I Postgres 9.4 eller nyere skal du bruge det samlede FILTER
mulighed. Typisk reneste og hurtigste:
SELECT category
, count(*) FILTER (WHERE question1 = 0) AS zero
, count(*) FILTER (WHERE question1 = 1) AS one
, count(*) FILTER (WHERE question1 = 2) AS two
FROM reviews
GROUP BY 1;
Detaljer for FILTER
klausul:
- Samlet kolonner med yderligere (særskilte) filtre
Hvis du vil have det kort :
SELECT category
, count(question1 = 0 OR NULL) AS zero
, count(question1 = 1 OR NULL) AS one
, count(question1 = 2 OR NULL) AS two
FROM reviews
GROUP BY 1;
Flere syntaksvarianter:
- Er SUM hurtigere eller COUNT for absolut ydeevne?
Korrekt krydstabuleringsforespørgsel
crosstab()
giver den bedste ydeevne og er kortere for lange lister af muligheder:
SELECT * FROM crosstab(
'SELECT category, question1, count(*) AS ct
FROM reviews
GROUP BY 1, 2
ORDER BY 1, 2'
, 'VALUES (0), (1), (2)'
) AS ct (category text, zero int, one int, two int);
Detaljeret forklaring:
- PostgreSQL krydstabulatorforespørgsel