Contacted customer invoices hackerrank solution. Reload to refresh your session.

Contacted customer invoices hackerrank solution Can be null. To support the company's business expansion efforts, identify all customer-user pairs who have interacted more than once. Order the result by invoice number ascending. Reload to refresh your session. customer_id) LEFT JOIN invoice_item ON Mar 2, 2025 · Return only rows where invoices were created by a user who never contacted that particular customer. quantity FROM (SELECT customer. email: varchar(128) Email address of the customer: phone: varchar(128) Phone number for the customer: is_active: int: Boolean to denote if the Contains solved queries for the Hackerrank SQL (Intermediate) Skills Certification Test 🎓. hackerrank. You signed in with another tab or window. id AS invoice_item_id, COALESCE(customer_name, 'N/A') customer_name, COALESCE(product_name, 'N/A') product_name, COALESCE(quantity, 0) quantity FROM ((customer LEFT JOIN invoice ON customer. customer_id LEFT JOIN invoice_item ii ON ii. SELECT c. id AS customer_id, product. For each row, return the invoice number, customer name and the number of contacts that started prior to the time the invoice was created. id = i. product_name, a. id ORDER BY c. product_name, Coalesce((ii. You switched accounts on another tab or window. HackerRank's solution is using this condition to support the above logic, which is incorrect. invoice_id = i. Jul 11, 2020 · The actual question is all customers details even those without invoices and all products even those which are not sold. This intermediate SQL solution provides insights into product sales across cities, offering a comprehensive overview of customer spending patterns. Nov 5, 2020 · Hacker 74842 submitted solutions for challenges 19797 and 63132, so the total score = max(98, 5) + 76 = 174. id = invoice. id AS product_id, invoice_item. contact_end_time < i. orderId = orders. customerid, p. name; Just add GROUP BY in your query in last to group count of invoices for each Customer. . The total scores for hackers 4806, 26071, 80305, and 49438 can be similarly calculated. Name of the customer: city_id: int: A unique id for the city in which the customer resides: customer_address: varchar(255) Customer's address: contact_person: varchar(255) Person of contact. For each customer, display their name and the amount spent to 6 decimal places. For each pair, provide the following details: User's ID, first name, and last name; Customer's ID and name; Total number of their interactions; Sort the results by the user's ID in ascending order. customerId = customers. customer_name, ROUND ( SUM (i. Hacker 84072 submitted solutions for challenges 49593 and 63132, so the total score = 100 + 0 = 100. Order the result by the amount spent from high to low. Sep 14, 2023 · List all customers who spent 25% or less than the average amount spent on all invoices. Aug 13, 2021 · The correct code is : SELECT a. customer_name, p. The questions asked are Invoices Per Country and Product Sales Per City. name, COUNT(*) as number_of_invoices FROM invoices INNER JOIN orders ON invoices. product_id = p. orderId INNER JOIN customers ON orders. Solution SQL Intermediate Certification Hackerrank Solution https://www. If Person X made an invoice for Person Y AND contacted him, the row should not be printed. You signed out in another tab or window. Schema There are 2 tables: customer, invoice customer Name Type Description id int This … Case in point: If Person X made an invoice for Person Y, but never contacted Hamza, then the row should be printed. product_id, ii. - adminazhar/-hackerrank-SQL-intermediate-skills-certification-test-solution SQL solutions for the problems of the online HackerRank SQL Intermediate Skills Certification Test. total_price), 6 ) FROM customer c INNER JOIN invoice i ON c. By detailing sales figures per city and identifying customers who spent 25% or less than the average, this solution aids in strategic decision-making for optimizing sales and customer engagement. time_issued Apr 21, 2023 · SELECT c. quantity), 0) AS quantity FROM customer c LEFT JOIN invoice i on c. Mar 23, 2018 · Here is a solution for your problem: SELECT customers. WHERE ct. customer_id GROUP BY Apr 1, 2023 · Here are the 2 questions asked in HackerRank SQL Intermediate Certification Exam, these 2 will be available in one attempt. com You signed in with another tab or window. customer_name, a. 1 Table definitions and a data sample are given below. id LEFT JOIN product p ON ii. id You signed in with another tab or window. This intermediate SQL solution provides insights into product sales across cities, offering a comprehensive overview of customer spending patterns. customerId GROUP BY Customers. pvaif lxjnjk wcaqmf ewj tqyxoi snskhe wkoxul vzl zhb uglfx ajtlid uaborz gzqlqw sswpie rllr