检测MySQL DB记录之间的关系

Question

假设我有3个数据库表：

1. 教师，包括以下字段：
   • ID
   • 名称
2. 科目，包括以下字段：
   • ID
   • 名称
3. 教师 - 科目，包括以下字段：
   • teacher_id
   • subject_id

因此，一些教师有一个共同的主题。

我已经考虑过用于查询的算法：

  

对于主题表中的每个记录：

     

    

在teachers_subjects表中选择计数     如果count大于1
    在teacher_subjects中，获取teacher_id，其中subject_id是s。

  

     

建立关系对。

我希望获得如下输出：

+----------+----------+---------------+
| Teacher1 | Teacher2 | Relationships |
+----------+----------+---------------+
| John     | Larry    |            10 |
| John     | Samantha |            12 |
| Samantha | Larry    |             9 |
| Ian      | Louis    |             3 |
+----------+----------+---------------+

如果我需要检索哪些科目建立了关系，我们需要得到类似的东西：

+----+----------+----------+
| id | teacher1 | teacher2 |
+----+----------+----------+
|  1 | John     | Larry    |
|  2 | John     | Samantha |
|  3 | Samantha | Larry    |
|  4 | Ian      | Louis    |
+----+----------+----------+

等等：

+-----+------------+
| id  |  subject   |
+-----+------------+
| 1   | math       |
| 1   | english    |
| 1   | science    |
| ... | ...        |
| ... | ...        |
| 4   | science    |
| 4   | literature |
| 4   | databases  |
+-----+------------+

以图形方式，我在通过网络浏览器查询的图表中表示这一点：



因此，当我悬停一条边时，它会显示关系信息。

我将通过php编程接收查询结果，这样的查询或存储过程会产生这样的必需输出？

Answer 1

试试这个（您应该在SQLFiddle中加载一些示例数据以便更好地进行测试）

SELECT t1.`name` AS teacher1, t2.`name` AS teacher2, count(*)
FROM teachers AS t1
JOIN teachers AS t2
  ON t1.id > t2.id
JOIN teacher_subjects AS ts1
  ON ts1.teacher_id = t1.id
JOIN teacher_subjects AS ts2
  ON ts2.teacher_id = t2.id
    AND ts2.subject_id = ts1.subject_id
GROUP BY teacher1, teacher2
ORDER BY COUNT(*) DESC;

Answer 2

这些是我对SQL的提议。

请记住，这些解决方案只是对以前的解决方案进行了一些修改。

第一张表

http://sqlfiddle.com/#!2/dc3f2/3

SELECT t1.name as 'Teacher1', t2.name as 'Teacher2', count(*) AS 'Relationships' 
FROM teachers t1 
JOIN teachers t2 ON t1.id > t2.id
JOIN teachers_subjects ts1 ON ts1.teacher_id = t1.id
JOIN teachers_subjects ts2 ON ts2.teacher_id = t2.id AND ts1.subject_id = ts2.subject_id
GROUP BY ts1.teacher_id, ts2.teacher_id;

第二张表

我认为它现在正在运作。

http://sqlfiddle.com/#!2/dc3f2/22

SELECT @rownum := @rownum + 1 as id, relations.Teacher1, relations.Teacher2
FROM (SELECT t1.name as 'Teacher1', t2.name as 'Teacher2'
FROM teachers t1 
JOIN teachers t2 ON t1.id > t2.id
JOIN teachers_subjects ts1 ON ts1.teacher_id = t1.id
JOIN teachers_subjects ts2 ON ts2.teacher_id = t2.id AND ts1.subject_id = ts2.subject_id
GROUP BY ts1.teacher_id, ts2.teacher_id) as relations,
(SELECT @rownum := 0) r;

第三张表

http://sqlfiddle.com/#!2/dc3f2/27

SELECT id, subject
FROM
(SELECT @rownum := @rownum + 1 as id, relations.Teacher1, relations.Teacher2, relations.id1, relations.id2
FROM (SELECT t1.name as 'Teacher1', t2.name as 'Teacher2', t1.id as 'id1', t2.id as 'id2'
FROM teachers t1 
JOIN teachers t2 ON t1.id > t2.id
JOIN teachers_subjects ts1 ON ts1.teacher_id = t1.id
JOIN teachers_subjects ts2 ON ts2.teacher_id = t2.id AND ts1.subject_id = ts2.subject_id
GROUP BY ts1.teacher_id, ts2.teacher_id) as relations,
(SELECT @rownum := 0) r) as rel
JOIN
(SELECT t1.id as 'id1', t2.id as 'id2', s.name as 'subject'
FROM teachers t1 
JOIN teachers t2 ON t1.id > t2.id
JOIN teachers_subjects ts1 ON ts1.teacher_id = t1.id
JOIN teachers_subjects ts2 ON ts2.teacher_id = t2.id AND ts1.subject_id = ts2.subject_id
JOIN subjects s ON ts1.subject_id = s.id) as rel_subjects
ON rel.id1 = rel_subjects.id1 AND rel.id2 = rel_subjects.id2
ORDER BY id

最后一件事。请记住，关系的伪键是非常多变的。这意味着如果您添加新行并删除一些，您将更改这些关系的ID。因此，不要出现任何错误，不能在数据库中的任何存储列中使用它们 - 只需在SELECT查询中使用它们。