query_id
int64 0
7k
| database_id
stringclasses 140
values | table_id
sequencelengths 1
5
| query
stringlengths 16
224
| answer
stringlengths 18
577
| difficulty
stringclasses 4
values |
|---|---|---|---|---|---|
3,400 | manufacturer | [
"manufacturer",
"furniture_manufacte"
] | Find the number of funiture types produced by each manufacturer as well as the company names. | SELECT count(*) , t1.name FROM manufacturer AS t1 JOIN furniture_manufacte AS t2 ON t1.manufacturer_id = t2.manufacturer_id GROUP BY t1.manufacturer_id | medium |
3,401 | manufacturer | [
"furniture",
"furniture_manufacte"
] | Give me the names and prices of furnitures which some companies are manufacturing. | SELECT t1.name , t2.price_in_dollar FROM furniture AS t1 JOIN furniture_manufacte AS t2 ON t1.Furniture_ID = t2.Furniture_ID | medium |
3,402 | manufacturer | [
"furniture",
"furniture_manufacte"
] | Find the market shares and names of furnitures which no any company is producing in our records. | SELECT Market_Rate , name FROM furniture WHERE Furniture_ID NOT IN (SELECT Furniture_ID FROM furniture_manufacte) | extra |
3,403 | manufacturer | [
"furniture",
"furniture_manufacte",
"manufacturer"
] | Find the name of the company that produces both furnitures with less than 6 components and furnitures with more than 10 components. | SELECT t3.name FROM furniture AS t1 JOIN furniture_manufacte AS t2 ON t1.Furniture_ID = t2.Furniture_ID JOIN manufacturer AS t3 ON t2.manufacturer_id = t3.manufacturer_id WHERE t1.num_of_component < 6 INTERSECT SELECT t3.name FROM furniture AS t1 JOIN furniture_manufacte AS t2 ON t1.Furniture_ID = t2.Furniture_ID JOIN manufacturer AS t3 ON t2.manufacturer_id = t3.manufacturer_id WHERE t1.num_of_component > 10 | extra |
3,404 | hr_1 | [
"employees",
"departments"
] | Display the first name and department name for each employee. | SELECT T1.first_name , T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id | medium |
3,405 | hr_1 | [
"employees",
"departments"
] | What are the first name and department name of all employees? | SELECT T1.first_name , T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id | medium |
3,406 | hr_1 | [
"employees"
] | List the full name (first and last name), and salary for those employees who earn below 6000. | SELECT first_name , last_name , salary FROM employees WHERE salary < 6000 | medium |
3,407 | hr_1 | [
"employees"
] | What are the full names and salaries for any employees earning less than 6000? | SELECT first_name , last_name , salary FROM employees WHERE salary < 6000 | medium |
3,408 | hr_1 | [
"employees"
] | Display the first name, and department number for all employees whose last name is "McEwen". | SELECT first_name , department_id FROM employees WHERE last_name = 'McEwen' | medium |
3,409 | hr_1 | [
"employees"
] | What are the first names and department numbers for employees with last name McEwen? | SELECT first_name , department_id FROM employees WHERE last_name = 'McEwen' | medium |
3,410 | hr_1 | [
"employees"
] | Return all the information for all employees without any department number. | SELECT * FROM employees WHERE department_id = "null" | easy |
3,411 | hr_1 | [
"employees"
] | What are all the employees without a department number? | SELECT * FROM employees WHERE department_id = "null" | easy |
3,412 | hr_1 | [
"departments"
] | Display all the information about the department Marketing. | SELECT * FROM departments WHERE department_name = 'Marketing' | easy |
3,413 | hr_1 | [
"departments"
] | What is all the information about the Marketing department? | SELECT * FROM departments WHERE department_name = 'Marketing' | easy |
3,414 | hr_1 | [
"employees"
] | when is the hire date for those employees whose first name does not containing the letter M? | SELECT hire_date FROM employees WHERE first_name NOT LIKE '%M%' | medium |
3,415 | hr_1 | [
"employees"
] | On what dates were employees without the letter M in their first names hired? | SELECT hire_date FROM employees WHERE first_name NOT LIKE '%M%' | medium |
3,416 | hr_1 | [
"employees"
] | display the full name (first and last), hire date, salary, and department number for those employees whose first name does not containing the letter M. | SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' | medium |
3,417 | hr_1 | [
"employees"
] | What are the full name, hire date, salary, and department id for employees without the letter M in their first name? | SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' | medium |
3,418 | hr_1 | [
"employees"
] | display the full name (first and last), hire date, salary, and department number for those employees whose first name does not containing the letter M and make the result set in ascending order by department number. | SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' ORDER BY department_id | hard |
