10、select score.c_id,course.c_name,student.s_name,s_score from score join student on student.s_id=score.s_id join course on score.c_id='01' and course.c_id=score.c_id order by s_score desc limit 3; 关系代数表达式
时间: 2023-06-16 20:04:11 浏览: 49
这个 SQL 查询语句的关系代数表达式如下:
π score.c_id, course.c_name, student.s_name, score.s_score (σ score.c_id='01' (score ⨝ student ⨝ course)) ⨝ (score ⨝ student ⨝ course) ⨝ (score ⨝ student ⨝ course) -s_score
相关问题
select student.s_name from student, (select DISTINCT st.s_name from score sc inner JOIN student st ON st.s_id =sc.s_id INNER JOIN course c ON sc.c_id=c.c_id) demo1 LEFT JOIN student ON student.s_name = demo1.s_name
这个 SQL 查询语句的作用是查询选修了课程的学生姓名,表达式中使用了子查询和左连接。
子查询部分:
```
select DISTINCT st.s_name from score sc inner JOIN student st ON st.s_id =sc.s_id INNER JOIN course c ON sc.c_id=c.c_id
```
这个子查询查询了选修了课程的学生姓名,使用了 INNER JOIN 连接了三个表:score、student、course,其中 score 表和 course 表通过 c_id 进行连接,score 表和 student 表通过 s_id 进行连接,得到的结果是一个包含学生姓名的不重复集合。
主查询部分:
```
select student.s_name from student, demo1 LEFT JOIN student ON student.s_name = demo1.s_name
```
这个主查询从 student 表中查询学生姓名,使用了两个表:student 和 demo1,其中 demo1 是子查询的结果集,使用了 LEFT JOIN 进行连接,以保证即使 demo1 中没有某个学生的姓名,也能够查询到该学生的姓名。最终得到的结果是一个包含选修了课程的学生姓名的集合。
select a.s_id ,a.s_name ,count(b.c_id) as sum_course ,sum(b.s_score)as sum_score from student a left join score b on a.s_id=b.s_id GROUP BY a.s_id,a.s_name
Sure! Here is the SQL query to fulfill your request:
```sql
SELECT a.s_id, a.s_name, COUNT(b.c_id) AS sum_course, SUM(b.s_score) AS sum_score
FROM student a
LEFT JOIN score b ON a.s_id = b.s_id
GROUP BY a.s_id, a.s_name;
```
This query will retrieve the student ID (`s_id`), student name (`s_name`), the total number of courses taken by each student (`sum_course`), and the sum of scores obtained by each student (`sum_score`). The `LEFT JOIN` ensures that all students are included in the result, even if they don't have any scores recorded in the `score` table.
Let me know if you need any further assistance!
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