在关系表中使用外键选择字段

Question

我想显示主题创建（用户名），但必须通过关系表来检索它。我已经创建了一个查询，显示答复的创建者（用户名），我相信我需要一个子查询，但以前从未使用过。

什么，我基本上试图做的是使用外键来获取用户名，我希望下面的解释它：

Forum_replies.topic_id >>>>> forum_topics.topic_id和forum_topics.user_id >>>>用户。用户名。

的表如下：

forum_replies

• reply_id
• topic_id
• USER_ID
• REPLY_TEXT
• 回复日期

个

forum_topics

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

用户

• user_id说明
• 用户名

这是我的代码目前显示 forum_topics.Topic_title，forum_replies.reply_date，forum_replies.user_id（示出了回复创建者的用户名），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_ *函数已被弃用，但我被要求uni工作人员使用它们。我会很乐意提供任何帮助。由于

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 

...并在您的其他问题中提到：使用别名

Answer 2

您可以加入一个表多次，因为你需要。因此，您可以更多地连接表格用户，但使用不同的ON语句。

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

另一个计算器用户发送我的代码，它现在的工作很好，感谢你们的帮助，虽然。我非常感谢你的建议。这里是代码，如果你的兴趣：

 $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 

         ";