sqlrvergroupby分组后，取每个分组的第⼀条数据
1 --建⽴测试环境 CREATE TABLE table1(a VARCHAR(10),b VARCHAR(10),c VARCHAR(10));
--插⼊数据 INSERT INTO table1 VALUES('2004','12','storea');
四川大学怎么样INSERT INTO table1 VALUES('2005','07','storea');
INSERT INTO table1 VALUES('2004','11','storeb');
INSERT INTO table1 VALUES('2005','06','storeb'); COMMIT;糖尿病人的饮食
2 取分组记录的第⼀条
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
最新域名升级解法⼆
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SELECT * FROM scott.table1 a WHERE NOT EXISTS (SELECT 1 FROM scott.table1 b WHERE a.a=b.a AND TO_NUMBER(a.b)
黄玫瑰花<TO_NUMBER(b.b));
解法三
道德经第九章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');檀字怎么读
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