Дз - 3 неделя

Week3/solutions/fifth_task.sql

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+
+-- Для каждого сотрудника подсчитать количество клиентов, количестов выполненных заказов,
+-- количество уникальных адресов доставки и общее число приобретенных треков и вывести полное имя сотрудника.
+-- (employee_full_name, invoices_count, customers_count, addresses_count, tracks_count)
+
+SELECT e.first_name || ' ' || e.last_name as employee_full_name,
+       SUM(cnt_inv) as invoices_count,
+       COUNT(c.customer_id) as customers_count,
+       SUM(cnt_addr) as addresses_count,
+       SUM(cnt_trc) as tracks_count
+FROM customer c JOIN (
+    SELECT customer_id,
+           count(distinct invoice_id) as cnt_inv,
+           count(distinct billing_address) as cnt_addr,
+           sum(quantity) as cnt_trc
+    FROM customer c
+             JOIN invoice i USING (customer_id)
+             JOIN invoice_line il USING (invoice_id)
+    GROUP BY customer_id) st USING (customer_id)
+        JOIN employee e on c.support_rep_id = e.employee_id
+GROUP BY e.employee_id;

Week3/solutions/first_task.sql

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+
+-- Вывести все альбомы со средней длинной трека, большей 250 секунд.
+-- Вывести название альбома и количество треков. (album_name, cnt).
+
+SELECT a.title as album_name,
+       COUNT(track_id) as cnt
+FROM album a JOIN track t USING (album_id)
+GROUP BY a.album_id
+HAVING AVG(t.milliseconds) > 250000;

Week3/solutions/fourth_task.sql

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+
+-- Для всех треков жанра Rock и Metal, для каждого артиста подсчитать количество треков.
+-- Вывести имя артиста, количество треков. (artist_name, qty)
+
+SELECT art.name as artist_name,
+       COUNT(track_id) as qty
+FROM artist art JOIN album a USING (artist_id)
+    JOIN track t USING (album_id)
+        JOIN (SELECT *
+              FROM genre
+              WHERE name IN ('Rock', 'Metal')) g USING (genre_id)
+GROUP BY art.artist_id;

Week3/solutions/ninth_task.sql

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+
+
+-- Для каждого альбома жанр определяется как наибольшая суммарная продолжительность треков по жанру.
+-- Вывести альбом, жанр, и общее количество купленных песен по нему. (album_name, genre_name, cnt)
+WITH stat AS (
+    SELECT title,
+           album_id,
+           genre_id,
+           SUM(milliseconds) as duration
+    FROM album al
+             JOIN artist a USING (artist_id)
+             JOIN track t USING (album_id)
+    GROUP BY album_id, genre_id
+), genre_stat AS (
+    SELECT album_id,
+           MAX(duration) as sum
+    FROM stat
+    GROUP BY album_id
+)
+
+SELECT title as album_name,
+       g.name as genre_name,
+       cnt
+FROM genre_stat gst JOIN stat st ON gst.album_id = st.album_id
+                                        AND gst.sum = st.duration
+JOIN (SELECT title,
+             SUM(quantity) as cnt
+      FROM album a JOIN track t on a.album_id = t.album_id
+          JOIN invoice_line il on t.track_id = il.track_id
+      GROUP BY title) res USING (title)
+JOIN genre g USING (genre_id);

Week3/solutions/second_task.sql

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+
+-- Для каждого набора жанр, тип медиа вывести количество треков.
+-- Если для набора жанр, тип медиа треков нет, то такой набор выводить не требуется.
+-- (genre_name, media_type_name, cnt)
+
+SELECT genre.name as genre_name,
+       media_type.name as media_type_name,
+       COUNT(track_id) as cnt
+FROM track JOIN media_type USING (media_type_id)
+    JOIN genre USING (genre_id)
+GROUP BY genre.name, media_type.name;

Week3/solutions/seventh_task.sql

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+
+-- Вывести компании, которые приобрели больше всех композиций по каждому из жанров.
+-- При этом указать самую популярную композицию по жанру у данной компании.
+-- (популярность определяется по количеству покупок) (company_name, track_name)
+
+
+WITH company_stat AS (
+    SELECT *
+    FROM invoice i
+             JOIN invoice_line il USING (invoice_id)
+             JOIN track t USING (track_id)
+             JOIN (SELECT customer_id,
+                          company
+                   FROM customer
+                   WHERE company IS NOT NULL) comp USING (customer_id)
+),
+     top_chart AS(
+         SELECT company,
+                genre_id,
+                SUM(quantity) as count
+         FROM company_stat
+         GROUP BY company, genre_id
+     ),
+     count_track AS (
+         SELECT company,
+                genre_id,
+                name,
+                SUM(quantity) as pop
+         FROM company_stat
+         GROUP BY company,genre_id,name
+     ),
+     top_track AS (
+         SELECT company,
+               genre_id,
+               name,
+               MAX(pop)
+        FROM count_track
+        GROUP BY company,genre_id,name
+     )
+
+SELECT DISTINCT ON(2)
+       top_chart.company,
+       tp.genre_id,
+       name
+FROM top_chart
+JOIN (SELECT genre_id,
+             MAX(count) as max
+      FROM top_chart
+      GROUP BY genre_id ) g_top ON top_chart.genre_id = g_top.genre_id
+                                       AND top_chart.count = g_top.max
+JOIN top_track tp ON top_chart.company = tp.company
+                         AND top_chart.genre_id = tp.genre_id;

