複雜的MYSQL從5個表中選擇

Question

我有五個表格存儲設施，員工，完成的單位，項目和班次。

我需要確定每個設施的項目的生產力（完成累計數和累積小時數），如果該設施有登錄項目的小時數。該設施存儲在員工表中。

這裏是我的表結構，（減少到必要的字段爲這個問題）：

facilities 
+------------+ 
| id   | 
+------------+ 
| label  | 
+------------+ 

employees 
+------------+ 
| id   | 
+------------+ 
| facility | 
+------------+ 

completes 
+------------+ 
| id   | 
+------------+ 
| project | 
+------------+ 
| completes | 
+------------+ 
| employee | 
+------------+ 

projects 
+------------+ 
| id   | 
+------------+ 
| title  | 
+------------+ 

shifts 
+------------+ 
| id   | 
+------------+ 
| project | 
+------------+ 
| length  | 
+------------+ 
| employee | 
+------------+ 

當我遇到麻煩的是選擇生產的工廠，當存儲與員工，不與轉變或＃完成。也許這是一個設計缺陷？班次和完成表是否也有設施專欄？當員工已經確定設施時，擁有設施欄似乎是多餘的。

SELECT p.title, SUM(c.completes), SUM(s.length) 
FROM projects as p 
LEFT JOIN completes as c ON c.project = p.id 
LEFT JOIN shifts as s ON s.project = p.id 
GROUP BY p.id, shifts.employee.facility; 

GROUP BY聲明當然是無效的。但它表明了我想要完成的事情。我還想在聲明中選擇設施標籤。我如何解決這個問題？

在此先感謝您提供的任何見解！

Answer 1

這是罕見的情況之一，我會找到一個CROSS JOIN是最好的辦法：

SELECT p.title, f.label, SUM(c.completes), SUM(s.length) 
FROM facilities as f   
CROSS JOIN projects as p 
JOIN employees as e ON f.id = e.facility 
LEFT JOIN completes as c ON c.employee = e.id AND c.project = p.id 
LEFT JOIN shifts as s ON s.employee = e.id AND s.project = p.id 
WHERE c.id IS NOT NULL OR s.id IS NOT NULL 
GROUP BY p.id, p.title, f.id, f.label 

這本身不是一個不好的設計，但它確實讓人很難，你可以」在這裏使用典型的親子關係，因爲completes和shifts表中實際上有2位父母。

另一種方法是使用沒有LEFT JOIN的UNION。這可能會返回一個更快的結果，這取決於行的數量目前在facilities和projects表：

SELECT p.title, 
     f.label, 
     SUM(CASE aggregated.type WHEN 'completes' THEN units ELSE 0 END), 
     SUM(CASE aggregated.type WHEN 'shifts' THEN units ELSE 0 END) 
FROM (
    SELECT 'completes' as type, project, employee, completes as units 
    FROM completes 
    UNION 
    SELECT 'shifts', project, employee, length 
    FROM shifts) aggregated 
JOIN projects as p ON aggregated.project = p.id 
JOIN employees as e ON aggregated.employee = e.id 
JOIN facilities as f ON e.facility = f.id 
GROUP BY p.id, p.title, f.id, f.label 

你可以從這個查詢看到，這個問題可以很解決很容易，如果completes和shifts分別存放在同一張桌子。根據您的需求，這可能不適合您，但可能需要考慮將其作爲替代方案。