1

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打印为我的网页上的表格：message.subject、user.username 和评论数。其他一切正常，但我还没有设法让 PHP 打印评论的数量。

到目前为止，这是我尝试做的：

<?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>
4

5 回答 5

2

难道你不能通过一个查询来获得所有这些信息吗？

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，您的链接没有结束</a>标签，您的脚本可能容易受到 XSS 攻击。然后是嵌套的 while 循环，这会导致不必要的复杂性和错误。

于 2012-05-17T17:26:55.913 回答
0

我看到的问题是您正在嵌套您的 while 循环。请参阅：http ://www.php.net/manual/en/control-structures.while.php#52733

使用一条 sql 语句。

于 2012-05-17T17:27:17.003 回答
0

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的评论，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;
于 2012-05-17T17:27:34.370 回答
0

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
";
于 2012-05-17T17:28:08.133 回答
0

为什么不做类似下面的事情呢？

$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东西？

手册是这样说的mysql_num_rows()：

mysql_num_rows — 获取结果中的行数

于 2012-05-17T17:28:13.950 回答