Week3/solutions/sixth_task.sql

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+
+-- Выполнить сегментацию клиентов по количеству букв а в его фамилии - Группа 'A' - 0 букв, 'B' - 1 буква,
+-- 'C' - 2 буквы, 'D' - 3 и более букв. Для каждой группы вывести количество уникальных типов медиа,
+-- купленных клиентами по трекам из различных жанров метала.
+-- Вывести имя группы и количество уникальных типов медиа
+-- (segment_group, cnt)
+
+select segment_group,
+       COUNT(distinct media_type_id)
+FROM invoice i JOIN invoice_line il on i.invoice_id = il.invoice_id
+    JOIN track ON il.track_id = track.track_id
+        JOIN genre g on track.genre_id = g.genre_id
+            JOIN (SELECT customer_id,
+                         last_name,
+                         CASE
+                            WHEN CHAR_LENGTH(last_name) - CHAR_LENGTH(REPLACE(last_name, 'a', '')) = 0 THEN 'A'
+                            WHEN CHAR_LENGTH(last_name) - CHAR_LENGTH(REPLACE(last_name, 'a', '')) = 1 THEN 'B'
+                            WHEN CHAR_LENGTH(last_name) - CHAR_LENGTH(REPLACE(last_name, 'a', '')) = 2 THEN 'C'
+                            WHEN CHAR_LENGTH(last_name) - CHAR_LENGTH(REPLACE(last_name, 'a', '')) > 2 THEN 'D'
+                        END as segment_group
+                 FROM customer) seg ON i.customer_id = seg.customer_id
+WHERE g.name LIKE '%Metal'
+GROUP BY segment_group;

Week3/solutions/tenth_task.sql

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+
+
+-- Для каждого альбома тип медиа определяется как наибольшая суммарный размер треков по жанру.
+-- Для каждого исполнителя вывести тип медиа, соответствующий наибольшему количеству типо медиа по альбомам,
+-- где они участвуют.
+-- Вывести имя исполнителя, тип медиа, количество альбомов с данными типом, общее количество альбомов.
+-- (artist_name, media_type_name, media_cnt, whole_cnt)
+
+WITH stat AS (
+    SELECT album_id,
+           media_type_id,
+           SUM(bytes) as size
+    FROM album al
+             JOIN artist a USING (artist_id)
+             JOIN track t USING (album_id)
+    GROUP BY album_id, media_type_id
+), media_stat AS (
+    SELECT album_id,
+           MAX(size) as sum
+    FROM stat
+    GROUP BY album_id
+), album_type AS (
+    SELECT st.album_id as album_id,
+           st.media_type_id as mt
+    FROM media_stat mst JOIN stat st ON mst.album_id = st.album_id
+                                            AND mst.sum = st.size
+), artist_stat AS (
+    SELECT artist_id,
+           mt,
+           count(album_id) as count
+    FROM album_type
+             JOIN album al USING (album_id)
+    GROUP BY artist_id, mt
+), artist_max AS (
+    SELECT artist_id,
+       MAX(count) as sum
+    FROM artist_stat
+    GROUP BY artist_id
+)
+
+SELECT art.name as artist_name,
+       met.name as media_type_name,
+       sum as media_cnt,
+       total as whole_cnt
+FROM artist_max am JOIN artist_stat ast ON am.artist_id = ast.artist_id
+                                               AND am.sum = ast.count
+JOIN (SELECT artist_id,
+             sum(count) as total
+      FROM artist_stat
+      GROUP BY artist_id) t ON am.artist_id = t.artist_id
+JOIN media_type met ON met.media_type_id = mt
+    JOIN artist art ON art.artist_id = am.artist_id;

Week3/solutions/third_task.sql

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+
+-- Вывести сотрудников, которые старше своего менеджера, но были наняты позже своего менеджера.
+-- Вывести полное имя сотрудника(Фамилия<Пробел>Имя), день рождения, дата найма,
+-- полное имя менеджера, день рождения, дата найма.
+-- (employee_full_name, employee_birth_date, employee_hire_date,
+-- manager_full_name, manager_birth_date, manager_hire_date)
+
+SELECT e.first_name || ' ' || e.last_name as employee_full_name,
+       e.birth_date as employee_birth_date,
+       e.hire_date as employee_hire_date,
+       m.first_name || ' ' || m.last_name as manager_full_name,
+       m.birth_date as manager_birth_date,
+       m.hire_date as manager_hire_date
+FROM employee e JOIN employee m ON m.employee_id = e.reports_to
+WHERE e.birth_date < m.birth_date AND e.hire_date > m.hire_date;

