Решение по теме агрегации и функции

ebelykh/postgresql/агрегации и группировки/агрегации.sql

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+--Задача 1
+В варианте А: WHERE фильтрует товары и убирает те, которые не в наличии, потом эти товары(строки)участвуют в группировке, в варианте В - HAVING фильтрует уже сгруппированные данные(категории)
+
+--Задача 2356
+SELECT teacher_id,
+       COUNT(DISTINCT subject_id) AS cnt
+FROM teacher
+GROUP BY teacher_id;
+
+--Задача 1581
+SELECT customer_id,
+       COUNT(v.visit_id) AS count_no_trans
+FROM visits v
+LEFT JOIN transactions t
+ON v.visit_id = t.visit_id
+WHERE transaction_id IS NULL
+GROUP BY customer_id;
+
+--Задача 1141 
+SELECT activity_date AS day,
+       COUNT(DISTINCT user_id) AS active_users
+FROM Activity
+WHERE activity_date BETWEEN '2019-06-28' AND '2019-07-27'
+GROUP BY activity_date;
+
+--Задача 596
+SELECT class
+FROM Courses
+GROUP BY class
+HAVING COUNT(student) >= 5;
+
+--Задача 1075
+SELECT project_id,
+       ROUND(AVG(experience_years), 2) AS average_years
+FROM Project p
+JOIN Employee e
+ON p.employee_id = e.employee_id
+GROUP BY project_id;
+
+--Задача 1729
+SELECT user_id, 
+       COUNT (follower_id) AS followers_count 
+FROM followers 
+GROUP BY user_id
+ORDER BY user_id;
+
+--Задача 1633
+SELECT
+    r.contest_id,
+    ROUND(
+        COUNT(r.user_id) * 100.0 / (SELECT COUNT(*) FROM Users),
+        2
+    ) AS percentage
+FROM Register r
+GROUP BY r.contest_id
+ORDER BY percentage DESC, r.contest_id ASC;
+
+--Задача 1251
+SELECT
+    p.product_id,
+    COALESCE(
+        ROUND(
+            SUM(u.units * p.price) * 1.0 / SUM(u.units),
+            2
+        ),
+        0
+    ) AS average_price
+FROM Prices p
+LEFT JOIN UnitsSold u
+    ON p.product_id = u.product_id
+    AND u.purchase_date BETWEEN p.start_date AND p.end_date
+GROUP BY p.product_id;