php - 如何让 PHP 打印来自 MySQL 数据库的评论数

Question

我有以下问题。我有三个表:消息、评论和用户。我希望 PHP 将 MySQL 数据库中以下字段的数据作为表格打印在我的网页上:message.subject、user.username，然后是评论数。其他一切正常，但我还没有设法让 PHP 打印评论数。

这是我到目前为止尝试做的:

<?php
 include("info.php");
 $connect;
 $sql="SELECT * FROM message, user
       WHERE message.userID = user.userID AND 
       ORDER BY message.messageID DESC";
 $result=mysql_query($sql) or die(mysql_error());

 $sql2="SELECT message.messageID, COUNT(*) as comments FROM comment 
        INNER JOIN
        message ON comment.messageID = message.messageID 
        GROUP BY comment.messageID";
 $result2=mysql_query($sql2) or die(mysql_error());

<table>
 <thead>
  <tr>
   <th>Subject</th>   
   <th>Sender</th>
   <th>Number of comments</th>
  </tr>
 </thead>
 <tbody>
  <?php
   while($rows=mysql_fetch_array($result)){
  ?>
  <tr>
   <td>
    <a href="php/message.php?id=<?php echo $rows['messageID']; ?>">
    <?php echo $rows['subject']; ?>
   </td>
   <td>
    <?php echo $rows['username']; ?>
   </td>                        
   <td>
    <?php while($rows2=mysql_fetch_array($result2)){ echo $rows2['comments'];}?>
   </td>
  </tr>
 <?php
 } mysql_close(); ?>
 </tbody>
</table>

Answer 1

您不能通过一次查询获取所有这些信息吗？

SELECT
    m.messageID,
    m.subject,
    u.username,
    c.numOfComments
FROM
    message m
    INNER JOIN user u ON m.userID = u.userID
    LEFT JOIN (SELECT COUNT(1) AS numOfComments, messageID FROM comments GROUP BY messageID) c ON m.messageID = c.messageID
ORDER BY
    m.messageID DESC

---

试试这个脚本:

<?php
    include("info.php");
    connect();
    $sql = "
        SELECT
            m.messageID,
            m.subject,
            u.username,
            c.numOfComments
        FROM
            message m
            INNER JOIN user u ON m.userID = u.userID
            LEFT JOIN (SELECT COUNT(1) AS numOfComments, messageID FROM comments GROUP BY messageID) c ON m.messageID = c.messageID
        ORDER BY
            m.messageID DESC
    ";
    $result = mysql_query($sql) or die(mysql_error());

    echo "<table>
    <thead>
        <tr>
            <th>Subject</th>
            <th>Sender</th>
            <th>Number of comments</th>
        </tr>
    </thead>
    <tbody>\n";
    while ($row = mysql_fetch_assoc($result))
    {
        $row = array_map("htmlspecialchars", $row); // sanitize to prevent XSS

        echo "      <tr>
            <td><a href=\"php/message.php?id={$row["messageID"]}\">{$row["subject"]}</a></td>
            <td>{$row["username"]}</td>
            <td>{$row["numOfComments"]}</td>
        </tr>\n";
    }
    echo "  </tbody>
</table>";
?>

还有一些其他的更正。例如，您的链接没有结尾 </a>标记，您的脚本可能容易受到 XSS 攻击。然后是嵌套的 while 循环，它导致了不必要的复杂化和错误。