id	uid	nick_name	achievement	level	job	register_time
1	1001	牛客1号	19	0	算法	2020-01-01 10:00:00
2	1002	牛客2号	1200	3	算法	2020-02-01 10:00:00
3	1003	牛客3号♂	22	0	算法	2020-01-02 10:00:00
4	1004	牛客4号	25	0	算法	2020-01-02 11:00:00
5	1005	牛客555号	4000	7	C++	2020-01-11 10:00:00
6	1006	666666	3000	6	C++	2020-11-01 10:00:00

请从中找到注册时间最早的3个人。由示例数据结果输出如下：

解释：按注册时间排序后选取前三名，输出其用户ID、昵称、注册时间。

示例1

输入:

drop table if exists user_info,exam_record;
CREATE TABLE user_info (
    id int PRIMARY KEY AUTO_INCREMENT COMMENT '自增ID',
    uid int UNIQUE NOT NULL COMMENT '用户ID',
    `nick_name` varchar(64) COMMENT '昵称',
    achievement int COMMENT '成就值',
    level int COMMENT '用户等级',
    job varchar(32) COMMENT '职业方向',
    register_time datetime COMMENT '注册时间'
)CHARACTER SET utf8 COLLATE utf8_general_ci;

INSERT INTO user_info(uid,`nick_name`,achievement,level,job,register_time) VALUES
  (1001, '牛客1', 19, 0, '算法', '2020-01-01 10:00:00'),
  (1002, '牛客2号', 1200, 3, '算法', '2020-02-01 10:00:00'),
  (1003, '牛客3号♂', 22, 0, '算法', '2020-01-02 10:00:00'),
  (1004, '牛客4号', 25, 0, '算法', '2020-01-02 11:00:00'),
  (1005, '牛客555号', 4000, 7, 'C++', '2020-01-11 10:00:00'),
  (1006, '666666', 3000, 6, 'C++', '2020-11-01 10:00:00');

输出:

1001|牛客1|2020-01-01 10:00:00
1003|牛客3号♂|2020-01-02 10:00:00
1004|牛客4号|2020-01-02 11:00:00

原站题解

SELECT  uid , nick_name,register_time from user_info order by register_time LIMIT 3;

select uid, nick_name, register_time
from
user_info
order by register_time
limit 3

select uid,nick_name,register_time from
(select uid,nick_name,register_time,
 ROW_NUMBER() over(order by register_time) rank_time from user_info) a
where rank_time<=3

select 
    uid,
    nick_name,
    register_time
from user_info
order by register_time
limit 3

select uid,nick_name,register_time from user_info 
order by register_time
limit 3