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经典SQL查询问题的练习第二天

news 来源：原创 2025/6/1 10:32:16

📝 第二天SQL打卡题目
表结构不变：
- `student(studentId, studentName)`
- `course(courseId, courseName, teacher)`
- `score(score, studentId, courseId)`

1. 查询未选修任何课程的学生姓名（考察：NOT EXISTS / LEFT JOIN + NULL）
```sql
-- 方案1：NOT EXISTS
SELECT studentName
FROM student s
WHERE NOT EXISTS (
SELECT 1 FROM score sc WHERE sc.studentId = s.studentId
);

-- 方案2：LEFT JOIN + IS NULL
SELECT s.studentName
FROM student s
LEFT JOIN score sc ON s.studentId = sc.studentId
WHERE sc.courseId IS NULL;
```

2. 查询每门课程成绩最高的学生姓名及分数（考察：窗口函数 RANK() / 子查询）
```sql
-- 窗口函数方案（推荐）
SELECT courseName, studentName, score
FROM (
SELECT
c.courseName,
s.studentName,
sc.score,
RANK() OVER (PARTITION BY sc.courseId ORDER BY sc.score DESC) AS rk
FROM score sc
JOIN student s ON sc.studentId = s.studentId
JOIN course c ON sc.courseId = c.courseId
) t
WHERE t.rk = 1;
```

3. 查询所有课程平均分高于全体学生总平均分的课程名称（考察：聚合函数嵌套 + HAVING）
```sql
SELECT
c.courseName,
AVG(sc.score) AS avg_score
FROM score sc
JOIN course c ON sc.courseId = c.courseId
GROUP BY c.courseId, c.courseName
HAVING AVG(sc.score) > (
SELECT AVG(score) FROM score -- 全体学生总平均分
);
```

4. 查询至少有一门课成绩超过90分的学生中，总成绩排名前5的学生（考察：子查询 + 复合条件排序）
```sql
SELECT
s.studentId,
s.studentName,
SUM(sc.score) AS total_score
FROM student s
JOIN score sc ON s.studentId = sc.studentId
WHERE s.studentId IN (
SELECT DISTINCT studentId
FROM score
WHERE score > 90 -- 至少一门课>90分
)
GROUP BY s.studentId, s.studentName
ORDER BY total_score DESC
LIMIT 5;
```

5. 查询每个学生成绩最高的课程及对应教师（考察：关联子查询 + 多表JOIN）
```sql
SELECT
s.studentName,
c.courseName,
c.teacher,
sc.score
FROM score sc
JOIN student s ON sc.studentId = s.studentId
JOIN course c ON sc.courseId = c.courseId
WHERE sc.score = (
SELECT MAX(score)
FROM score
WHERE studentId = sc.studentId -- 关联当前学生
);
```

6. 优化题：如何为`score`表设计索引以加速以下查询？（考察：索引设计原则）
```sql
-- 查询语句：
SELECT studentId, AVG(score)
FROM score
WHERE courseId = '0006'
GROUP BY studentId;
```
参考答案：
```sql
CREATE INDEX idx_course_student ON score(courseId, studentId, score);
```
原理：
- `WHERE courseId` 作为第一条件用等值查询
- `GROUP BY studentId` 需有序访问，故作为第二列
- 覆盖索引（包含`score`）避免回表查数据

题目一解释：
找那些"啥课都没选"的学生。想象一个班级点名册（student表），我们要找出名字从来没出现在任何课程成绩单（score表）上的学生。