| jablan |
30. 09. 2009. 18:42 |
Brojanje grupisanih rekorda
Evo skripta za kreiranje tabele:
Kôd:
CREATE TABLE items (
a character varying(10),
b character varying(10),
total integer,
id integer NOT NULL
);
INSERT INTO items (a, b, total, id) VALUES ('a', 'b', 10, 1);
INSERT INTO items (a, b, total, id) VALUES ('a', 'c', 10, 2);
INSERT INTO items (a, b, total, id) VALUES ('a', 'd', 10, 3);
INSERT INTO items (a, b, total, id) VALUES ('b', 'd', 10, 4);
INSERT INTO items (a, b, total, id) VALUES ('b', 'd', 10, 5);
INSERT INTO items (a, b, total, id) VALUES ('c', 'd', 10, 6);
INSERT INTO items (a, b, total, id) VALUES ('c', 'c', 10, 7);
INSERT INTO items (a, b, total, id) VALUES ('c', 'b', 10, 8);
INSERT INTO items (a, b, total, id) VALUES ('c', 'd', 10, 9);
INSERT INTO items (a, b, total, id) VALUES ('c', 'b', 10, 10);
INSERT INTO items (a, b, total, id) VALUES ('a', 'b', 10, 11);
I upit:
Kôd:
test=# select a, b, sum(total) from items group by a, b;
a | b | sum
---+---+-----
b | d | 20
a | c | 10
c | b | 20
a | d | 10
c | d | 20
a | b | 20
c | c | 10
(7 rows)
Elem, meni treba upit koji vraća COUNT(*) za prethodni upit (dakle 7), ali bez korišćenja podupita, tj. bez da se ceo gornji upit wrapuje u SELECT COUNT(*) FROM ... |