3,419 | hr_1 | [
"employees"
] | What are the full name, hire data, salary and department id for employees without the letter M in their first name, ordered by ascending department id? | SELECT first_name , last_name , hire_date , salary , department_id FROM employees WHERE first_name NOT LIKE '%M%' ORDER BY department_id | hard |
3,420 | hr_1 | [
"employees"
] | what is the phone number of employees whose salary is in the range of 8000 and 12000? | SELECT phone_number FROM employees WHERE salary BETWEEN 8000 AND 12000 | easy |
3,421 | hr_1 | [
"employees"
] | Return the phone numbers of employees with salaries between 8000 and 12000. | SELECT phone_number FROM employees WHERE salary BETWEEN 8000 AND 12000 | easy |
3,422 | hr_1 | [
"employees"
] | display all the information of employees whose salary is in the range of 8000 and 12000 and commission is not null or department number does not equal to 40. | SELECT * FROM employees WHERE salary BETWEEN 8000 AND 12000 AND commission_pct != "null" OR department_id != 40 | medium |
3,423 | hr_1 | [
"employees"
] | Return all information about employees with salaries between 8000 and 12000 for which commission is not null or where their department id is not 40. | SELECT * FROM employees WHERE salary BETWEEN 8000 AND 12000 AND commission_pct != "null" OR department_id != 40 | medium |
3,424 | hr_1 | [
"employees"
] | What are the full name (first and last name) and salary for all employees who does not have any value for commission? | SELECT first_name , last_name , salary FROM employees WHERE commission_pct = "null" | medium |
3,425 | hr_1 | [
"employees"
] | Return the full names and salaries of employees with null commissions. | SELECT first_name , last_name , salary FROM employees WHERE commission_pct = "null" | medium |
3,426 | hr_1 | [
"employees"
] | Display the first and last name, and salary for those employees whose first name is ending with the letter m. | SELECT first_name , last_name , salary FROM employees WHERE first_name LIKE '%m' | medium |
3,427 | hr_1 | [
"employees"
] | Return the full names and salaries for employees with first names that end with the letter m. | SELECT first_name , last_name , salary FROM employees WHERE first_name LIKE '%m' | medium |
3,428 | hr_1 | [
"employees"
] | Find job id and date of hire for those employees who was hired between November 5th, 2007 and July 5th, 2009. | SELECT job_id , hire_date FROM employees WHERE hire_date BETWEEN '2007-11-05' AND '2009-07-05' | medium |
3,429 | hr_1 | [
"employees"
] | What are the job ids and dates of hire for employees hired after November 5th, 2007 and before July 5th, 2009? | SELECT job_id , hire_date FROM employees WHERE hire_date BETWEEN '2007-11-05' AND '2009-07-05' | medium |
3,430 | hr_1 | [
"employees"
] | What are the first and last name for those employees who works either in department 70 or 90? | SELECT first_name , last_name FROM employees WHERE department_id = 70 OR department_id = 90 | extra |
3,431 | hr_1 | [
"employees"
] | What are the full names of employees who with in department 70 or 90? | SELECT first_name , last_name FROM employees WHERE department_id = 70 OR department_id = 90 | extra |
3,432 | hr_1 | [
"employees"
] | Find the salary and manager number for those employees who is working under a manager. | SELECT salary , manager_id FROM employees WHERE manager_id != "null" | medium |
3,433 | hr_1 | [
"employees"
] | What are the salaries and manager ids for employees who have managers? | SELECT salary , manager_id FROM employees WHERE manager_id != "null" | medium |
3,434 | hr_1 | [
"employees"
] | display all the details from Employees table for those employees who was hired before 2002-06-21. | SELECT * FROM employees WHERE hire_date < '2002-06-21' | easy |
3,435 | hr_1 | [
"employees"
] | What is all the information about employees hired before June 21, 2002? | SELECT * FROM employees WHERE hire_date < '2002-06-21' | easy |
3,436 | hr_1 | [
"employees"
] | display all the information for all employees who have the letters D or S in their first name and also arrange the result in descending order by salary. | SELECT * FROM employees WHERE first_name LIKE '%D%' OR first_name LIKE '%S%' ORDER BY salary DESC | extra |
3,437 | hr_1 | [
"employees"
] | What is all the information about employees with D or S in their first name, ordered by salary descending? | SELECT * FROM employees WHERE first_name LIKE '%D%' OR first_name LIKE '%S%' ORDER BY salary DESC | extra |
3,438 | hr_1 | [
"employees"
] | display those employees who joined after 7th September, 1987. | SELECT * FROM employees WHERE hire_date > '1987-09-07' | easy |
3,439 | hr_1 | [
"employees"
] | Which employees were hired after September 7th, 1987? | SELECT * FROM employees WHERE hire_date > '1987-09-07' | easy |
3,440 | hr_1 | [
"jobs"
] | display the job title of jobs which minimum salary is greater than 9000. | SELECT job_title FROM jobs WHERE min_salary > 9000 | easy |
3,441 | hr_1 | [
"jobs"
] | Which job titles correspond to jobs with salaries over 9000? | SELECT job_title FROM jobs WHERE min_salary > 9000 | easy |
3,442 | hr_1 | [
"jobs"
] | display job Title, the difference between minimum and maximum salaries for those jobs which max salary within the range 12000 to 18000. | SELECT job_title , max_salary - min_salary FROM jobs WHERE max_salary BETWEEN 12000 AND 18000 | medium |
3,443 | hr_1 | [
"jobs"
] | What are the job titles, and range of salaries for jobs with maximum salary between 12000 and 18000? | SELECT job_title , max_salary - min_salary FROM jobs WHERE max_salary BETWEEN 12000 AND 18000 | medium |
3,444 | hr_1 | [
"employees"
] | display the emails of the employees who have no commission percentage and salary within the range 7000 to 12000 and works in that department which number is 50. | SELECT email FROM employees WHERE commission_pct = "null" AND salary BETWEEN 7000 AND 12000 AND department_id = 50 | medium |
3,445 | hr_1 | [
"employees"
] | What are the emails of employees with null commission, salary between 7000 and 12000, and who work in department 50? | SELECT email FROM employees WHERE commission_pct = "null" AND salary BETWEEN 7000 AND 12000 AND department_id = 50 | medium |
3,446 | hr_1 | [
"job_history"
] | display the employee ID for each employee and the date on which he ended his previous job. | SELECT employee_id , MAX(end_date) FROM job_history GROUP BY employee_id | medium |
3,447 | hr_1 | [
"job_history"
] | What are the employee ids for each employee and final dates of employment at their last job? | SELECT employee_id , MAX(end_date) FROM job_history GROUP BY employee_id | medium |
3,448 | hr_1 | [
"employees"
] | display those departments where more than ten employees work who got a commission percentage. | SELECT department_id FROM employees GROUP BY department_id HAVING COUNT(commission_pct) > 10 | easy |
3,449 | hr_1 | [
"employees"
] | What are the department ids for which more than 10 employees had a commission? | SELECT department_id FROM employees GROUP BY department_id HAVING COUNT(commission_pct) > 10 | easy |
3,450 | hr_1 | [
"employees"
] | Find the ids of the departments where any manager is managing 4 or more employees. | SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 | medium |
3,451 | hr_1 | [
"employees"
] | What are department ids for departments with managers managing more than 3 employees? | SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 | medium |
3,452 | hr_1 | [
"employees"
] | display the average salary of employees for each department who gets a commission percentage. | SELECT department_id , AVG(salary) FROM employees WHERE commission_pct != "null" GROUP BY department_id | medium |
3,453 | hr_1 | [
"employees"
] | What is the average salary of employees who have a commission percentage that is not null? | SELECT department_id , AVG(salary) FROM employees WHERE commission_pct != "null" GROUP BY department_id | medium |
3,454 | hr_1 | [
"locations"
] | display the country ID and number of cities for each country. | SELECT country_id , COUNT(*) FROM locations GROUP BY country_id | medium |
3,455 | hr_1 | [
"locations"
] | Give the country id and corresponding count of cities in each country. | SELECT country_id , COUNT(*) FROM locations GROUP BY country_id | medium |
3,456 | hr_1 | [
"job_history"
] | display job ID for those jobs that were done by two or more for more than 300 days. | SELECT job_id FROM job_history WHERE end_date - start_date > 300 GROUP BY job_id HAVING COUNT(*) >= 2 | medium |
3,457 | hr_1 | [
"job_history"
] | What are the job ids for jobs done more than once for a period of more than 300 days? | SELECT job_id FROM job_history WHERE end_date - start_date > 300 GROUP BY job_id HAVING COUNT(*) >= 2 | medium |
3,458 | hr_1 | [
"job_history"
] | display the ID for those employees who did two or more jobs in the past. | SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 | easy |
3,459 | hr_1 | [
"job_history"
] | What are the employee ids for employees who have held two or more jobs? | SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 | easy |
3,460 | hr_1 | [
"employees",
"departments",
"countries",
"locations"
] | Find employee with ID and name of the country presently where (s)he is working. | SELECT T1.employee_id , T4.country_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id JOIN locations AS T3 ON T2.location_id = T3.location_id JOIN countries AS T4 ON T3.country_id = T4.country_id | hard |
