在学生信息数据库中查询各门课程的最高成绩的学生的姓名及其成绩

这是一个技术问题，可以通过编写SQL语句进行查询。以下是一种可能的SQL语句：

SELECT students.name, MAX(scores.score)
FROM students
JOIN scores ON students.id = scores.student_id
GROUP BY scores.course_id

请注意，此SQL查询假定您具有名为“students”的表，其中包含学生的姓名和ID，并且具有名为“scores”的表，其中包含学生ID、课程ID和成绩。你可以使用以上SQL语句来查找最高分，并且会将其与每个课程ID相关联的学生姓名返回。

相关问题

MySQL以数据库user_db及数据表student、teacher、class、course、sc为基础（11）查询选修“数据库原理”的学生的最高成绩、最低成绩、平均成绩； （12）查询平均成绩在60分以上的课程及其平均成绩； （13）查询平均成绩大于“高莉”的平均成绩的学生学号、姓名及平均成绩； （14）查询每门课程中成绩最低的学生的学号、姓名、性别、院系； （15）查询只被一名学生选修的课程的课程号、课程名。

11.

SELECT MAX(score) AS max_score, MIN(score) AS min_score, AVG(score) AS avg_score 
FROM sc 
WHERE course_name = '数据库原理';

12.

SELECT course_id, course_name, AVG(score) AS avg_score 
FROM sc 
GROUP BY course_id, course_name 
HAVING AVG(score) > 60;

13.

SELECT s.stu_id, s.stu_name, AVG(score) AS avg_score 
FROM student s, sc c 
WHERE s.stu_id = c.stu_id 
GROUP BY s.stu_id, s.stu_name 
HAVING AVG(score) > (SELECT AVG(score) FROM sc WHERE stu_name = '高莉');

14.

SELECT s.stu_id, s.stu_name, s.gender, s.department, c.course_name, MIN(score) AS min_score 
FROM student s, sc c 
WHERE s.stu_id = c.stu_id 
AND score = (SELECT MIN(score) FROM sc WHERE course_name = c.course_name)
GROUP BY c.course_name, s.stu_id, s.stu_name, s.gender, s.department;

15.

SELECT course_id, course_name 
FROM sc 
GROUP BY course_id, course_name 
HAVING COUNT(DISTINCT stu_id) = 1;

(1)查询选修课程号为C01，并且成绩及格的学生学号、姓名和系别名称。 (2)查询计算机学院选修C03课程的学生姓名、课程名称以及这门课程的成绩。(3)在Reports表中查询至少选修课了程号为C01和C02的学生学号和姓名。(4)查询既没有选修课程号C01，也没有选修课程号C02的学生的学号和姓名。(5) 查询每个同学基本信息及其选修课程的情况。(6)查询每个学生及其选修课程的情况(若某一个学生没有选课，只输出学生的基本情况信息，其选课信息为空值)。 (7)查询选修了课程名为“GIS开发”的学生学号和姓名。(8) 查询没有选修C02课程的学生学号和姓名。 (9)查询所有课程成绩均大于90分的学生学号和姓名。(10)查询选修了4门及以上课程的学生的学号、姓名和选修课程数。(11) 查询选修全部课程的学生姓名，相当于查询这样的学生，没有一门课程是他不选的。 (12) 查询平均成绩大于等于黄河同学的平均成绩的学生的学号、姓名和平均成绩。(13)查询每门课程获得最高分的同学的学号、姓名、课程名称和成绩，并按照最高成绩从高到低进行排序。

很抱歉，我是一个语言模型AI，无法查询数据库。但是我可以为您提供一些SQL语句示例，您可以根据需要自行修改：

1. SELECT student.id, student.name, department.name
FROM student, department, score, course
WHERE student.id = score.id AND score.course_id = course.id AND
course.id = 'C01' AND score.score >= 60 AND student.dept_id = department.id;

2. SELECT student.name, course.name, score.score
FROM student, course, score
WHERE student.id = score.id AND score.course_id = course.id AND
course.id = 'C03' AND student.dept_id = '计算机学院';

3. SELECT DISTINCT student.id, student.name
FROM student, score
WHERE student.id = score.id AND
score.course_id IN ('C01', 'C02');

4. SELECT student.id, student.name
FROM student
WHERE student.id NOT IN
(SELECT id FROM score WHERE course_id IN ('C01', 'C02'));

5. SELECT student.id, student.name, department.name, course.name, score.score
FROM student
LEFT JOIN score ON student.id = score.id
LEFT JOIN course ON score.course_id = course.id
LEFT JOIN department ON student.dept_id = department.id;

6. SELECT student.id, student.name, department.name, course.name, score.score
FROM student
LEFT JOIN score ON student.id = score.id
LEFT JOIN course ON score.course_id = course.id
LEFT JOIN department ON student.dept_id = department.id
GROUP BY student.id;

7. SELECT student.id, student.name
FROM student, score, course
WHERE student.id = score.id AND score.course_id = course.id AND
course.name = 'GIS开发';

8. SELECT student.id, student.name
FROM student
WHERE student.id NOT IN
(SELECT id FROM score WHERE course_id = 'C02');

9. SELECT student.id, student.name
FROM student, score
WHERE student.id = score.id
GROUP BY student.id
HAVING MIN(score.score) > 90;

10. SELECT student.id, student.name, COUNT(*) AS course_num
FROM student, score
WHERE student.id = score.id
GROUP BY student.id
HAVING course_num >= 4;

11. SELECT student.name
FROM student, course
WHERE NOT EXISTS
(SELECT * FROM course
WHERE NOT EXISTS
(SELECT * FROM score
WHERE score.course_id = course.id AND score.id = student.id));

12. SELECT s1.id, s1.name, AVG(s1.score)
FROM student s1, student s2, score sc1, score sc2
WHERE s1.id = sc1.id AND s2.id = sc2.id AND
s2.name = '黄河' AND AVG(sc2.score) <= AVG(sc1.score)
GROUP BY s1.id
HAVING AVG(sc1.score) >= AVG(sc2.score);

13. SELECT score.id, student.name, course.name, score.score
FROM score, student, course
WHERE score.id = student.id AND score.course_id = course.id AND
score.score = (SELECT MAX(score.score) FROM score WHERE score.course_id = course.id)
ORDER BY score.score DESC;