在多個表上使用一對多來構建MySQL查詢問題

Question

我有一個博客樣式的應用程序，允許用戶使用主題標記每個帖子。我將這些數據保存在三個獨立的表格中：帖子（用於實際的博客文章），主題（用於各種標籤）和posts_topics（存儲兩者之間關係的表格）。

爲了保持MVC結構（我使用Codeigniter）儘可能乾淨，我想運行一個MySQL查詢來抓取所有發佈數據和關聯主題數據，並將其返回到一個數組或對象中。到目前爲止，我沒有任何運氣。

表結構是這樣的：

Posts 
+--------+---------------+------------+-----------+--------------+---------------+ 
|post_id | post_user_id | post_title | post_body | post_created | post_modified | 
+--------+---------------+------------+-----------+--------------+---------------+ 
|  1 | 1   | Post 1  | Body 1 | 00-00-00 | 00-00-00  | 
|  2 | 1   | Post 1  | Body 1 | 00-00-00 | 00-00-00  | 
+--------+---------------+------------+-----------+--------------+---------------+ 

// this table governs relationships between posts and topics 
Posts_topics 
+--------------+---------------------+-------------------------+-----------------+ 
|post_topic_id | post_topic_post_id | post_topic_topic_id | post_topic_created | 
+--------------+---------------------+-------------------------+-----------------+ 
|  1  | 1    | 1     |   00-00-00 | 
|  2  | 1    | 2     |   00-00-00 | 
|  3  | 2    | 2     |   00-00-00 | 
|  4  | 2    | 3     |   00-00-00 | 
+--------------+---------------------+-------------------------+-----------------+ 

Topics 
+---------+-------------+-----------+----------------+ 
|topic_id | topic_name | topic_num | topic_modified | 
+---------+-------------+-----------+----------------+ 
|  1 | Politics | 1   | 00-00-00  | 
|  2 | Religion | 2   | 00-00-00  | 
|  3 | Sports  | 1   | 00-00-00  | 
+---------+-------------+-----------+----------------+ 

我曾嘗試與正成功這個簡單的查詢：

select * from posts as p inner join posts_topics as pt on pt.post_topic_post_id = post_id join topics as t on t.topic_id = pt.post_topic_topic id 

我使用GROUP_CONCAT也嘗試過，但是這給了我兩個問題： 1）我需要主題中的所有字段，而不僅僅是名稱，以及2）我的MySQL中有一個小故障，因此所有GROUP_CONCAT數據都以BLOB形式返回（請參閱here）。

我也樂於聽到任何建議，我運行兩個查詢，並嘗試爲每個結果構建數組;我想，下面，但失敗的碼（此代碼還包括加入用戶表，這將是巨大的，以保持這一以及）：

$this->db->select('u.username,u.id,s.*'); 
    $this->db->from('posts as p'); 
    $this->db->join('users as u', 'u.id = s.post_user_id'); 
    $this->db->order_by('post_modified', 'desc'); 
    $query = $this->db->get(); 
    if($query->num_rows() > 0) 
    { 
     $posts = $query->result_array(); 
     foreach($posts as $p) 
     { 
      $this->db->from('posts_topics as p'); 
      $this->db->join('topics as t','t.topic_id = p.post_topic_topic_id'); 
      $this->db->where('p.post_topic_post_id',$p['post_id']); 
      $this->db->order_by('t.topic_name','asc'); 
      $query = $this->db->get(); 
      if($query->num_rows() > 0) 
      { 
       foreach($query->result_array() as $t) 
       { 
        $p['topics'][$t['topic_name']] = $t; 
       } 
      } 
     } 
     return $posts; 
    } 

任何幫助極大的讚賞。

Answer 1

此查詢應該做的伎倆。只需將*更改爲您所需的字段列表，以便您不會在每次運行查詢時拉取多餘的數據。

Select 
    * 
FROM 
    posts, 
    post_topics, 
    topics 
WHERE 
    post_topic_topic_id = topic_id AND 
    post_topic_post_id = post_id 
ORDER BY 
post_id, topic_id; 

Select 
    * 
FROM 
    posts, 
    post_topics, 
    topics, 
    users 
WHERE 
    post_topic_topic_id = topic_id AND 
    post_topic_post_id = post_id AND 
    post_user_id = user_id 
ORDER BY 
post_id, topic_id; 

Answer 2

聖牛，你可以做到！看到它有助於幫助。從來不知道，試試這個 選擇 POST_ID， GROUP_CONCAT（DISTINCT TOPIC_NAME）姓名 FROM 帖子， post_topics， 主題 WHERE post_topic_topic_id = topic_id和 post_topic_post_id = POST_ID

GROUP BY POST_ID;

你得到 1，「政治，relligion」 2，「體育，relligion」