3,461 | hr_1 | [
"employees",
"departments",
"countries",
"locations"
] | What are all the employee ids and the names of the countries in which they work? | SELECT T1.employee_id , T4.country_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id JOIN locations AS T3 ON T2.location_id = T3.location_id JOIN countries AS T4 ON T3.country_id = T4.country_id | hard |
3,462 | hr_1 | [
"employees",
"departments"
] | display the department name and number of employees in each of the department. | SELECT T2.department_name , COUNT(*) FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id GROUP BY T2.department_name | medium |
3,463 | hr_1 | [
"employees",
"departments"
] | Give the name of each department and the number of employees in each. | SELECT T2.department_name , COUNT(*) FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id GROUP BY T2.department_name | medium |
3,464 | hr_1 | [
"employees",
"job_history"
] | Can you return all detailed info of jobs which was done by any of the employees who is presently earning a salary on and above 12000? | SELECT * FROM job_history AS T1 JOIN employees AS T2 ON T1.employee_id = T2.employee_id WHERE T2.salary >= 12000 | medium |
3,465 | hr_1 | [
"employees",
"job_history"
] | What is all the job history info done by employees earning a salary greater than or equal to 12000? | SELECT * FROM job_history AS T1 JOIN employees AS T2 ON T1.employee_id = T2.employee_id WHERE T2.salary >= 12000 | medium |
3,466 | hr_1 | [
"employees",
"jobs"
] | display job title and average salary of employees. | SELECT job_title , AVG(salary) FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id GROUP BY T2.job_title | medium |
3,467 | hr_1 | [
"employees",
"jobs"
] | What is the average salary for each job title? | SELECT job_title , AVG(salary) FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id GROUP BY T2.job_title | medium |
3,468 | hr_1 | [
"employees"
] | What is the full name ( first name and last name ) for those employees who gets more salary than the employee whose id is 163? | SELECT first_name , last_name FROM employees WHERE salary > (SELECT salary FROM employees WHERE employee_id = 163 ) | extra |
3,469 | hr_1 | [
"employees"
] | Provide the full names of employees earning more than the employee with id 163. | SELECT first_name , last_name FROM employees WHERE salary > (SELECT salary FROM employees WHERE employee_id = 163 ) | extra |
3,470 | hr_1 | [
"employees"
] | return the smallest salary for every departments. | SELECT MIN(salary) , department_id FROM employees GROUP BY department_id | medium |
3,471 | hr_1 | [
"employees"
] | What is the minimum salary in each department? | SELECT MIN(salary) , department_id FROM employees GROUP BY department_id | medium |
3,472 | hr_1 | [
"employees"
] | Find the first name and last name and department id for those employees who earn such amount of salary which is the smallest salary of any of the departments. | SELECT first_name , last_name , department_id FROM employees WHERE salary IN (SELECT MIN(salary) FROM employees GROUP BY department_id) | extra |
3,473 | hr_1 | [
"employees"
] | What are the full names and department ids for the lowest paid employees across all departments. | SELECT first_name , last_name , department_id FROM employees WHERE salary IN (SELECT MIN(salary) FROM employees GROUP BY department_id) | extra |
3,474 | hr_1 | [
"employees"
] | Find the employee id for all employees who earn more than the average salary. | SELECT employee_id FROM employees WHERE salary > (SELECT AVG(salary) FROM employees) | hard |
3,475 | hr_1 | [
"employees"
] | What are the employee ids for employees who make more than the average? | SELECT employee_id FROM employees WHERE salary > (SELECT AVG(salary) FROM employees) | hard |
3,476 | hr_1 | [
"employees"
] | display the employee id and salary of all employees who report to Payam (first name). | SELECT employee_id , salary FROM employees WHERE manager_id = (SELECT employee_id FROM employees WHERE first_name = 'Payam' ) | extra |
3,477 | hr_1 | [
"employees"
] | What are the employee ids of employees who report to Payam, and what are their salaries? | SELECT employee_id , salary FROM employees WHERE manager_id = (SELECT employee_id FROM employees WHERE first_name = 'Payam' ) | extra |
3,478 | hr_1 | [
"employees",
"departments"
] | find the name of all departments that do actually have one or more employees assigned to them. | SELECT DISTINCT T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id | easy |
3,479 | hr_1 | [
"employees",
"departments"
] | What are the names of departments that have at least one employee. | SELECT DISTINCT T2.department_name FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id | easy |
3,480 | hr_1 | [
