如何让PHP打印来自MySQL数据库的注释数 - How the get PHP to print the number of comments from MySQL database

我有以下问题。我有三个表：消息，评论和用户。我希望PHP将MySQL数据库中的以下字段的数据打印为我的网页上的表格：message.subject，user.username ，然后是注释数。其他一切正常，但我还没有设法让 PHP 打印评论数量。

这是我到目前为止尝试做的事情：

<?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>

您不能通过一个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

<小时 />

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>";
?>

还有其他一些更正。例如，您的链接没有结束</a>标记，并且您的脚本可能容易受到 XSS 攻击。然后是嵌套的while循环，这导致了不必要的复杂性和错误。

我看到的问题是你正在嵌套你的 while 循环。请参阅：http://www.php.net/manual/en/control-structures.while.php#52733

使用一个 sql 语句。

这是最好的方法(我认为(：

$sql = "
SELECT message.*, user.*, commentsCounter.comments FROM message
INNER JOIN user
ON user.userID = message.userID
INNER JOIN ( SELECT COUNT( 1 ) as comments, comment.messageID FROM comment GROUP BY comment.messageID ) commentsCounter
ON commentsCounter.messageID = message.messageID
ORDER BY message.messageID DESC
";

为什么不做下面这样的事情呢？

$query = "SELECT comment FROM message where userId = $userId";
$result = mysql_query($query);
$numberOfComments = mysql_num_rows($result);
echo "$numberOfComments";

至少我相信这就是你要找的？

手册是这样说的 mysql_num_rows((：

mysql_num_rows — 获取结果中的行数

http://www.w3schools.com/sql/sql_func_count.asp

$result3=mysql_query("COUNT(*) FROM comment") or die(mysql_error());
$row=mysql_fetch_array($result3);
$numcomments = $row[0];
echo $numcomments;

我在几个站点上使用它(使用不同的表名(。

答案 2

假设您想要对特定帖子发表评论，请尝试以下操作：

$result3=mysql_query("SELECT DISTINCT message.messageID FROM comment ") or die(mysql_error());
$row=mysql_fetch_array($result3);
$numcomments = count($row);
echo $numcomments;