Combinarea tabelelor

Raportare în SQL

Tyler Pernes

Learning & Development Consultant

Raportul țintă

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     |
+----------+--------------------+-------+

Tabele relevante

Opțiunea A: JOIN primul, UNION al doilea

Pasul 1: Configurați interogarea de sus cu 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!

Opțiunea A: JOIN primul, UNION al doilea

Pasul 2: Configurați interogarea de jos + UNION cele două

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;

Opțiunea B: UNION primul, JOIN al doilea

Pasul 1: Creați UNION-ul inițial

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

Opțiunea B: UNION primul, JOIN al doilea

Pasul 2: Convertiți în subinterogare + 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;

Comparație

Opțiunea 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;

Opțiunea 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;

Concluzii cheie

Mai multe modalități de a crea același raport
Pas cu pas = mai ușor de depanat

Timp pentru interogări!

Raportare în SQL