Question

如何加入以下两个视图？我自己尝试过，但他们总是打破，并显示重复的数据。我能想到的唯一工作是创建两个视图，然后是第三个视图来加入两个视图。问题是，对于我的应用程序，我将有大约10个视图，这些视图效率不高，设计不正确。

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

查看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

这是我的原始视图，它无法正常工作。值会重复。

原创 - 视图破碎 选择e.id，e.name，GROUP_CONCAT（电子邮件SEPARATOR'，'）AS电子邮件， GROUP_CONCAT（web_site SEPARATOR'，'）AS web_sites 来自实体AS e LEFT JOIN电子邮件ON e.id = emails.entity_id LEFT JOIN web_sites ON e.id = web_sites.entity_id GROUP BY id;

编辑：我已经在select语句中探索了distinct参数，该参数非常有效。但是在一个表中我有一个二进制列，并且distinct参数“擦除”该值，使列无用。关于此事的任何建议？

编辑＃2（FIX）：我设法为可能遇到类似问题的任何人解决了这个问题。

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

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 ;

如果您使用这些视图，则不需要上一个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，可以使用一个查询编写它 - 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 ;

SQL如何将这三个视图合并为一个？

1 个答案: