【问渠哪得清如许-数据分析】学习笔记-下

【问渠哪得清如许】是我的一个系列，数据分析的内容，随着我的学习会陆续把我的笔记上传，如果你也正在学习相关内容，可以参考我的学习路径。

【问渠哪得清如许-数据分析】学习笔记-上-CSDN博客

另外还有【问渠哪得清如许-产品经理】是阅读产品经理相关书籍的笔记。

【问渠哪得清如许-产品经理】阅读笔记《产品经理方法论------构建完整的产品知识体系（第2版）》上-CSDN博客

表名称和字段介绍

接着上次每天写完的题目

改正如下：

复制代码

SELECT *
FROM (
    SELECT sc.*,
           ROW_NUMBER() OVER(PARTITION BY cid ORDER BY score DESC) AS rank_s
    FROM sc
) t
WHERE rank_s < 4
ORDER BY cid, rank_s;

6、列出每个学生（sid和性别）的：上课总时长，数学上课总时长，理科上课总时长，男生上课总时长。

使用CASE WHEN + 聚合函数

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SELECT 
    s.sid,
    s.s_sex,
    -- 上课总时长（分钟）
    SUM(EXTRACT(EPOCH FROM (c.end_time - c.create_time))/60) AS total_minutes,
    -- 数学上课总时长（课程编号02或课程名称包含数学）
    SUM(CASE WHEN c.cid = '02' OR c.c_name LIKE '%数学%' 
             THEN EXTRACT(EPOCH FROM (c.end_time - c.create_time))/60 
             ELSE 0 END) AS math_minutes,
    -- 理科上课总时长（物理、化学、生物等）
    SUM(CASE WHEN c.c_type = '理科' OR c.c_name IN ('物理', '化学', '生物')
             THEN EXTRACT(EPOCH FROM (c.end_time - c.create_time))/60 
             ELSE 0 END) AS science_minutes,
    -- 男生上课总时长（学生性别为男生）
    SUM(CASE WHEN s.s_sex = 'M' 
             THEN EXTRACT(EPOCH FROM (c.end_time - c.create_time))/60 
             ELSE 0 END) AS male_minutes
FROM stu s
LEFT JOIN cour_t ct ON s.sid = ct.sid
LEFT JOIN cour c ON ct.cid = c.cid
GROUP BY s.sid, s.s_sex
ORDER BY s.sid;

7、统计每门课程的学生选修人数（超过5人的才统计）。

使用GROUP BY + HAVING

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SELECT 
    c.cid,
    c.c_name,
    COUNT(DISTINCT sc.sid) AS student_count
FROM cour c
LEFT JOIN sc ON c.cid = sc.cid
GROUP BY c.cid, c.c_name
HAVING COUNT(DISTINCT sc.sid) > 5
ORDER BY student_count DESC;

8、查询任何一门课程成绩在70分以上的学生姓名、课程名称和分数。

使用INNER JOIN + WHERE

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SELECT 
    s.s_name AS student_name,
    c.c_name AS course_name,
    sc.score
FROM sc
JOIN stu s ON sc.sid = s.sid
JOIN cour c ON sc.cid = c.cid
WHERE sc.score > 70
ORDER BY s.s_name, sc.score DESC;

9、查询每个学生的月平均上课时间

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SELECT 
    s.sid,
    s.s_name,
    ROUND(AVG(monthly_hours), 2) AS avg_monthly_hours
FROM (
    SELECT 
        ct.sid,
        DATE_TRUNC('month', ct.create_time) AS month,
        SUM(EXTRACT(EPOCH FROM (ct.end_time - ct.create_time))/3600) AS monthly_hours
    FROM cour_t ct
    GROUP BY ct.sid, DATE_TRUNC('month', ct.create_time)
) t
JOIN stu s ON t.sid = s.sid
GROUP BY s.sid, s.s_name
ORDER BY avg_monthly_hours DESC;

10、查询每个学生月最大连续上课天数

使用窗口函数 + 分组计算

复制代码

WITH daily_attendance AS (
    -- 先按天去重，避免同一天多次上课
    SELECT DISTINCT
        sid,
        DATE(create_time) AS attend_date,
        DATE_TRUNC('month', create_time) AS month
    FROM cour_t
),
date_gaps AS (
    -- 使用窗口函数标记连续分组
    SELECT 
        sid,
        month,
        attend_date,
        attend_date - ROW_NUMBER() OVER (
            PARTITION BY sid, month 
            ORDER BY attend_date
        )::int AS date_group
    FROM daily_attendance
),
continuous_days AS (
    -- 计算每个连续组的天数
    SELECT 
        sid,
        month,
        COUNT(*) AS continuous_days,
        MIN(attend_date) AS start_date,
        MAX(attend_date) AS end_date
    FROM date_gaps
    GROUP BY sid, month, date_group
)
-- 取每个学生每月的最大连续天数
SELECT 
    s.sid,
    s.s_name,
    TO_CHAR(c.month, 'YYYY-MM') AS month,
    MAX(c.continuous_days) AS max_continuous_days,
    -- 可选：显示最长连续上课的起止日期
    MAX(c.continuous_days) || '天 (' || 
    MIN(c.start_date) FILTER (WHERE c.continuous_days = MAX(c.continuous_days) OVER (PARTITION BY c.sid, c.month)) || 
    ' 至 ' || 
    MAX(c.end_date) FILTER (WHERE c.continuous_days = MAX(c.continuous_days) OVER (PARTITION BY c.sid, c.month)) || 
    ')' AS details
FROM continuous_days c
JOIN stu s ON c.sid = s.sid
GROUP BY s.sid, s.s_name, c.month
ORDER BY s.sid, c.month;