synthetic_text2sql.csv

text_query,sql_command
Show all customer names and emails,"SELECT first_name, last_name, email FROM customers"
List phone numbers of all customers,SELECT phone FROM customers
Find the email of customer named Alice Smith,SELECT email FROM customers WHERE first_name = 'Alice' AND last_name = 'Smith'
Get all product names and prices,"SELECT name, price FROM products"
Which products belong to the Electronics category?,SELECT name FROM products WHERE category = 'Electronics'
List products that cost more than 500,"SELECT name, price FROM products WHERE price > 500"
Show all orders with their status,"SELECT order_id, status FROM orders"
Find all orders placed by customer with ID 1,SELECT * FROM orders WHERE customer_id = 1
Get order IDs and dates for delivered orders,"SELECT order_id, order_date FROM orders WHERE status = 'Delivered'"
List all customer names with their order status,"SELECT c.first_name, c.last_name, o.status FROM customers c JOIN orders o ON c.customer_id = o.customer_id"
Which customers placed orders in September 2025?,"SELECT DISTINCT c.first_name, c.last_name FROM customers c JOIN orders o ON c.customer_id = o.customer_id WHERE o.order_date LIKE '2025-09%'"
Get all products and quantities for order 1,"SELECT p.name, oi.quantity FROM order_items oi JOIN products p ON oi.product_id = p.product_id WHERE oi.order_id = 1"
List total quantity of items per order,"SELECT order_id, SUM(quantity) AS total_qty FROM order_items GROUP BY order_id"
Show all employees and their departments,"SELECT first_name, last_name, department FROM employees"
List employees with salary above 5000,"SELECT first_name, last_name FROM employees WHERE salary > 5000"
Get all departments and their locations,"SELECT name, location FROM departments"
Which department is located in Chicago?,SELECT name FROM departments WHERE location = 'Chicago'
List employees with their department location,"SELECT e.first_name, e.last_name, d.location FROM employees e JOIN departments d ON e.department = d.name"
Which customers ordered more than 2 different products?,"SELECT c.first_name, c.last_name, COUNT(DISTINCT oi.product_id) AS product_count FROM customers c JOIN orders o ON c.customer_id = o.customer_id JOIN order_items oi ON o.order_id = oi.order_id GROUP BY c.customer_id HAVING COUNT(DISTINCT oi.product_id) > 2"
"List the total amount spent by each customer, ordered by amount descending.","SELECT c.first_name, c.last_name, SUM(oi.quantity * p.price) AS total_spent FROM customers c JOIN orders o ON c.customer_id = o.customer_id JOIN order_items oi ON o.order_id = oi.order_id JOIN products p ON oi.product_id = p.product_id GROUP BY c.customer_id ORDER BY total_spent DESC"
Show the most expensive product in each category.,"SELECT category, name, price FROM products p1 WHERE price = (     SELECT MAX(p2.price) FROM products p2 WHERE p2.category = p1.category)"
Which employees earn more than the average salary of their department?,"SELECT e.first_name, e.last_name, e.department, e.salary FROM employees e JOIN departments d ON e.department = d.name WHERE e.salary > (     SELECT AVG(e2.salary)     FROM employees e2     WHERE e2.department = e.department)"
Find customers who have never placed an order.,"SELECT c.first_name, c.last_name FROM customers c LEFT JOIN orders o ON c.customer_id = o.customer_id WHERE o.order_id IS NULL"
Which products were ordered the most times (by total quantity)?,"SELECT p.name, SUM(oi.quantity) AS total_quantity FROM products p JOIN order_items oi ON p.product_id = oi.product_id GROUP BY p.product_id ORDER BY total_quantity DESC"
Get the number of orders for each order status.,"SELECT status, COUNT(*) AS order_count FROM orders GROUP BY status"
Find employees working in departments located in New York.,"SELECT e.first_name, e.last_name, e.department FROM employees e JOIN departments d ON e.department = d.name WHERE d.location = 'New York'"
Which customers ordered Electronics products in September 2025?,"SELECT DISTINCT c.first_name, c.last_name FROM customers c JOIN orders o ON c.customer_id = o.customer_id JOIN order_items oi ON o.order_id = oi.order_id JOIN products p ON oi.product_id = p.product_id WHERE p.category = 'Electronics' AND o.order_date LIKE '2025-09%'"
"Show average product price per category, sorted highest to lowest.","SELECT category, AVG(price) AS avg_price FROM products GROUP BY category ORDER BY avg_price DESC"
"List customers and their total number of orders, but only if they have more than 1 order.","SELECT c.first_name, c.last_name, COUNT(o.order_id) AS total_orders FROM customers c JOIN orders o ON c.customer_id = o.customer_id GROUP BY c.customer_id HAVING COUNT(o.order_id) > 1"
Which employees earn above 5500 and are not in IT?,"SELECT first_name, last_name, department, salary FROM employees WHERE salary > 5500 AND department <> 'IT'"
"For each customer, list their most recent order date.","SELECT c.first_name, c.last_name, MAX(o.order_date) AS latest_order FROM customers c JOIN orders o ON c.customer_id = o.customer_id GROUP BY c.customer_id"
What is the average quantity of products per order?,"SELECT AVG(total_items) AS avg_items_per_order FROM (     SELECT order_id, SUM(quantity) AS total_items     FROM order_items     GROUP BY order_id ) t"
List the departments along with the number of employees in each.,"SELECT d.name AS department, COUNT(e.employee_id) AS employee_count FROM departments d LEFT JOIN employees e ON e.department = d.name GROUP BY d.department_id"
List order IDs that contain products from at least 2 different categories., SELECT o.order_id FROM orders o JOIN order_items oi ON o.order_id = oi.order_id JOIN products p ON oi.product_id = p.product_id GROUP BY o.order_id HAVING COUNT(DISTINCT p.category) >= 2;
Find the department(s) with the highest average employee salary., SELECT d.name, AVG(e.salary) AS avg_salary FROM departments d JOIN employees e ON e.department = d.name GROUP BY d.name HAVING AVG(e.salary) = (SELECT MAX(avg_salary) FROM (SELECT AVG(e2.salary) AS avg_salary FROM employees e2 JOIN departments d2 ON e2.department = d2.name GROUP BY d2.name) sub);
List customer names and order IDs where the total order cost exceeds $500, including total order cost., SELECT c.first_name || ' ' || c.last_name AS customer_name, o.order_id, SUM(oi.quantity * p.price) AS 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 JOIN products p ON oi.product_id = p.product_id GROUP BY o.order_id, c.first_name, c.last_name HAVING SUM(oi.quantity * p.price) > 500;
Find all employees whose salary is above the average salary of their department, showing employee name, department name, and salary., SELECT e.first_name, e.last_name, d.name AS department_name, e.salary FROM employees e JOIN departments d ON e.department = d.name WHERE e.salary > (SELECT AVG(e2.salary) FROM employees e2 WHERE e2.department = e.department);