Week4/solutions/fifth_task.sql

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+
+-- Задание 5
+-- Для каждого сотрудника вывести общее количество купленных треков, проведенных им в заказе,
+-- самый часто используемый жанр треков, и общее число треков по минимальной стоимости.
+-- Если у сотрудника есть подчиненные, то все величины должны быть подсчитаны с учетом подчинения
+-- (то есть все заказы, которые обслужил сам сотрудник + заказы его подчиненных).
+-- Учесть потенциальную глубину вложенности.
+-- (employee_name, tracks_count, most_popular_genre, tracks_minimal_price).
+WITH source_date AS (
+    SELECT employee_id,
+           genre_id,
+           t.track_id as track_id,
+           t.unit_price
+    FROM employee e JOIN customer c
+    ON e.employee_id = c.support_rep_id
+        JOIN invoice i on c.customer_id = i.customer_id
+            JOIN invoice_line il on i.invoice_id = il.invoice_id
+                JOIN track t on il.track_id = t.track_id
+), price_amnt AS (
+    SELECT employee_id,
+           unit_price,
+           COUNT(track_id) as count
+    FROM source_date
+    GROUP BY employee_id, unit_price
+), rank_price AS (
+    SELECT employee_id,
+           unit_price,
+           count as count_price,
+           row_number() over (PARTITION BY employee_id ORDER BY unit_price, count DESC) as rn
+    FROM price_amnt
+), genre_amnt AS (
+    SELECT employee_id,
+           genre_id,
+           COUNT(track_id) as count_genre
+    FROM source_date
+    GROUP BY employee_id, genre_id
+), rank_genre AS (
+    SELECT
+           employee_id,
+           genre_id,
+           count_genre,
+           row_number() over (PARTITION BY employee_id ORDER BY count_genre DESC) as rn
+    FROM genre_amnt
+), worker_stat AS (SELECT DISTINCT ON(1, 2) employee_id as eid,
+                                            COUNT(track_id) OVER (PARTITION BY employee_id) as tracks_count,
+                                            rg.genre_id as most_popular_genre,
+                                            count_price as tracks_minimal_price
+                   FROM source_date sd
+    JOIN (SELECT * FROM rank_price WHERE rn = 1) rp USING (employee_id)
+        JOIN (SELECT * FROM rank_genre WHERE rn = 1) rg USING (employee_id)
+), manager_stat AS (
+    SELECT reports_to as eid,
+           SUM(tracks_count) as tracks_count,
+           mode() within group (order by most_popular_genre) most_popular_genre,
+           SUM(tracks_minimal_price) as tracks_minimal_price
+    FROM worker_stat ws JOIN employee e1 ON ws.eid = e1.employee_id
+    GROUP BY reports_to
+), directory_stat AS (
+    SELECT e.reports_to as eid,
+       SUM(tracks_count) as tracks_count,
+       mode() within group (order by most_popular_genre) most_popular_genre,
+       SUM(tracks_minimal_price) as tracks_minimal_price
+    FROM manager_stat ms
+        JOIN employee e ON e.employee_id = ms.eid
+    GROUP BY employee_id
+), general_report AS (
+    SELECT *
+    FROM manager_stat
+    UNION
+    SELECT *
+    FROM directory_stat
+    UNION
+    SELECT *
+    FROM worker_stat
+)
+
+SELECT first_name || ' ' || last_name as employee_name,
+       tracks_count,
+       g.name as most_popular_genre,
+       tracks_minimal_price
+FROM general_report gr JOIN employee e
+    ON gr.eid = e.employee_id
+        JOIN genre g ON gr.most_popular_genre = g.genre_id

Week4/solutions/first_task.sql

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+-- Для каждого артиста вывести все жанры, которые есть в его песнях, и для каждого жанра вывести
+-- наиболее продолжительную песню. (artist_name, genre_name, track_name, track_length)
+
+WITH stat AS (
+    SELECT art.name as artist_name,
+           g.name as genre_name,
+           t.name as track_name,
+           milliseconds as track_length,
+           row_number() over (partition by art.artist_id, genre_id order by milliseconds DESC) as rn
+    FROM artist art JOIN album a USING (artist_id)
+        JOIN track t USING (album_id)
+            JOIN genre g USING (genre_id)
+)
+
+SELECT artist_name,
+       genre_name,
+       track_name,
+       track_length
+FROM stat
+WHERE rn = 1