Practice exercise #3此頁上組的最基本的例子說明如何做到這一點。
CREATE TABLE customers
( customer_id number(10) not null,
customer_name varchar2(50) not null,
city varchar2(50),
CONSTRAINT customers_pk PRIMARY KEY (customer_id)
);
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7001, 'Microsoft', 'New York');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7002, 'IBM', 'Chicago');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7003, 'Red Hat', 'Detroit');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7004, 'Red Hat', 'New York');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7005, 'Red Hat', 'San Francisco');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7006, 'NVIDIA', 'New York');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7007, 'NVIDIA', 'LA');
INSERT INTO customers (customer_id, customer_name, city)
VALUES (7008, 'NVIDIA', 'LA');
解決方案:
下面的SQL語句將返回不同的城市數量在客戶表中的每個CUSTOMER_NAME:
SELECT customer_name, COUNT(DISTINCT city) as "Distinct Cities"
FROM customers
GROUP BY customer_name;
這將返回以下結果集:
CUSTOMER_NAME Distinct Cities
IBM 1
Microsoft 1
NVIDIA 2
Red Hat 3
這聽起來很像功課...如果是這樣,請標記爲這樣。 – 2010-06-15 16:04:39
不,它不是。我正在研究一個項目,現在。 – Mikae 2010-06-15 16:55:45