Firefox                                     https://www.codecademy.com/learn/learn-sql/modules/learn-sql-aggregat...
●
                COUNT()
           ●
                SUM()
           ●
                MAX()
           ●
                MIN()
           ●
                AVG()
          COUNT()
               COUNT()
                                             SELECT COUNT(*)
                                             FROM employees
                                             WHERE experience < 5;
                   *      COUNT(*)
          COUNT(column)              NULL
          SUM()
               SUM()
                                             SELECT SUM(salary)
                                             FROM salary_disbursement;
          AVG()
               AVG()
                                             SELECT AVG(salary)
          salary                             FROM employees
                                             WHERE experience < 5;
1 of 3                                                                                          6/22/2020, 7:52 PM
Firefox                                                https://www.codecademy.com/learn/learn-sql/modules/learn-sql-aggregat...
          ROUND()
               ROUND()
                                                        SELECT year,
                                                            ROUND(AVG(rating), 2)
                                                        FROM movies
                                                        WHERE year = 2015;
          GROUP BY
               GROUP BY
                                                        SELECT rating,
                                                            COUNT(*)
                                           GROUP BY     FROM movies
                         FROM      WHERE                GROUP BY rating;
                    ORDER BY        LIMIT
               GROUP BY          ORDER BY
                                                        SELECT COUNT(*) AS 'total_movies',
                        SELECT                             rating
                                                        FROM movies
                                                        GROUP BY 2
           ●
                GROUP BY            2   rating          ORDER BY 1;
           ●
                ORDER BY            1   total_movies
          HAVING
               HAVING
                                                        SELECT year,
                                   GROUP BY
                                                           COUNT(*)
          HAVING
                                                        FROM movies
                                                        GROUP BY year
                                                        HAVING COUNT(*) > 5;
          MAX()
               MAX()
                                                        SELECT MAX(amount)
                                                        FROM transactions;
                   amount
2 of 3                                                                                                     6/22/2020, 7:52 PM
Firefox                  https://www.codecademy.com/learn/learn-sql/modules/learn-sql-aggregat...
          MIN()
            MIN()
                          SELECT MIN(amount)
              amount      FROM transactions;
          transactions
3 of 3                                                                       6/22/2020, 7:52 PM