数据库的 SQL 查询找到学生的 nr

Question

我是数据库的初学者，我有以下几点：我需要展示那些不正确学生数量最多的课程（所以那些还没有带来这本书的人以及那些迟于 date_to_return 带来的人）

1. 表格课程：课程代码、课程名称。

2. 表格pupils_on_courses：学生id、课程代码。

3. 表格学生：学生id

4. 表books_out_on_loan：学生id、isbn、date_borrowed、date_to_return_date_returned

我试图找到每个部分的不正确学生的数量，但现在我坚持找到最大值。

我认为错误在于我计算date_returned和date_to_return的条件

select 
    c.course_code, c.course_name, count (bb.pupil_id) 
from 
    courses c 
inner join 
    pupils_on_courses ps on c.course_code = ps.course_code
inner join 
    pupils p on p.pupil_id = ps.pupil_id
inner join 
    books_out_on_loan bb on p.pupil_id = bb.pupil_id
group by 
    c.course_code, c.course_name
having 
    count (bb.pupil_id) = (select max(count(*))
                           from pupil p 
                           inner join books_out_loan bb on p.pupil_id = bb.pupil_id
                           where bb.date_returned is null 
                              or bb.date_returned > date_to_return
                           group by c.course_code, c.course_name) 

感谢您的帮助

编辑： 一些示例数据将是：

Answer 1

select 
    c.course_code, min(c.course_name) course_name,
    count(distinct pc.pupil_id) as max_pupil_cnt
from 
    courses c inner join pupils_on_courses pc
        on ps.course_code = c.course_code
    inner join books_out_on_loan b
        on b.pupil_id = pc.pupil_id
where b.date_returned is null or b.date_returned > b.date_to_return
group by c.course_code
having count (distinct pc.pupil_id) >= all (
    select count(distinct pc2.pupil_id)
    from pupils_on_courses pc2 inner join books_out_loan b2. 
        on b2.pupil_id = pc2.pupil_id
    where b2.date_returned is null or b2.date_returned > b2.date_to_return
    group by pc2.course_code
);

有现代方法可以做到这一点，而无需重复逻辑。我的猜测是你需要坚持基本的东西，因为这看起来像是家庭作业。

Answer 2

如果您发布示例数据，我们可以更轻松地提供帮助。

如果您想要计算学生人数而不是未购买书籍或延迟退货的数量，这可能对您有用：

select 
    c.course_code, c.course_name, 
    count (distinct case when bb.date_returned is null or bb.date_returned > date_to_return
      then p.pupil_id else null end) as NoncorrectStudents
from 
    courses c 
inner join 
    pupils_on_courses ps on c.course_code = ps.course_code
inner join 
    pupils p on p.pupil_id = ps.pupil_id
left join 
    books_out_on_loan bb on p.pupil_id = bb.pupil_id
group by 
    c.course_code, c.course_name

一旦你有了这个，你应该能够按不正确学生的数量进行排序。请在有机会时发布数据，以便验证。

Answer 3

您可能已经在WHERE 子句中过滤了“不正确”。

那么它不会返回那些只有“正确”的。
但是，由于您只想要那些具有最大“不正确”的人，因此应该足够公平。

而且您可能还需要仅选择那些必须在当前日期之前返回的那些。

SELECT c.course_code, c.course_name
, COUNT(DISTINCT bb.pupil_id) AS non_correct_pupils
FROM books_out_on_loan bb 
INNER JOIN pupils_on_courses ps 
  ON ps.pupil_id = bb.pupil_id
INNER JOIN courses c 
  ON c.course_code = ps.course_code
WHERE bb.date_to_return < CAST(current_timestamp AS DATE)
AND (bb.date_returned IS NULL OR bb.date_returned > bb.date_to_return)
GROUP BY c.course_code, c.course_name
ORDER BY COUNT(DISTINCT bb.pupil_id) DESC

DISTINCT 仅用于计算唯一的学生 ID。

要获得最高排名，试试这个

SELECT * 
FROM
(
SELECT c.course_code, c.course_name
, COUNT(DISTINCT bb.pupil_id) AS non_correct_pupils
, DENSE_RANK() OVER (ORDER BY COUNT(DISTINCT bb.pupil_id) DESC) AS Rnk
FROM books_out_on_loan bb 
INNER JOIN pupils_on_courses ps 
  ON ps.pupil_id = bb.pupil_id
INNER JOIN courses c 
  ON c.course_code = ps.course_code
WHERE bb.date_to_return < CAST(current_timestamp AS DATE)
AND (bb.date_returned IS NULL OR bb.date_returned > bb.date_to_return)
GROUP BY c.course_code, c.course_name
)
WHERE Rnk = 1