小让の码场SQL226 将employees表的所有员工的last_name和first_name拼接起来作为Name
题目描述
现有员工表employees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 | Chirstian | 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 |
请将employees表的所有员工的last_name和first_name拼接起来作为Name，中间以一个空格区分。
(注：sqllite,字符串拼接为 || 符号，不支持concat函数，mysql支持concat函数)。
输入描述：
无
输出描述：
| Name |
|---|
| Facello Georgi |
| Simmel Bezalel |
| Bamford Parto |
| Koblick Chirstian |
| Maliniak Kyoichi |
| Preusig Anneke |
| Zielinski Tzvetan |
| Kalloufi Saniya |
| Peac Sumant |
| Piveteau Duangkaew |
| Sluis Mary |
SQL Schema
sql
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');答案
sql
SELECT CONCAT(last_name, ' ', first_name) AS `Name`
FROM employees;