Unterabfragen in SELECT

Datenbearbeitung in SQL

Mona Khalil

Data Scientist, Greenhouse Software

SELECT

Gibt einen einzelnen Wert zurück
- Aggregierte Werte für den Vergleich mit einzelnen Werten hinzufügen

Wird für Rechenaufgaben eingesetzt
- Abweichung vom Durchschnitt

Unterabfragen in SELECT

Bestimme Gesamtzahl der Spiele über alle Saisons hinweg

SELECT COUNT(id) FROM match;

Unterabfragen in SELECT

SELECT
  season,
  COUNT(id) AS matches,
  12837 as total_matches
FROM match
GROUP BY season;

| season    | matches | total_matches |
|-----------|---------|---------------|
| 2011/2012 | 3220    | 12837         |
| 2012/2013 | 3260    | 12837         |
| 2013/2014 | 3032    | 12837         |
| 2014/2015 | 3325    | 12837         |

Unterabfragen in SELECT

SELECT
  season,
  COUNT(id) AS matches,
  (SELECT COUNT(id) FROM match) as total_matches
FROM match
GROUP BY season;

| season    | matches | total_matches |
|-----------|---------|---------------|
| 2011/2012 | 3220    | 12837         |
| 2012/2013 | 3260    | 12837         |
| 2013/2014 | 3032    | 12837         |
| 2014/2015 | 3325    | 12837         |

Rechnen mit SELECT-Unterabfragen

SELECT AVG(home_goal + away_goal) 
FROM match
WHERE season = '2011/2012';

2.72

SELECT
  date,
  (home_goal + away_goal) AS goals,
  (home_goal + away_goal) - 2.72 AS diff
FROM match
WHERE season = '2011/2012';

Unterabfragen in SELECT

SELECT
  date,
  (home_goal + away_goal) AS goals,
  (home_goal + away_goal) - 
     (SELECT AVG(home_goal + away_goal) 
      FROM match
      WHERE season = '2011/2012') AS diff
FROM match
WHERE season = '2011/2012';

| date       | goals | diff              |
|------------|-------|-------------------|
| 2011-07-29 | 3     | 0.28354037267081  |
| 2011-07-30 | 2     | -0.71645962732919 |
| 2011-07-30 | 4     | 1.28354037267081  |
| 2011-07-30 | 1     | -1.71645962732919 |

SELECT-Unterabfragen – was du beachten solltest

Muss einen EINZELNEN Wert zurückgeben
- Sonst wird Fehler erzeugt
Prüfe die Position deiner Filter
- Filtere Hauptanfrage und Unterabfrage richtig!

SELECT
    date,
    (home_goal + away_goal) AS goals,
    (home_goal + away_goal) - 
       (SELECT AVG(home_goal + away_goal) 
        FROM match
        WHERE season = '2011/2012') AS diff
FROM match
WHERE season = '2011/2012';

Lass uns üben!

Datenbearbeitung in SQL