LEFT JOIN不适用于2张桌子

Question

Basically I have 2 tables

• Topics
• Users

I am trying to use a left join so that I can link the "posted_by" in "topics" with "user_id" in "users", so that I can output both the users.username for display, as well as users.profile(avatar picture).

Here is my current code, which is giving me boolean errors.

        <?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 you know this will produce ON topics.posted_by = 1 users.user_id for example which is invalid SQL syntax. Use WHERE instead

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

PS: Using mysql_ is highly not recommended. You should change the API.

Answer 2

You need to provide how users relates to topics in your join clause. You have provided a outside variable to your join clause. This doesn't allow the database engine to establish a relationship on how the two tables should be joined.

I believe the query you are expecting should be something like this

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

Answer 3

I believe you are using something like

$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";

Try adding

$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, topics WHERE topics.posted_by = users.user_id ORDER BY topics.posted DESC";