【051】row_number() over() 查询各个用户最长的连续登陆天数？

♣题目部分       查询各个用户最长的连续登陆天数？♣答案部分
row_number() over() 解说例一：      row_number() over() :简单的说row_number()从1开始，为每一条分组记录返回一个数字。例如首先在在线数据库中创建表，插入上图左边源表数据CREATE TABLE IF NOT EXISTS `loadrecord` (  `uid` int(6) unsigned NOT NULL,  `loadtime` varchar(200) NOT NULL,  PRIMARY KEY (`uid`,`loadtime`)) DEFAULT CHARSET=utf8;INSERT INTO `loadrecord` (`uid`, `loadtime`) VALUES  ('201', '2017/1/1'),  ('201', '2017/1/2'),  ('202', '2017/1/2'),  ('202', '2017/1/3'),  ('203', '2017/1/3'),  ('201', '2017/1/4'),  ('202', '2017/1/4'),  ('201', '2017/1/5'),  ('202', '2017/1/5'),  ('201', '2017/1/6'),  ('203', '2017/1/6'),  ('203', '2017/1/7');    其中，select uid,(row_number() over(partition by uid order by uid ) ) AS row_number_uid from loadrecord;返回为升序以后的每条uid记录返回一个序号：row_number() over()分组排序功能：在使用 row_number() over()函数时候，over()里头的分组以及排序的执行晚于 where 、group by、  order by 的执行。
例二：
create table TEST_ROW_NUMBER_OVER(       id varchar(10) not null,       name varchar(10) null,       age varchar(10) null,       salary int null);select * from TEST_ROW_NUMBER_OVER t; insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(1,'a',10,8000);insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(1,'a2',11,6500);insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(2,'b',12,13000);insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(2,'b2',13,4500);insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(3,'c',14,3000);insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(3,'c2',15,20000);insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(4,'d',16,30000);insert into TEST_ROW_NUMBER_OVER(id,name,age,salary) values(5,'d2',17,1800);
对查询结果进行排序（无分组）：
select id,name,age,salary,row_number()over(order by salary desc) rnfrom TEST_ROW_NUMBER_OVER t
根据id分组排序：
select id,name,age,salary,row_number()over(partition by id order by salary desc) `rank`from TEST_ROW_NUMBER_OVER t;
找出每一组中序号为一的数据？
select * from (select id,name,age,salary,row_number()over(partition by id order by salary desc) `rank`from TEST_ROW_NUMBER_OVER ) t where `rank` <2;
排序找出年龄在13岁到16岁数据，按salary排序：
select id,name,age,salary,row_number()over(order by salary desc)  `rank`from TEST_ROW_NUMBER_OVER t where age between '13' and '16';
#结果中 rank 的序号，其实就表明了 over(order by salary desc) 是在where age between and 后执行的
题目答案解析：
#第一步#判断连续的核心是row_number,因为row_number是连续的#所以day-row_number，如果值是恒定的，说明也是连续的，反之一定会变化#简单的说row_number()从1开始，为每一条分组记录返回一个数字，#这里的row_number() over(partition by uid uid 是先把uid列升序，#再为升序以后的每条uid记录返回一个序号。select uid,(day(loadtime) - row_number() over(partition by uid order by uid ) ) AS cnt from loadrecord;
#第二步#第一步得到的结果还不是很明显，需要分组用count()计数，取得不同连续值的次数SELECT uid,count(*) FROM(SELECT uid,(DAY(loadtime)-row_number() over(PARTITION BY uid ORDER BY uid) ) AS cnt FROM loadrecord) AS a GROUP BY uid,cnt;
#第三步#取得最大连续值#采用类似部门最高薪的方法 select max() group by SELECT uid, MAX(cnt) FROM(SELECT uid,count(*) AS cnt FROM(SELECT uid,(DAY(loadtime)-row_number() over(PARTITION BY uid ORDER BY uid) ) AS cnt FROM loadrecord) AS aGROUP BY uid,cnt) AS b GROUP BY uid;SELECT * from loadrecord WHERE row_number() over(partition by uid order by uid) ) AS cnt ;

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