human
Schema:
regions (REGION_ID, REGION_NAME)
countries (COUNTRY_ID, COUNTRY_NAME, REGION_ID)
departments (DEPARTMENT_ID, DEPARTMENT_NAME, MANAGER_ID, LOCATION_ID)
jobs (JOB_ID, JOB_TITLE, MIN_SALARY, MAX_SALARY)
employees (EMPLOYEE_ID, FIRST_NAME, LAST_NAME, EMAIL, PHONE_NUMBER, HIRE_DATE, JOB_ID, SALARY, COMMISSION_PCT, MANAGER_ID, DEPARTMENT_ID)
job_history (EMPLOYEE_ID, START_DATE, END_DATE, JOB_ID, DEPARTMENT_ID)
locations (LOCATION_ID, STREET_ADDRESS, POSTAL_CODE, CITY, STATE_PROVINCE, COUNTRY_ID)
Please answer with SQL.
assistant
Sure, I understand.
human
Hello! Can you tell me how many total employees are on this list?
assistant
SELECT COUNT( distinct EMPLOYEE_ID ) FROM employees;
human
Very good! Can display a list of all of the employees first names which contain a letter Z in their first name? | which colums to return? | Oh thanks! You went ahead and added the last name and city! Could you also add to this list which country each of these people live in?
assistant
SELECT T1.first_name , T1.last_name , T3.city , T3.COUNTRY_ID 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 WHERE T1.first_name LIKE "%z%" ;