SQL如何將這三個視圖合併爲一個？

Question

如何加入以下兩個視圖？我自己嘗試過，但他們總是打破，並顯示重複的數據。我能想到的唯一工作是創建兩個視圖，然後是第三個視圖來加入這兩個視圖。問題是，對於我的應用程序，我將有大約10個視圖，這是不高效，也沒有正確設計。

視圖1

SELECT e.id, e.name, GROUP_CONCAT(email SEPARATOR ', ') AS array_emails 
FROM entity AS e 
LEFT JOIN emails ON e.id=emails.entity_id 
GROUP BY id; 

視圖2

SELECT e.id, e.name, GROUP_CONCAT(web_site SEPARATOR ', ') AS array_web_sites 
FROM entity AS e 
LEFT JOIN web_sites ON e.id=web_sites.entity_id 
GROUP BY id 

視圖3（此視圖聯接視圖1和2一起）

SELECT e.id, e.name, view_web_sites.array_web_sites, view_emails.array_emails 
FROM entity AS e 
LEFT JOIN view_web_sites ON e.id=view_web_sites.id 
LEFT JOIN view_emails ON e.id=view_emails.id 
GROUP BY id 

這是我原來的看法，它沒有正常工作。這些值會重複。

原始 - 碎視圖 SELECT e.id，e.name，GROUP_CONCAT（電子郵件分離器 ' '）如電子郵件， GROUP_CONCAT（WEB_SITE分離器'，'）AS web_sites 從實體爲E LEFT JOIN電子郵件ON e.id = emails.entity_id LEFT JOIN web_sites ON e.id = web_sites.entity_id GROUP BY id;

編輯： 我已經在select語句中探索了獨特的參數，這非常有用。然而，在一張表中我有一個二進制列，不同的參數「擦除」這個值，使列無效。對此有何建議？

編輯＃2（修正）： 我設法解決任何人可能面臨類似問題的問題。

SELECT entity.id, entity.name, 
(SELECT GROUP_CONCAT(emails.email) FROM emails WHERE entity.id=emails.entity_id) AS emails_array, 
(SELECT GROUP_CONCAT(web_sites.web_site) FROM web_sites WHERE 
entity.id=web_sites.entity_id) AS websites_array 
FROM entity 
ORDER BY id 

Answer 1

SELECT e.id, 
     e.name, 
     GROUP_CONCAT(DISTINCT m.email SEPARATOR ', ') AS array_emails, 
     GROUP_CONCAT(DISTINCT w.web_site SEPARATOR ', ') AS array_web_sites 
FROM entity AS e 
    LEFT JOIN emails AS m ON e.id = m.entity_id 
    LEFT JOIN web_sites AS w ON e.id = w.entity_id 
GROUP by e.id ; 

如果您使用的意見，你並不需要最後GROUP BY id：

SELECT e.id, 
     e.name, 
     m.array_emails, 
     w.array_web_sites 
FROM entity AS e 
    LEFT JOIN view_emails AS m ON e.id = m.entity_id 
    LEFT JOIN view_web_sites AS w ON e.id = w.entity_id ; 

如果你不想嵌套的意見，你可以使用一個查詢寫它 - 通過刪除派生表中的連接來簡化它：

SELECT e.id, 
     e.name, 
     m.array_emails, 
     w.array_web_sites 
FROM entity AS e 
    LEFT JOIN 
    (SELECT entity_id, 
      GROUP_CONCAT(email SEPARATOR ', ') AS array_emails 
     FROM emails 
     GROUP BY entity_id 
    ) AS m ON e.id = m.entity_id 
    LEFT JOIN 
    (SELECT entity_id, 
      GROUP_CONCAT(web_site SEPARATOR ', ') AS array_web_sites 
     FROM web_sites 
     GROUP BY entity_id 
    ) AS w ON e.id = w.entity_id ;