customer_number) as eorder by e.cnt desclimit 1;select customer_numberfrom ( select customer_number, count() as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_number select customer_number, count() as cnt from orders group by customer_number) as eorder by e.cnt desclimit 1;select customer_numberfrom ( select customer_number, count() as cnt from orders group by customer_number) as elimit 1;
Solution
The correct SQL query to find the customer who has placed the largest number of orders is:
select customer_number
from (
select customer_number, count(*) as cnt
from orders
group by customer_number
) as e
order by e.cnt desc
limit 1;
Here's how this query works:
-
The inner query
select customer_number, count(*) as cnt from orders group by customer_numbercounts the number of orders for each customer. Thegroup byclause groups the orders by thecustomer_number, and thecount(*)function counts the number of orders for each group. -
The outer query
select customer_number from (...) as e order by e.cnt desc limit 1selects thecustomer_numberfrom the result of the inner query. Theorder by e.cnt descclause sorts the customers in descending order by the count of their orders, and thelimit 1clause returns only the first row, which is the customer with the largest number of orders.
For the follow-up question, if more than one customer have the largest number of orders, you can find all their customer numbers by removing the limit 1 clause:
select customer_number
from (
select customer_number, count(*) as cnt
from orders
group by customer_number
) as e
order by e.cnt desc;
This query will return all customers sorted by the number of their orders in descending order. The customers with the largest number of orders will be at the top of the result.
Similar Questions
A) We have following relation orders(order_id,customer_id,order_date,amount). Create table with Primary Key, otherrelevant constraints in given schema.1) Find out the number of orders for each customer by customer_id and show only customer_id with number of orders above 5.2) Find out the total amount by order_id and order_date.B) Find the sum of a user-inserted number's first and last digits using PL/pgSQL function
Table: OrdersOrderIDCustomerIDOrderDateAmount11012023-01-01500.0021022023-01-03150.0031012023-01-04200.0041032023-01-07700.0051022023-01-10300.00i. Write an SQL query to count the number of orders.ii. Write an SQL query to calculate the total amount of all orders.iii.Write an SQL query to find the total amount of orders for each customer.iv. Write an SQL query to find the largest order amount.v. Write an SQL query to find customers with a total order amount greater than 500.Answer text Question 3EditViewInsertFormatToolsTableHelp
Create a table named orders with columns "order_id" (int), "customer_id" (int), "order_date" (date), "total_amount" (decimal), and "status" (varchar). The table should be configured with constraints such that "order_id" and "customer_id" cannot be empty, and "order_id" serves as the primary key.
From the following table, write an SQL query to retrieve the customer ID, order date, and maximum purchase amount for each customer on each order date where the maximum purchase amount exceeds 2000.00. The results should be sorted by customer ID.
We have following relationorders(order_id,customer_id,order_date,amount)1) Find out the number of orders for each customer by customer_id.2) Find out the total amount by order_id and order_date.3) Find out the number of orders for each customer by customer_id. Show onlycustomer_id with number of orders above 5
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