Question

我有两张桌子：

帖子

+--------------------------------------------------------------------+
|id          ¦status¦type ¦url                                       |
+------------+------+-----+------------------------------------------+
|25949664    ¦80    ¦link ¦http://example.com/25949664               |
|25777570    ¦80    ¦photo¦http://example.com/25777570               |
+--------------------------------------------------------------------+

属性

╔════════════╦════════════╦══════════════════════════════════════════╗
║id          ║attr        ║value                                     ║
╠════════════╬════════════╬══════════════════════════════════════════╣
║25949664    ║timestamp   ║1430836105                                ║
║25949664    ║tag         ║red                                       ║
║25949664    ║tag         ║yellow                                    ║
║25949664    ║tag         ║brown                                     ║
║25949664    ║source      ║http://example.com/wallin/                ║
║25949664    ║source_title║wallin                                    ║
║25949664    ║state       ║published                                 ║
║25949664    ║format      ║html                                      ║
║25777570    ║timestamp   ║1430836105                                ║
║25777570    ║tag         ║red                                       ║
║25777570    ║tag         ║yellow                                    ║
║25777570    ║tag         ║brown                                     ║
║25777570    ║tag         ║black                                     ║
║25777570    ║tag         ║orange                                    ║
╚════════════╩════════════╩══════════════════════════════════════════╝

执行此查询：

  SELECT posts.id, GROUP_CONCAT(attributes.value) as tags
    FROM posts
    JOIN attributes ON attributes.id = posts.id
   WHERE ( attributes.attr = 'tag' )
     AND ( attributes.value IN ('red','brown') )
GROUP BY posts.id
  HAVING COUNT(DISTINCT attributes.value) = 2

我有这个结果：

╔════════════╦════════════╗
║id          ║tags        ║
╠════════════╬════════════╣
║25949664    ║red,brown   ║
║25777570    ║red,brown   ║
╚════════════╩════════════╝

我宁愿拥有这个：

╔════════════╦════════════════════════════════╗
║id          ║tags                            ║
╠════════════╬════════════════════════════════╣
║25949664    ║red,yellow,brown                ║
║25777570    ║red,yellow,brown,black,orange   ║
╚════════════╩════════════════════════════════╝

实际上，拥有 n 标记后，我会检索只执行一次查询的所有帖子标记。

任何人都有建议，或者这完全不可能？

Answer 1

  SELECT posts.id, GROUP_CONCAT(attributes.value) as tags
    FROM posts
    JOIN attributes ON attributes.id = posts.id
   WHERE ( attributes.attr = 'tag' )
     AND ( attributes.value IN ('red','brown') )
GROUP BY posts.id

使用和条件，仅获取红色和棕色属性。因此，您需要删除此条件并重写该条件。

我认为这可以帮到你：

 SELECT posts.id, GROUP_CONCAT(attributes.value) as tags
    FROM posts
    JOIN attributes ON attributes.id = posts.id
   WHERE ( attributes.attr = 'tag' )     
GROUP BY posts.id
  HAVING Find_In_Set('red',tags)>0 AND Find_In_Set('brown',tags)>0

Online Demo

GROUP_CONCAT：搜索子集并检索所有值

1 个答案: