如何向这个庞大的 SQL 联合语句添加连接？答案

【问题标题】：How can I add a join to this bulky SQL union statement?如何向这个庞大的 SQL 联合语句添加连接？
【发布时间】：2011-07-05 12:02:31
【问题描述】：

我有一个语句，根据链接表的共享标签和扬声器的数量对项目进行排名，但我希望它也能够从主 tbl_talks 中提取 talk_title 和 talk_brief 信息，这实际上不在查询中现在（仅使用链接表）。基本上链接表有talk_id和tag_id或talk_id和speaker_id。

我尝试了几种尝试 LEFT JOIN 的方法，或者在应用联合但没有运气的成对 SELECT 中添加 tbl_talks。我需要将 tbl_linking_talk_tag.talk_id 和 tbl_linking_speakers_talk 与 tbl_talks.talk_id 相关联，然后从 tbl_talks 中获取talk_title 和talk_brief

SELECT talk_id, count(*) AS common_term_count FROM 
      (
      SELECT talk_id FROM tbl_linking_talk_tag WHERE tag_id IN 
         (SELECT tag_id FROM tbl_linking_talk_tag WHERE talk_id = 1)
      UNION ALL 
      SELECT talk_id FROM tbl_linking_speakers_talk WHERE speaker_id IN
         (SELECT speaker_id FROM tbl_linking_speakers_talk WHERE talk_id = 1)
       ) AS accumulator_table

 WHERE talk_id != 1 

 GROUP BY talk_id

 ORDER BY common_term_count DESC

【问题讨论】：

标签： mysql join union relational-database inner-join

【解决方案1】：

在这种情况下，您不需要工会。如果您只需要计数，那么为什么不尝试类似..

选择talk_id， (SELECT count(*) FROM tbl_linking_talk_tag sub WHERE tag_id IN (SELECT tag_id FROM tbl_linking_talk_tag WHERE talk_id = 1) and maintalk.talk_id=sub.talk_id) + (SELECT count(*) FROM tbl_linking_speakers_talk sub1 WHERE speaker_id IN (SELECT speaker_id FROM tbl_linking_speakers_talk WHERE talk_id = 1) 和 maintalk.talk_id=sub1.talk_id ),talk_title,talk_brief 来自主讲其中talk_id != 1 按talk_id分组

语法可能需要一些修正，但方法是

从主表中选择列

【讨论】：

【解决方案2】：

我得到了它与这个代码一起工作：

 SELECT a.talk_id, a.talk_image, a.talk_title, a.talk_brief, count(b.talk_id)  AS common_term_count 
        FROM tbl_talks a LEFT OUTER JOIN
             (
             SELECT talk_id FROM tbl_linking_talk_tag WHERE tag_id IN 
                (SELECT tag_id FROM tbl_linking_talk_tag WHERE talk_id = 1)
             UNION ALL 
             SELECT talk_id FROM tbl_linking_speakers_talk WHERE speaker_id IN
                (SELECT speaker_id FROM tbl_linking_speakers_talk WHERE talk_id = 1) 
             ) AS b
             ON a.talk_id = b.talk_id       
        WHERE a.talk_id != 1        
        GROUP BY b.talk_id      
        ORDER BY common_term_count DESC

【讨论】：