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查找最晚入职员工的所有信息

[编程题]查找最晚入职员工的所有信息

热度指数：799700 时间限制：C/C++ 1秒，其他语言2秒空间限制：C/C++ 32M，其他语言64M
算法知识视频讲解

有一个员工employees表简况如下:

emp_no	birth_date	first_name	last_name	gender	hire_date
10001	1953-09-02	Georgi	Facello	M	1986-06-26
10002	1964-06-02	Bezalel	Simmel	F	1985-11-21
10003	1959-12-03	Parto	Bamford	M	1986-08-28
10004	1954-05-01	Christian	Koblick	M	1986-12-01
10005	1955-01-21	Kyoichi	Maliniak	M	1989-09-12'
10006	1953-04-20	Anneke	Preusig	F	1989-06-02
10007	1957-05-23	Tzvetan	Zielinski	F	1989-02-10
10008	1958-02-19	Saniya	Kalloufi	M	1994-09-15
10009	1952-04-19	Sumant	Peac	F	1985-02-18
10010	1963-06-01	Duangkaew	Piveteau	F	1989-08-24
10011	1953-11-07	Mary	Sluis	F	1990-01-22

请你查找employees里最晚入职员工的所有信息，以上例子输出如下:

emp_no	birth_date	first_name	last_name	gender	hire_date
10008	1958-02-19	Saniya	Kalloufi	M	1994-09-15

示例1

输入

drop table if exists  `employees` ; 
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');
INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');
INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10');
INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15');
INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18');
INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24');
INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');

输出

10008|1958-02-19|Saniya|Kalloufi|M|1994-09-15

算法知识视频讲解

Mysql

可以k

SELECT

emp_no,

birth_date,

first_name,

last_name,

gender,

hire_date

FROM employees

ORDER BY hire_date DESC

LIMIT 1;

发表于 2025-03-06 16:19:40 回复(0)

Mysql

offer多多的小刺猬很奔放

drop table if exists `employees` ;
CREATE TABLE `employees` (
`emp_no` int(11) NOT NULL,
`birth_date` date NOT NULL,
`first_name` varchar(14) NOT NULL,
`last_name` varchar(16) NOT NULL,
`gender` char(1) NOT NULL,
`hire_date` date NOT NULL,
PRIMARY KEY (`emp_no`));
INSERT INTO employees VALUES(10001,'1953-09-02','Georgi','Facello','M','1986-06-26');
INSERT INTO employees VALUES(10002,'1964-06-02','Bezalel','Simmel','F','1985-11-21');
INSERT INTO employees VALUES(10003,'1959-12-03','Parto','Bamford','M','1986-08-28');
INSERT INTO employees VALUES(10004,'1954-05-01','Chirstian','Koblick','M','1986-12-01');
INSERT INTO employees VALUES(10005,'1955-01-21','Kyoichi','Maliniak','M','1989-09-12');
INSERT INTO employees VALUES(10006,'1953-04-20','Anneke','Preusig','F','1989-06-02');
INSERT INTO employees VALUES(10007,'1957-05-23','Tzvetan','Zielinski','F','1989-02-10');
INSERT INTO employees VALUES(10008,'1958-02-19','Saniya','Kalloufi','M','1994-09-15');
INSERT INTO employees VALUES(10009,'1952-04-19','Sumant','Peac','F','1985-02-18');
INSERT INTO employees VALUES(10010,'1963-06-01','Duangkaew','Piveteau','F','1989-08-24');
INSERT INTO employees VALUES(10011,'1953-11-07','Mary','Sluis','F','1990-01-22');

发表于 2025-03-04 20:54:33 回复(0)

Mysql

牛客606568169号

sql200

select * from employees where hire_date=(select max(hire_date) from employees);

发表于 2025-02-18 10:52:31 回复(0)

Mysql

牛客239692899号

用基础函数，当然也可以子查询，两种方法都试一试

SELECT

emp_no

,birth_date

,first_name

,last_name

,gender

,hire_date

FROM employees

ORDER BY hire_date DESC

LIMIT 1;

发表于 2025-02-14 22:51:02 回复(0)

Mysql

徐文达

防止有多人同一天入职的情况，使用了 IN 而不是 =

SELECT
    *
FROM
    employees
WHERE
    hire_date IN
    (
        SELECT
            MAX(hire_date)
        FROM
            employees
    );

发表于 2025-01-04 20:04:00 回复(0)

Mysql

牛客943742290号

SELECT a.emp_no,

a.birth_date,

a.first_name,

a.last_name,

a.gender,

a.hire_date

FROM(SELECT

emp_no,

birth_date,

first_name,

last_name,

gender,

hire_date,

RANK() OVER(ORDER BY hire_date DESC) rank_date

FROM employees) a

WHERE a.rank_date =1

想复杂了，用窗口函数做出来的

发表于 2024-11-17 17:05:48 回复(0)

Mysql

Wuppny

select
    *
from
    employees
order by
    hire_date desc
limit
    1

发表于 2024-10-05 13:44:58 回复(0)

Mysql

lynn_z

select *
from employees
where hire_date in (
    select max(hire_date) as last_hire_date
    from employees)

关键点：找到最晚入职时间

因无法确认最晚入职是否只有1人，不能使用group by limit

发表于 2024-09-02 15:40:25 回复(0)

Mysql

牛客292332323号

select *
from employees
where hire_date=
(
select hire_date
from employees
order by hire_date desc
limit 1
)

发表于 2024-08-02 16:32:38 回复(0)

Mysql

牛客574028452号

select * from employees where hire_date=(select hire_date from employees order by hire_date desc limit 1);

发表于 2024-07-24 14:11:19 回复(0)

Mysql

刷牛客的孤勇者很野蛮

select *
from employees
where hire_date=(
    select max(hire_date) from employees
)

发表于 2024-07-17 16:22:26 回复(0)

Mysql

是汽水a

窗口函数方法：

with a as(
	select *,
	rank() over(order by hire_date desc) as rn
	from employees
)

select emp_no, birth_date, first_name, 
last_name, gender, hire_date
from a
where rn = 1

编辑于 2024-03-21 16:10:18 回复(0)

Mysql

小谷围做题家

select * from employees 
order by hire_date desc 
limit 1;

发表于 2024-02-05 11:37:01 回复(0)

Mysql

要暴富的牛肉丸很孤独

select

emp_no

,birth_date,first_name,last_name,gender,hire_date

from(

select *

,rank()over(partition by emp_no order by hire_date desc) posn

from employees

) re

where posn=1

limit 1

发表于 2024-01-26 11:59:12 回复(0)

Mysql

DylanWu92

select * from employees
where hire_date =
(select max(hire_date) from employees)

编辑于 2024-01-10 16:06:31 回复(0)

Mysql

牛客898716324号

遇事不决，开窗函数

select e.emp_no,
e.birth_date,
e.first_name,
e.last_name,
e.gender,
e.hire_date
from
(select emp_no,
birth_date,
first_name,
last_name,
gender,
hire_date,
dense_rank() over(order by hire_date desc) as rk
from employees) as e
where e.rk = 1

编辑于 2024-01-03 16:10:32 回复(0)