Php 在关系表中使用外键选择字段

我试图显示主题create（username），但必须通过一个关系表来检索它。我已经创建了一个显示回复创建者（用户名）的查询，我相信我需要一个子查询，但以前从未使用过

     $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

                       ";

我基本上是想用外键检索用户名，希望下面能解释一下：

     $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

                       ";

Forum\u Responses.topic\u id>>>>>Forum\u topics.topic\u id和Forum\u topics.user\u id>>>>>>users.user\u id

     $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

                       ";

各表如下：

     $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

                       ";

论坛的答复

     $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

                       ";

• 答复:
• 主题id
• 用户id
• 答复(文本)
• 答复日期

论坛主题

     $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

                       ";

• 主题id
• 类别识别码
• 用户\u id
• 题目
• 主题描述
• 主题日期

使用者

     $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

                       ";

• 用户id
• 用户名

这是我当前显示的代码 论坛主题。主题标题，论坛回复。回复日期，论坛回复。用户id（显示回复创建者的用户名），论坛回复。回复文本

    $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>

     $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

                       ";

$queryreply=“选择论坛\u represponse.reply\u id，forum_urepresponse.topic\u id，forum\u represponse.user\u id，
论坛回复。回复文本，论坛回复。回复日期，用户。用户id，用户。用户名，论坛主题。用户id，
论坛主题。主题id、论坛主题。主题标题、论坛主题。主题日期
来自论坛的回复
左键加入论坛主题
在论坛上回复.topic\u id=论坛主题.topic\u id
左加入用户
在论坛上\u回复.user\u id=users.user\u id
";
$result=mysql\u query（$queryreply）或die（mysql\u error（））；
$row=mysql\u fetch\u数组（$result）；
if（空（$row['topic_id']））{
echo“此主题中未发布任何回复，请首先使用下面的表格发表您的意见。”；}？>

我知道mysql_*函数已被弃用，但uni工作人员要求我使用它们。如果有任何帮助，我将不胜感激。谢谢

添加另一个加入：

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

     $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

                       ";

---

…正如您在另一个问题中提到的：使用别名可以根据需要多次加入表。所以您可以使用不同的ON语句将表用户连接到更多的表用户

     $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

                       ";

选择
论坛回复。回复id，论坛回复。主题id，论坛回复。用户id，论坛回复。回复文本，
论坛回复。回复日期，用户。用户id，用户。用户名，论坛主题。用户id，
论坛主题。主题id，论坛主题。主题标题，论坛主题。主题日期，
topic_creators.username
来自论坛的回复
左侧加入论坛\u论坛上的主题\u回复。主题\u id=论坛\u主题。主题\u id
左加入论坛上的用户。\u回复。用户\u id=用户。用户\u id
左键作为论坛上的主题创建者加入用户。用户id=主题创建者。用户id另一个stackoverflow用户向我发送了代码，现在它工作得很好，不过谢谢你们的帮助。我非常感谢你的建议。以下是您感兴趣的代码：

     $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

                       ";

不行-你有两个名为username的列我已经测试过了，效果很好。在你的回答中，你有两列名为user_id。我也是，如果要澄清的话）这两列我都有别名。另一个stackoverflow用户给我发送了代码，它现在工作得很好，不过谢谢你们的帮助。非常感谢您的建议。感谢您的帮助，我使用别名添加了另一个加入，但当我尝试呼出主题创建者的用户名时，它会显示回复创建者。你知道我应该使用什么代码吗？当然你必须使用
echo$row['topic_username']我在声明中使用的名称。顺便说一句-你应该编辑你的问题并添加你现在所做的事情另一个stackoverflow用户向我发送了代码，它现在工作得很好，谢谢你们的帮助。我非常感谢你的建议。以下是您感兴趣的代码：$queryreply=“选择a.reply\u id、a.reply\u text、a.reply\u date、b.topic\u title、c.username作为reply\u用户（从user\u id=b.user\u id的用户中选择用户名）作为论坛的主题创建者，论坛答复左加入论坛主题b。主题id=b。主题id左加入用户c。用户id=c。用户id“；