使用SQLite和内部联接查找唯一对和对匹配数

Question

如果我们有以下两个表，“教师”和“类”，我们如何使用INNER JOIN语句来生成最终表

1. “teacher1_id”和“teacher2_id”代表教授同一班级至少2次的教师的所有独特组合
2. “pair_count”表示“teacher1_id”和“teacher2_id”教过的相同类的数量
3. “average_teacher_rating”表示“teacher1_id”和“teacher2_id”教授的那些相同课程的平均评分？

“老师”表

teachers

class_id    teacher_id    teacher_name    class_count
1111        1234          Smith           2
1111        2345          Jones           4
2222        1234          Smith           2
2222        2345          Jones           4
1111        5678          Taylor          2
3333        2345          Jones           4
4444        5678          Taylor          2
4444        2345          Jones           4

“班级”表

classes

id        class_name      rating
1111      Math            60
2222      Biology         70
3333      Psychology      50
4444      Physics         80
5555      Chemistry       30
6666      Economics       60
7777      Computing       70
8888      Statistics      90

我们希望决赛桌看起来像：

teacher1_id      teacher2_id     pair_count   average_teacher_rating
1234             2345            2            65           
2345             5678            2            70

我在sqlite中尝试了以下代码，但它没有给出所需的结果（如上面的最终表中所示）。

SELECT 
    t.teacher1_id
    , t.teacher2_id
    , COUNT(t.teacher1_id = t.teacher2_id)
    , (c.rating1 + c.rating2) / COUNT(t.teacher1_id = t.teacher2_id)
FROM teachers t
INNER JOIN classes c on c.id = t.class_id
WHERE COUNT(t.teacher1_id = t.teacher2_id) > 1;

Answer 1

尝试：

select 
    t1.teacher_id as t1
    , t2.teacher_id as t2
    , count()
    , avg(c.rating)
from teachers t1 inner join teachers as t2
on t1.class_id = t2.class_id and t1.teacher_id < t2.teacher_id
left join classes c on c.id = t1.class_id
group by t1, t2
having count()>1;

见sqlfiddle。

Answer 2

我相信以下内容可能符合您的要求： -

WITH cte1 AS
    (
        SELECT 
            t1.teacher_id AS teacher1_id, 
            t1.teacher_name AS t1name,
            t2.teacher_id AS teacher2_id, 
            t2.teacher_name AS t2name, 
            c.class_name,
            avg(c.rating) AS average_teacher_rating,
            count(c.class_name) AS pair_count
        FROM teachers AS t1
        LEFT JOIN classes AS c ON t1.class_id = c.id
        INNER JOIN teachers AS t2 ON t1.class_id = t2.class_id AND t1.teacher_id < t2.teacher_id
        GROUP BY teacher1_id,teacher2_id
        HAVING pair_count > 1
    )
SELECT teacher1_id, teacher2_id, pair_count, average_teacher_rating FROM cte1
;

这导致： -

• 由于不需要子查询，因此可以简化上述内容。但是，附加值可能证明是有用的。

即它可能是： -

SELECT 
    t1.teacher_id AS teacher1_id, 
    t2.teacher_id AS teacher2_id, 
    avg(c.rating) AS average_teacher_rating,
    count(c.class_name) AS pair_count
FROM teachers AS t1
JOIN teachers AS t2 ON t1.class_id = t2.class_id AND t1.teacher_id < t2.teacher_id
JOIN classes AS c ON t1.class_id = c.id
GROUP BY teacher1_id,teacher2_id
HAVING pair_count > 1
;