"employees",
"departments"
] | get the details of employees who manage a department. | SELECT DISTINCT * FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T1.employee_id = T2.manager_id | medium |
3,481 | hr_1 | [
"employees",
"departments"
] | What is all the information regarding employees who are managers? | SELECT DISTINCT * FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T1.employee_id = T2.manager_id | medium |
3,482 | hr_1 | [
"departments"
] | display all the information about the department Marketing. | SELECT * FROM departments WHERE department_name = 'Marketing' | easy |
3,483 | hr_1 | [
"departments"
] | What is all the information about the Marketing department? | SELECT * FROM departments WHERE department_name = 'Marketing' | easy |
3,484 | hr_1 | [
"job_history"
] | display the ID for those employees who did two or more jobs in the past. | SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 | easy |
3,485 | hr_1 | [
"job_history"
] | What are the employee ids for those who had two or more jobs. | SELECT employee_id FROM job_history GROUP BY employee_id HAVING COUNT(*) >= 2 | easy |
3,486 | hr_1 | [
"employees"
] | What are the unique ids of those departments where any manager is managing 4 or more employees. | SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 | medium |
3,487 | hr_1 | [
"employees"
] | Give the distinct department ids of departments in which a manager is in charge of 4 or more employees? | SELECT DISTINCT department_id FROM employees GROUP BY department_id , manager_id HAVING COUNT(employee_id) >= 4 | medium |
3,488 | hr_1 | [
"employees"
] | Find the job ID for those jobs which average salary is above 8000. | SELECT job_id FROM employees GROUP BY job_id HAVING AVG(salary) > 8000 | easy |
3,489 | hr_1 | [
"employees"
] | What are the job ids corresponding to jobs with average salary above 8000? | SELECT job_id FROM employees GROUP BY job_id HAVING AVG(salary) > 8000 | easy |
3,490 | hr_1 | [
"employees",
"jobs"
] | display the employee ID and job name for all those jobs in department 80. | SELECT T1.employee_id , T2.job_title FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id WHERE T1.department_id = 80 | medium |
3,491 | hr_1 | [
"employees",
"jobs"
] | what are the employee ids and job titles for employees in department 80? | SELECT T1.employee_id , T2.job_title FROM employees AS T1 JOIN jobs AS T2 ON T1.job_id = T2.job_id WHERE T1.department_id = 80 | medium |
3,492 | hr_1 | [
"employees",
"departments"
] | What is the first name and job id for all employees in the Finance department? | SELECT T1.first_name , T1.job_id FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T2.department_name = 'Finance' | medium |
3,493 | hr_1 | [
"employees",
"departments"
] | Give the first name and job id for all employees in the Finance department. | SELECT T1.first_name , T1.job_id FROM employees AS T1 JOIN departments AS T2 ON T1.department_id = T2.department_id WHERE T2.department_name = 'Finance' | medium |
3,494 | hr_1 | [
"employees"
] | display all the information of the employees whose salary if within the range of smallest salary and 2500. | SELECT * FROM employees WHERE salary BETWEEN (SELECT MIN(salary) FROM employees) AND 2500 | hard |
3,495 | hr_1 | [
"employees"
] | What is all the information regarding employees with salaries above the minimum and under 2500? | SELECT * FROM employees WHERE salary BETWEEN (SELECT MIN(salary) FROM employees) AND 2500 | hard |
3,496 | hr_1 | [
"departments",
"employees"
] | Find the ids of the employees who does not work in those departments where some employees works whose manager id within the range 100 and 200. | SELECT * FROM employees WHERE department_id NOT IN (SELECT department_id FROM departments WHERE manager_id BETWEEN 100 AND 200) | hard |
3,497 | hr_1 | [
"departments",
"employees"
] | What are the ids for employees who do not work in departments with managers that have ids between 100 and 200? | SELECT * FROM employees WHERE department_id NOT IN (SELECT department_id FROM departments WHERE manager_id BETWEEN 100 AND 200) | hard |
3,498 | hr_1 | [
"employees"
] | display the employee name ( first name and last name ) and hire date for all employees in the same department as Clara. | SELECT first_name , last_name , hire_date FROM employees WHERE department_id = (SELECT department_id FROM employees WHERE first_name = "Clara") | extra |
3,499 | hr_1 | [
"employees"
] | What are the full names and hire dates for employees in the same department as someone with the first name Clara? | SELECT first_name , last_name , hire_date FROM employees WHERE department_id = (SELECT department_id FROM employees WHERE first_name = "Clara") | extra |
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