MYSQL加入三張表

我有一個簡單的多學校管理系統，我正在努力獲得教師總數以及特定學校的學生總數。我的表結構如下：MYSQL加入三張表

 teachers 
-------------------------- 
id | schoolid | Name | etc... 
-------------------------- 
1 | 1  | Bob | 
2 | 1  | Sarah| 
3 | 2  | John | 


     students 
-------------------------- 
id | schoolid | Name | etc... 
-------------------------- 
1 | 1  | Jack | 
2 | 1  | David| 
3 | 2  | Adam | 

     schools 
-------------------------- 
id |  Name  | etc... 
--------------------------- 
1 | River Park High | 
2 | Stirling High |

我可以指望剛剛全部用下面的查詢教師：

SELECT COUNT(a.id) AS `totalteachers` 
FROM teachers a 
LEFT JOIN schools b ON a.schoolid = b.id WHERE b.id = '1'

，同樣我可以用下面的查詢次數，教師數量：

SELECT COUNT(a.id) AS `totalstudents` 
FROM students a 
LEFT JOIN schools b ON a.schoolid = b.id WHERE b.id = '1'

但是，我試圖結合這兩個查詢來得到這樣一個簡單的結果：

totalstudents | totalteachers 
-------------------------------- 
     2  |  2

我曾嘗試以下：

SELECT COUNT(a.id) as `totalteachers`, COUNT(c.id) as `totalstudents` 
FROM teachers a 
LEFT JOIN schools b ON a.schoolid = b.id 
LEFT JOIN students c ON c.schoolid=b.id WHERE b.id = '5'

來源

2017-06-22 Bruno

多個左連接的問題是它爲每個學生生成每個老師的附加記錄;人爲地誇大你的計數

有四種方式來解決這個問題：（IMO最好什麼是安德魯骨所做的那樣）

只需選擇直列沒有連接，因此該計數不膨脹。（最希望在我的腦海裏，因爲它很容易維護）

SELECT (SELECT COUNT(a.id) AS `totalteachers` 
     FROM teachers a 
     WHERE A.SchoolID = '1') as TotalTeachers 
    , (SELECT COUNT(a.id) AS `totalstudents` 
     FROM students a 
     WHERE a.SchoolID = '1') as TotalStudents

使用子查詢，首先得到計數加入之前，然後再加入。由於count總是1，因此交叉連接可以工作。

SELECT totalTeachers, totalStudents 
FROM (SELECT COUNT(a.id) AS `totalteachers` 
     FROM teachers a 
     LEFT JOIN schools b 
     ON a.schoolid = b.id 
     WHERE b.id = '1') 
CROSS JOIN (SELECT COUNT(a.id) AS `totalstudents` 
      FROM students a 
      LEFT JOIN schools b ON a.schoolid = b.id 
      WHERE b.id = '1')

使用關鍵字的計數範圍內不同以免重複計數和否定人工通貨膨脹（至少希望在我的腦海，因爲這隱藏了人工計數增加）

SELECT COUNT(distinct a.id) as `totalteachers`, COUNT(distinct c.id) as `totalstudents` 
FROM teachers a 
LEFT JOIN schools b ON a.schoolid = b.id 
LEFT JOIN students c ON c.schoolid=b.id WHERE b.id = '5'

另一種方式要使用窗口函數，但是這些在mySQL中不可用。

來源

2017-06-22 13:29:21 xQbert

這是我正面臨的確切問題，人爲誇大計數！我選擇了你的第三個解決方案（使用不同的），它的功能就像一個魅力！感謝您抽出寶貴時間回答我的問題，並幫助我理解:)我很感激。 – Bruno

SELECT COUNT(t.id) AS TotalTeachers, COUNT(st.id) AS TotalStudents 
FROM schools s 
INNER JOIN teachers t 
ON s.id = t.schoolid 
INNER JOIN students st 
ON s.id = st.schoolid

試試這個SQL。我沒有嘗試，但它應該工作。

來源

2017-06-22 13:26:51

你可以做這樣的事情

SELECT 
    id, name, s.total AS totalstudents, t.total AS totalteachers 
FROM schools 
    JOIN (SELECT schoolid, COUNT(id) AS total FROM teachers GROUP BY schoolid) 
    AS t ON t.schoolid = id   
    JOIN (SELECT schoolid, COUNT(id) AS total FROM students GROUP BY schoolid) 
    AS s ON s.schoolid = id

那麼你可以添加where id = 2或任何限制的學校。

來源

2017-06-22 13:31:11

感謝您的回答安德魯，我欣賞它:) – Bruno

很好的答案。這使得學生可以輕鬆地設置where子句。 – xQbert

MYSQL加入三張表

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