用sql写“在选课表中查询选修课程门数最多的学生学号及选修课门数最少的学生学号。各列标题显示为学号、选修课程数。”
时间: 2023-06-08 21:07:30 浏览: 1712
你好,根据你的要求,以下是使用 SQL 语句查询选修课程门数最多的学生学号及选修课门数最少的学生学号的过程:
```
SELECT 学号, COUNT(选修课程) AS 选修课程数
FROM 选课表
GROUP BY 学号
HAVING COUNT(选修课程) = (SELECT MAX(选修课程数) FROM (SELECT COUNT(选修课程) AS 选修课程数 FROM 选课表 GROUP BY 学号) AS T)
OR COUNT(选修课程) = (SELECT MIN(选修课程数) FROM (SELECT COUNT(选修课程) AS 选修课程数 FROM 选课表 GROUP BY 学号) AS T)
```
这条 SQL 语句中,首先使用 SELECT 语句查询出每个学生的选修课门数,并使用 GROUP BY 语句对结果进行分组。随后,使用 HAVING 语句查询出选修课门数最多和最少的学生学号。为了避免重复查询,使用了嵌套查询查询出选修课门数的最大值和最小值。最终,SELECT 语句的结果中仅包含学号和选修课程数两列,并按照选修课程数进行排序。
希望能对您有所帮助。
相关问题
sql查询选修课最多的学生(学号)
这里假设有三个表:students(学生表)、courses(课程表)和选课表(course_selection)。选课表包含学生学号和课程编号。
可以使用以下SQL查询选修课最多的学生(学号):
```
SELECT student_id
FROM course_selection
GROUP BY student_id
ORDER BY COUNT(course_id) DESC
LIMIT 1;
```
这个查询会按照选修课程数量倒序排列学生,并返回选修课程数量最多的学生的学号。如果有多个学生选修课程数量相同,则会返回其中任意一个学生的学号。
MySQL以数据库user_db及数据表student、teacher、class、course、sc为基础(11)查询信息技术学院的必修课和限选课的数量及平均学分; (12)查询每名学生的选课数量及平均成绩,按照平均成绩降序排列; (13)查询各门课程编号、选课人数及其及格率; (14)查询选课人数前五名的课程; (15)查询选修课程数最多和最少的学生的学号及选课数; 2、SQL子查询的应用; (1)查询信息技术学院年龄最大的学生的学号、姓名、性别; (2)查询每个院系年龄最大的学生的学号、姓名、性别、院系及年龄; (3)查询和“崔晓”一个院系的其他学生的学号、姓名、院系; (4)查询平均成绩在60分以上的课程及其平均成绩; (5)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; (6)查询课程的学分高于全体课程平均学分的课程编号、名称、学分及所属院系,查询结果按照学分降序排列; (7)查询至少有一名学生选修的所有课程的课程号及课程名;
11. 查询信息技术学院的必修课和限选课的数量及平均学分:
```
SELECT
SUM(CASE WHEN course_type = '必修' THEN 1 ELSE 0 END) AS required_course_count,
SUM(CASE WHEN course_type = '限选' THEN 1 ELSE 0 END) AS elective_course_count,
AVG(CASE WHEN course_type IN ('必修', '限选') THEN credit END) AS avg_credit
FROM
class
WHERE
department = '信息技术学院';
```
12. 查询每名学生的选课数量及平均成绩,按照平均成绩降序排列:
```
SELECT
sc.student_id,
COUNT(*) AS course_count,
AVG(score) AS avg_score
FROM
sc
JOIN
student ON student.id = sc.student_id
GROUP BY
sc.student_id
ORDER BY
avg_score DESC;
```
13. 查询各门课程编号、选课人数及其及格率:
```
SELECT
course_id,
COUNT(*) AS student_count,
SUM(CASE WHEN score >= 60 THEN 1 ELSE 0 END) / COUNT(*) AS pass_rate
FROM
sc
GROUP BY
course_id;
```
14. 查询选课人数前五名的课程:
```
SELECT
course_id,
COUNT(*) AS student_count
FROM
sc
GROUP BY
course_id
ORDER BY
student_count DESC
LIMIT 5;
```
15. 查询选修课程数最多和最少的学生的学号及选课数:
```
SELECT
student_id,
COUNT(*) AS course_count
FROM
sc
GROUP BY
student_id
HAVING
COUNT(*) = (SELECT MAX(cnt) FROM (SELECT COUNT(*) AS cnt FROM sc GROUP BY student_id) AS t)
OR COUNT(*) = (SELECT MIN(cnt) FROM (SELECT COUNT(*) AS cnt FROM sc GROUP BY student_id) AS t);
```
1. SQL子查询的应用:
(1) 查询信息技术学院年龄最大的学生的学号、姓名、性别:
```
SELECT
id, name, gender
FROM
student
WHERE
department = '信息技术学院' AND age = (SELECT MAX(age) FROM student WHERE department = '信息技术学院');
```
(2) 查询每个院系年龄最大的学生的学号、姓名、性别、院系及年龄:
```
SELECT
s.id, s.name, s.gender, s.age, s.department
FROM
student s
JOIN
(SELECT department, MAX(age) AS max_age FROM student GROUP BY department) t
ON
s.department = t.department AND s.age = t.max_age;
```
(3) 查询和“崔晓”一个院系的其他学生的学号、姓名、院系:
```
SELECT
id, name, department
FROM
student
WHERE
department = (SELECT department FROM student WHERE name = '崔晓') AND name != '崔晓';
```
(4) 查询平均成绩在60分以上的课程及其平均成绩:
```
SELECT
course_id,
AVG(score) AS avg_score
FROM
sc
WHERE
course_id IN (SELECT course_id FROM sc GROUP BY course_id HAVING AVG(score) >= 60)
GROUP BY
course_id;
```
(5) 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:
```
SELECT
id, name, AVG(score) AS avg_score
FROM
student
JOIN
sc ON student.id = sc.student_id
WHERE
AVG(score) > (SELECT AVG(score) FROM student JOIN sc ON student.id = sc.student_id WHERE name = '高莉')
GROUP BY
id, name;
```
(6) 查询课程的学分高于全体课程平均学分的课程编号、名称、学分及所属院系,查询结果按照学分降序排列:
```
SELECT
course.id, course.name, course.credit, course.department
FROM
course
WHERE
credit > (SELECT AVG(credit) FROM course)
ORDER BY
credit DESC;
```
(7) 查询至少有一名学生选修的所有课程的课程号及课程名:
```
SELECT
id, name
FROM
course
WHERE
id IN (SELECT DISTINCT course_id FROM sc);
```
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