Duplicatie in rapporten

Rapporteren in SQL

Tyler Pernes

Learning & Development Consultant

Waardoor ontstaat duplicatie?

SELECT p.id, SUM(points) AS points, SUM(matches_win) AS matches_won
FROM points AS p
JOIN matches AS m ON p.id = m.id
GROUP BY p.id;

Waardoor ontstaat duplicatie?

+-----+--------+--------------+
| id  | points | matches_won  |
|-----|--------|--------------|
| 1   | 156    | 10           |
+-----+--------+--------------+

¹ zorgt voor een points-waarde die drie keer te hoog is: hier 156.

Waardoor ontstaat duplicatie?

Tussentabel
+-----+------+--------------+---------+ 
| id  | year | matches_won  | points  |
|-----|------|--------------|---------|    
| 1   | 2016 | 5            | 52      |
| 1   | 2017 | 2            | 52      |  
| 1   | 2018 | 3            | 52      |
+-----+------+--------------+---------+

Waardoor ontstaat duplicatie?

Tussentabel
+-----+------+--------------+---------+ 
| id  | year | matches_won  | points  |
|-----|------|--------------|---------|    
| 1   | 2016 | 5            | 52      | <--
| 1   | 2017 | 2            | 52      | <-- SOM(points) = 52 x 3 = 156
| 1   | 2018 | 3            | 52      | <-- 
+-----+------+--------------+---------+

Manieren om duplicatie op te lossen

1. Verwijder aggregaties

SELECT p.id, points, SUM(matches_won) AS matches_won
FROM points AS p
JOIN matches AS m ON p.id = m.id
GROUP BY p.id, points;

+-----+--------+--------------+
| id  | points | matches_won  |
|-----|--------|--------------|
| 1   | 52     | 10           |
+-----+--------+--------------+

Manieren om duplicatie op te lossen

2. Voeg veld toe aan JOIN

SELECT p.id, SUM(points) AS points, SUM(matches_win) AS matches_won
FROM points AS p
JOIN matches AS m ON p.id = m.id AND p.year = m.year
GROUP BY p.id;

Manieren om duplicatie op te lossen

2. Voeg veld toe aan JOIN

SELECT p.id, SUM(points) AS points, SUM(matches_win) AS matches_won
FROM points AS p
JOIN matches AS m ON p.id = m.id AND p.year = m.year
GROUP BY p.id;

Manieren om duplicatie op te lossen

SELECT id, SUM(matches_won)
FROM matches
GROUP BY id;

+-----+--------------+
| id  | matches_won  |
|-----|--------------|
| 1   | 10           |
| 2   | 7            |
+-----+--------------+

Manieren om duplicatie op te lossen

3. Rollup met subquery

SELECT p.id, points, matches_won
FROM points AS p
JOIN 
    (SELECT id, SUM(matches_won) AS matches_won
    FROM matches
    GROUP BY id) AS m 
ON p.id = m.id;

Manieren om duplicatie op te lossen

Verwijder aggregaties
Voeg veld toe aan JOIN
Rollup met subquery

Duplicatie herkennen

Waarde in originele tabel:

SELECT SUM(points) AS total_points
FROM points;

total_points = 52

Waarde in query:

SELECT SUM(points) AS total_points
FROM   
    (SELECT p.id, SUM(points) AS points
    FROM points AS p
    JOIN matches AS m ON p.id = m.id
    GROUP BY p.id) AS subquery;

total_points = 156

Doel van de hoofdstuk

Laten we oefenen!

Rapporteren in SQL