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LeetCode_sql_day22（1112.每位学生的最高成绩）

news 2025/6/4 2:32:47

描述：1112.每位学生的最高成绩

表：Enrollments

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| student_id    | int     |
| course_id     | int     |
| grade         | int     |
+---------------+---------+
(student_id, course_id) 是该表的主键（具有唯一值的列的组合）。
grade 不会为 NULL。

编写解决方案，找出每位学生获得的最高成绩和它所对应的科目，若科目成绩并列，取 course_id 最小的一门。查询结果需按 student_id 增序进行排序。

查询结果格式如下所示。

示例 1：

输入：
Enrollments 表：
+------------+-------------------+
| student_id | course_id | grade |
+------------+-----------+-------+
| 2          | 2         | 95    |
| 2          | 3         | 95    |
| 1          | 1         | 90    |
| 1          | 2         | 99    |
| 3          | 1         | 80    |
| 3          | 2         | 75    |
| 3          | 3         | 82    |
+------------+-----------+-------+
输出：
+------------+-------------------+
| student_id | course_id | grade |
+------------+-----------+-------+
| 1          | 2         | 99    |
| 2          | 2         | 95    |
| 3          | 3         | 82    |
+------------+-----------+-------+

数据准备：

Create table If Not Exists Enrollments (student_id int, course_id int, grade int)
Truncate table Enrollments
insert into Enrollments (student_id, course_id, grade) values ('2', '2', '95')
insert into Enrollments (student_id, course_id, grade) values ('2', '3', '95')
insert into Enrollments (student_id, course_id, grade) values ('1', '1', '90')
insert into Enrollments (student_id, course_id, grade) values ('1', '2', '99')
insert into Enrollments (student_id, course_id, grade) values ('3', '1', '80')
insert into Enrollments (student_id, course_id, grade) values ('3', '2', '75')
insert into Enrollments (student_id, course_id, grade) values ('3', '3', '82');

分析：

①根据每个学生分组根据分数降序分数相同根据课程号升序排名

select *,
row_number()
over (partition by student_id order by grade desc,course_id) r1
from Enrollments

②取排名为1 的数据
select *, row_number() over (partition by student_id order by grade desc,course_id) r1
from Enrollments;

代码：

with t1 as (select *, row_number() over (partition by student_id order by grade desc,course_id) r1 from Enrollments)
select student_id, course_id, grade
from t1
where r1 = 1;