Question

到目前为止，我在搜索组合4个表时没有成功。我有4张桌子：

内容

content_id | source_id | content_title   | content_date
--------------------------------------------------------
1          | 1         | The factory     | 189982398300
2          | 2         | Cold and cloudy | 189982398299
3          | 2         | Green tea       | 189982398298

来源

source_id | source_name
-------------------
1         | BBC
2         | Reuters

设置

setting_id | setting_name
-------------------------------
1          | bbc_iplayer_string
2          | reuters_video_id
3          | reuters_category

content_join_settings

content_id | setting_id | setting_data
------------------------------------------
1          | 1          | Js88sdhjsd0gDS09
2          | 2          | video_8AJK3ADJD8
2          | 3          | weather
3          | 2          | video_K7KD8N2ND9
3          | 3          | food and drinks

因为您可能已经注意到每个内容记录都有自己的源链接，并根据源提供额外的设置。 BBC帖子有bbc_iplayer_string，路透社的帖子有reuters_video_id和reuters_category。当然这只是一个例子。

我希望将结果合并，并将setting_name作为列名。我不确定如何解释它所以我会展示它：

content_id | source_name | content_title   | content_date | bbc_iplayer_string | reuters_video_id | reuters_category
--------------------------------------------------------------------------------------------------------
1          | BBC         | The factory     | 189982398300 | Js88sdhjsd0gDS09   | NULL             | NULL
2          | Reuters     | Cold and cloudy | 189982398299 | NULL               | video_8AJK3ADJD8 | weather
3          | Reuters     | Green tea       | 189982398298 | NULL               | video_K7KD8N2ND9 | food and drinks

我对复杂的MySQL查询不是很熟练。我不知道从哪里开始寻找，更不用说搜索什么了。我从这个查询开始，但我不知道如何继续。我可能不是很亲密。

SELECT 
  * 
FROM
  content,
  sources,
  settings,
  content_join_settings
WHERE
  content.content_id = content_join_settings.content_id AND
  settings.setting_id = content_join_settings.setting_id AND
  sources.source_id = content.source_id
GROUP BY
  content.content_id
ORDER BY
  content.content_date
  DESC

Answer 1

您需要加入表格，然后转动结果以将行移动到列中。

SELECT c.content_id, s.source_name, c.content_title, c.content_date,
       MAX(CASE setting_name WHEN 'bbc_iplayer_string' THEN setting_data END) AS bbc_iplayer_string,
       MAX(CASE setting_name WHEN 'reuters_video_id' THEN setting_data END) AS reuters_video_id,
       MAX(CASE setting_name WHEN 'reuters_category' THEN setting_data END) AS reuters_category
FROM content AS c
INNER JOIN sources AS s ON s.source_id = c.source_id
INNER JOIN content_join_settings AS cjs ON cjs.content_id = c.content_id
INNER JOIN settings as st ON st.setting_id = cjs.setting_id
GROUP BY c.content_id

DEMO

Answer 2

这会有帮助吗？

SELECT 
  * 
FROM
  content as ct inner join sources as ss on ct.source_id = ss.source_id
  inner join content_join_settings as cst on ct.setting_id = cst.setting_id
  inner join settings as st on st.setting_id = cst.setting_id
GROUP BY
  content.content_id
ORDER BY
  content.content_date
  DESC

来自4个表的复杂MySQL查询

2 个答案: