在“sc”表中查询选修了两门及以上课程的学生学号和选课数

查询语句如下：

SELECT sc.sno, COUNT(*) as num
FROM sc
GROUP BY sc.sno
HAVING COUNT(*) >= 2;

解释：

首先从 `sc` 表中选取学号 `sno` 和对应的选课成绩信息，然后按照学号分组，统计每个学生选课数目（即分组后每个学生对应的记录数），并将其命名为 `num`。最后通过 `HAVING` 子句筛选出选课数目大于等于 2 的学生，即选修了两门及以上课程的学生。

相关问题

sql查询选修了2以上课程的学生学号及选课门数。

以下是SQL查询选修了2以上课程的学生学号及选课门数的语句：

SELECT stu_id, COUNT(*) AS course_num
FROM sc
GROUP BY stu_id
HAVING COUNT(*) >= 2
ORDER BY stu_id ASC;

该语句首先从sc表中选取学生学号和课程号，然后按照学生学号进行分组，统计每个学生选修的课程门数，最后筛选出选修课程门数大于等于2的学生，并按照学生学号升序排序。

(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;