1 --建立測試環境 CREATE TABLE table1(a VARCHAR(10),b VARCHAR(10),c VARCHAR(10));
--插入數據 INSERT INTO table1 VALUES('2004','12','storea');
INSERT INTO table1 VALUES('2005','07','storea');
INSERT INTO table1 VALUES('2004','11','storeb');
INSERT INTO table1 VALUES('2005','06','storeb'); COMMIT;
2 取分組記錄的第一條
解法一
SELECT a.* FROM table1 a INNER JOIN ( SELECT MAX(a + b)TIME FROM table1 GROUP BY c ) b ON a.a + a.b = b.TIME
解法二
SELECT * FROM scott.table1 a WHERE NOT EXISTS (SELECT 1 FROM scott.table1 b WHERE a.a=b.a AND TO_NUMBER(a.b)<TO_NUMBER(b.b));
解法三
SELECT * FROM (SELECT t.*, RANK() OVER (PARTITION BY t.a ORDER BY t.b DESC) AS drank FROM table1 t) a WHERE drank=1
引申
SELECT t.*,SUM(b) OVER (PARTITION BY t.a) AS aaa, SUM(b) OVER () AS bbb FROM table1 t ORDER BY t.a,t.b
－－－－－－
SELECT d.department_id , e.last_name, e.salary, RANK() OVER (PARTITION BY e.department_id ORDER BY e.salary) AS drank FROM employees e, departments d WHERE e.department_id = d.department_id AND d.department_id IN ('60', '90');