在關係表中使用外鍵選擇字段

Question

我想顯示主題創建（用戶名），但必須通過關係表來檢索它。我已經創建了一個查詢，顯示答覆的創建者（用戶名），我相信我需要一個子查詢，但以前從未使用過。

什麼，我基本上試圖做的是使用外鍵來獲取用戶名，我希望下面的解釋它：

Forum_replies.topic_id >>>>> forum_topics.topic_id和forum_topics.user_id >>>>用戶。用戶名。

的表如下：

forum_replies

• reply_id
• topic_id
• USER_ID
• REPLY_TEXT
• 回覆日期

個

forum_topics

• topic_id
• CATEGORY_ID
• USER_ID
• TOPIC_TITLE
• topic_description
• topic_date

用戶

• user_id說明
• 用戶名

這是我的代碼目前顯示 forum_topics.Topic_title，forum_replies.reply_date，forum_replies.user_id（示出了回覆創建者的用戶名），forum_replies.reply_text。

$queryreply = "SELECT forum_replies.reply_id, forum_replies.topic_id, forum_replies.user_id, 
         forum_replies.reply_text, forum_replies.reply_date, users.user_id, users.username, forum_topics.user_id, 
         forum_topics.topic_id,forum_topics.topic_title, forum_topics.topic_date 
         FROM forum_replies 
         LEFT JOIN forum_topics 
         ON forum_replies.topic_id = forum_topics.topic_id 
         LEFT JOIN users 
         ON forum_replies.user_id = users.user_id 

         "; 


     $result = mysql_query($queryreply) or die (mysql_error()); 
     $row = mysql_fetch_array($result); 

      if(empty($row['topic_id'])){ 
      echo "No replies have been posted in this Topic, be the first to have your say using form below.";} ?> 
     <table id="categorytable"> 

       <tr><td><?php echo '<b>'.$row['topic_title'].'</b>';?></b><br><br></td></tr> 
       <tr><td><?php echo $row['reply_date'].' - '.$row['username'].' Replied with: ';?><br><br></td></tr> 
       <tr><td><?php echo $row['reply_text'];?></td></tr> 

我知道mysql_ *函數已被棄用，但我被要求uni工作人員使用它們。我會很樂意提供任何幫助。由於

Answer 1

再添JOIN：

SELECT forum_replies.reply_id, forum_replies.topic_id, forum_replies.user_id, 
    forum_replies.reply_text, forum_replies.reply_date, users.user_id, users.username, forum_topics.user_id, 
    forum_topics.topic_id,forum_topics.topic_title, forum_topics.topic_date, 
    users.user_id as topic_user_id, users.username as topic_username, 
FROM forum_replies 
    LEFT JOIN forum_topics ON forum_replies.topic_id = forum_topics.topic_id 
    LEFT JOIN users ON forum_replies.user_id = users.user_id 
    LEFT JOIN users u2 ON forum_topics.user_id = u2.user_id 

...並在您的其他問題中提到：使用別名

Answer 2

您可以加入一個表多次，因爲你需要。因此，您可以更多地連接表格用戶，但使用不同的ON語句。

SELECT 
 
    forum_replies.reply_id, forum_replies.topic_id, forum_replies.user_id, forum_replies.reply_text, 
 
    forum_replies.reply_date, users.user_id, users.username, forum_topics.user_id, 
 
    forum_topics.topic_id,forum_topics.topic_title, forum_topics.topic_date, 
 
    topic_creators.username 
 
FROM forum_replies 
 
LEFT JOIN forum_topics ON forum_replies.topic_id = forum_topics.topic_id 
 
LEFT JOIN users ON forum_replies.user_id = users.user_id 
 
LEFT JOIN users as topic_creators ON forum_topics.user_id = topic_creators.user_id;

Answer 3

另一個計算器用戶發送我的代碼，它現在的工作很好，感謝你們的幫助，雖然。我非常感謝你的建議。這裏是代碼，如果你的興趣：

 $queryreply = "SELECT a.reply_id,a.reply_text, a.reply_date, b.topic_title, c.username AS reply_user, (SELECT username FROM users 
     WHERE user_id=b.user_id) AS topic_creator FROM forum_replies a 
     LEFT JOIN forum_topics b ON a.topic_id=b.topic_id 
     LEFT JOIN users c ON a.user_id=c.user_id 

         ";