Question

基本上我有2张桌子

主题
用户

我正在尝试使用左连接，以便我可以将“主题”中的“posted_by”与“users”中的“user_id”相关联，这样我就可以同时输出users.username用于显示，以及用户.profile（头像图片）。

这是我当前的代码，它给出了布尔错误。

        <?php
    include 'core/init.php';
    include 'includes/overall/header.php';

    $sql = " SELECT *, users.id, users.username, users.profile
        FROM `topics` 
        LEFT JOIN
        users ON topics.posted_by = " . mysql_real_escape_string($_GET['topic_id']) . " users.user_id ORDER BY `posted` DESC";

    $result = mysql_query($sql);

    // Start looping table row
    while($rows = mysql_fetch_array($result)){
    ?>
    <table>
      <tr> 
        <td rowspan="4"> Avatar code to go here<br>
           <? echo $rows['username']; ?></td>
        <td><? echo $rows['category']; ?> > <? echo $rows['sub_category']; ?> </td>
      </tr>
      <tr>
        <td><? echo $rows['posted']; ?></td>
      </tr>
      <tr>
        <td><? echo $rows['topic_data']; ?></td>
      </tr>
      <tr>
        <td><a href="view_topic.php?id=<? echo $rows['topic_id']; ?>">Reply</a> (<? echo $rows['reply']; ?>) Replies</td>
      </tr>
    </table>

    <?php
    // Exit looping and close connection 
    }
    mysql_close();
    ?>

Answer 1

ON topics.posted_by = " . mysql_real_escape_string($_GET['topic_id']) . " users.user_id你知道这会产生ON topics.posted_by = 1 users.user_id例如无效的SQL语法。改用WHERE

ON topics.posted_by = users.user_id WHERE topics.id = (topic_id_variable)

P.S。：强烈建议不要使用mysql_。您应该更改API。

Answer 2

您需要在join子句中提供users与topics的关联方式。您已为join子句提供了外部变量。这不允许数据库引擎建立关于如何连接两个表的关系。

我相信您期望的查询应该是这样的

$sql = " SELECT *, users.id, users.username, users.profile
        FROM `topics` 
        LEFT JOIN
        users ON topics.posted_by = users.user_id 
        WHERE topics.id = '" . mysql_real_escape_string($_GET['topic_id']) . "'
        ORDER BY `posted` DESC";

Answer 3

我相信你正在使用像

这样的东西

$sql = "SELECT users.user_id, users.username, users.profile, topics.topic_id,     topics.category, topics.sub_category, 
topics.topic_data, topics.posted_by, topics.posted, topics.view, topics.reply 
FROM users WHERE topics.posted_by = users.user_id ORDER BY topics.posted DESC";

尝试添加

$ sql =“SELECT users.user_id，users.username，users.profile，topics.topic_id，topics.category，topics.sub_category， topics.topic_data，topics.posted_by，topics.posted，topics.view，topics.reply FROM 用户，主题 WHERE topics.posted_by = users.user_id ORDER BY topics.posted DESC“;

LEFT JOIN不适用于2个表

3 个答案: