SQL连接+计数

我已经很困在大楼的查询，这是统计的特殊情况SQL连接+计数

我有appearences：

teachers {id, email, first_name, last_name,...} 
faculties {id, name} 
pupils {id, name, teacher_id} //every pupil has their own curator 
faculties_teachers {id, teacher_id, faculty_id}

查询尝试检查数据的完整性。应该没有重复。

我需要得到一个查询与每老师在学院的学生数量。

这个查询工作绝对没问题，直到我说我试图计算每个院系不同的学生（他们确实是，在我的情况下，唯一的:)）

SELECT u1.id, u2.id, u1.owner, u2.owner, u1.email, u1.first_name, u2.last_name, u1.name_of_site, faculties_teachers.faculty_id, faculties.name, 
     (SELECT COUNT(pupils.*) FROM pupils as p WHERE faculties.id = pupils.faculty_id) 
FROM `teachers` AS u1 
JOIN `teachers` AS u2 ON u1.email = u2.email 
JOIN `faculties_teachers` AS ft ON u1.id = ft.teacher_id 
JOIN `faculties` ON faculties.id = ft.faculty_id 
JOIN `pupils` ON faculties.id = pupils.faculty_id 
WHERE u1.id < u2.id 
ORDER BY `u1`.`id` ASC

来源

2016-05-31 Alexey Abraham

你加入'老师'两次？样本数据和期望的结果将有所帮助，并说明你想要做什么。 –

嗨戈登，其实我们在我们的系统中有几个问题： - 不当的'老师删除'功能，只删除老师，没有依赖关系（faculties_teachers）。因此，当有人想要获得教师的所有教师时，他们会看到很多空行，这是错误的。 - 不当的“导师导师”的功能......这使教师的数据翻倍...... 因此，在1个查询中，我想“全部杀死”:) 现在我有一些大量的工作，但后来我会解决这个问题。我认为它很有用。 –

呵呵 - 感谢我的同事

SELECT u1.id, u2.id, u1.owner, u2.owner, u1.email, u1.first_name, u2.last_name, u1.name_of_site, faculties_teachers.faculty_id, faculties.name, COALESCE(pupil_counts_by_faculties.pupil_cnt, 0) AS pupil_cnt 
FROM `teachers` AS u1 
JOIN `teachers` AS u2 ON u1.email = u2.email 
JOIN `faculties_teachers` ON u1.id = faculties_teachers.urid 
JOIN `faculties` ON faculties.id = faculties_teachers.faculty_id 
LEFT JOIN (
    SELECT `faculty_id`, COUNT(*) AS pupil_cnt 
    FROM `pupils` 
    GROUP BY `faculty_id` 
) AS pupil_counts_by_faculties ON `faculties`.`id` = pupil_counts_by_faculties.faculty_id 
WHERE u1.id < u2.id

来源

2016-05-31 17:32:09

SQL连接+计数

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