SQL連接+計數

我已經很困在大樓的查詢，這是統計的特殊情況SQL連接+計數

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}

查詢嘗試檢查數據的完整性。應該沒有重複。

我需要得到一個查詢與每老師在學院的學生數量。

這個查詢工作絕對沒問題，直到我說我試圖計算每個院系不同的學生（他們確實是，在我的情況下，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

來源

2016-05-31 Alexey Abraham

你加入'老師'兩次？樣本數據和期望的結果將有所幫助，並說明你想要做什麼。 –

嗨戈登，其實我們在我們的系統中有幾個問題： - 不當的'老師刪除'功能，只刪除老師，沒有依賴關係（faculties_teachers）。因此，當有人想要獲得教師的所有教師時，他們會看到很多空行，這是錯誤的。 - 不當的「導師導師」的功能......這使教師的數據翻倍...... 因此，在1個查詢中，我想「全部殺死」:) 現在我有一些大量的工作，但後來我會解決這個問題。我認爲它很有用。 –

呵呵 - 感謝我的同事

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

來源

2016-05-31 17:32:09

SQL連接+計數

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