Об'єднання таблиць

Звітування в SQL

Tyler Pernes

Learning & Development Consultant

Звіт за метою

Gold Medals by Demographic Group
(Western European Countries Only)
+----------+--------------------+-------+
| season   |  demographic_group | golds |
|----------+--------------------+-------|
| Winter   | Male Age 26+       | 13    |
| Winter   | Female Age 26+     | 8     |
| Summer   | Male Age 13-25     | 7     |
| Summer   | Female Age 13-25   | 6     |
| Winter   | Male Age 13-25     | 4     |
| Summer   | Male Age 26+       | 4     |
| Winter   | Female Age 13-25   | 4     |
| Summer   | Female Age 26+     | 2     |
+----------+--------------------+-------+

Доречні таблиці

Варіант A: спочатку JOIN, потім UNION

Крок 1: Налаштуйте верхній запит із JOIN

SELECT 
    athlete_id, 
    gender, 
    age, 
    gold  
FROM summer_games AS sg
JOIN athletes AS a 
ON sg.athlete_id = a.id;

Query ran successfully!

Варіант A: спочатку JOIN, потім UNION

Крок 2: Додайте нижній запит і UNION між ними

SELECT 
    athlete_id, 
    gender, 
    age, 
    gold  
FROM summer_games AS sg
JOIN athletes AS a 
ON sg.athlete_id = a.id
UNION ALL
SELECT 
    athlete_id, 
    gender, 
    age, 
    gold  
FROM winter_games AS wg
JOIN athletes AS a 
ON wg.athlete_id = a.id;

Варіант B: спочатку UNION, потім JOIN

Крок 1: Створіть початковий UNION

SELECT 
    athlete_id, 
    gold  
FROM summer_games AS sg
UNION
SELECT 
    athlete_id, 
    gold  
FROM winter_games AS wg;

Варіант B: спочатку UNION, потім JOIN

Крок 2: Перетворіть на підзапит і додайте JOIN

SELECT 
    athlete_id, 
    gender, 
    age, 
    gold  
FROM
    (SELECT 
         athlete_id, 
         gold  
    FROM summer_games AS sg
    UNION ALL
    SELECT athlete_id, gold  
    FROM winter_games AS wg) AS g
JOIN athletes AS a 
ON g.athlete_id = a.id;

Порівняння

Варіант A

SELECT 
    athlete_id, 
    gender, 
    age, 
    gold  
FROM summer_games AS sg
JOIN athletes AS a 
ON sg.athlete_id = a.id
UNION ALL
SELECT 
    athlete_id, 
    gender, 
    age, 
    gold  
FROM winter_games AS wg
JOIN athletes AS a 
ON wg.athlete_id = a.id;

Варіант B

SELECT 
    athlete_id, 
    gender, 
    age, 
    gold  
FROM
    (SELECT 
         athlete_id, 
         gold  
    FROM summer_games AS sg
    UNION ALL
    SELECT athlete_id, gold  
    FROM winter_games AS wg) AS g
JOIN athletes AS a 
ON g.athlete_id = a.id;

Висновки

Є кілька способів отримати той самий звіт
Покроковий підхід = легше знаходити помилки

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Звітування в SQL