id	device_id	gender	age	university	gpa
1	2138	male	21	北京大学	3.4
2	3214	male	23	复旦大学	4
3	6543	female	20	北京大学	3.2
4	2315	female	23	浙江大学	3.6
5	5432	male	25	山东大学	3.8
6	2131	male	28	北京师范大学	3.3

你的查询应返回以下结果：

device_id	gpa	age
6534	3.2	20
2131	3.3	28
2138	3.4	21
2315	3.6	23
5432	3.8	25
3214	4	23

示例1

输入:

drop table if exists user_profile;
CREATE TABLE `user_profile` (
`id` int NOT NULL,
`device_id` int NOT NULL,
`gender` varchar(14) NOT NULL,
`age` int ,
`university` varchar(32) NOT NULL,
`gpa` float);
INSERT INTO user_profile VALUES(1,2138,'male',21,'北京大学',3.4);
INSERT INTO user_profile VALUES(2,3214,'male',23,'复旦大学',4.0);
INSERT INTO user_profile VALUES(3,6543,'female',20,'北京大学',3.2);
INSERT INTO user_profile VALUES(4,2315,'female',23,'浙江大学',3.6);
INSERT INTO user_profile VALUES(5,5432,'male',25,'山东大学',3.8);
INSERT INTO user_profile VALUES(6,2131,'male',28,'北京师范大学',3.3);

输出:

6543|3.200|20
2131|3.300|28
2138|3.400|21
2315|3.600|23
5432|3.800|25
3214|4.000|23

原站题解

select device_id,gpa,age from user_profile order by gpa,age;

select device_id, gpa, age from user_profile order by gpa ,age;

select device_id,gpa,age from user_profile order by gpa,age asc

select device_id,gpa,age from user_profile order by gpa,age;

select device_id,gpa,age FROM user_profile ORDER BY gpa,age