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如何加入以下两个视图？我自己尝试过，但它们总是会中断，并显示重复的数据。我能想到的唯一解决方法是创建两个视图，然后创建第三个视图来加入这两个视图。问题是，对于我的应用程序，我将拥有大约 10 个视图，这既不高效也不正确设计。

查看 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;

查看 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

视图 3（此视图将视图 1 和 2 连接在一起）

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

这是我原来的观点，没有正常工作。这些值会重复。

原始 - 损坏的视图 SELECT e.id, e.name, GROUP_CONCAT(email SEPARATOR ',') AS emails, GROUP_CONCAT(web_site SEPARATOR ',') AS web_sites FROM entity AS e LEFT JOIN emails ON e.id=emails.entity_id LEFT在 e.id=web_sites.entity_id GROUP BY id 上加入网站；

编辑：我已经探索了 select 语句中的不同参数，效果很好。但是，在一个表中，我有一个二进制列，并且 distinct 参数“擦除”该值，使该列无用。关于这个问题有什么建议吗？

编辑＃2（FIX）：我设法为任何可能面临类似问题的人解决了这个问题。

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
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1 回答 1

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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 ;

如果您使用视图，则不需要最后一个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 ;

如果您不想要嵌套视图，您可以使用一个查询来编写它 - 并通过删除派生表中的连接来简化它：

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 ;
于 2013-10-19T18:46:01.543 回答