php - 我不知道的 SQL 错误消息 : "Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource"

Question

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Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
mysql_num_rows(): supplied argument is not a valid MySQL result resource
Warning: mysql_fetch_* expects parameter 1 to be resource, boolean given error

我正在尝试通过 url 传递一个 id，在该 url 中我收到一条我不知道的错误消息。我的代码如下:

 <?php

$result = mysql_query("SELECT * FROM Agendas INNER JOIN Meetings ON Meetings.meeting_id = Agendas.meeting_id WHERE approval = 'pending' ORDER BY Agendas.meeting_id DESC")
or die(mysql_error());

if (mysql_num_rows($result) == 0) {
       echo '<h2>No Action Is Required Yet</h2>';
    } else {

while($info = mysql_fetch_array($result))
{

        echo "<tr>";
        echo "<td><br/>" .'Meeting Title: '. $info['title']." </td>";
        echo "<td><br/><br/>" .'<a href="viewagenda.php?agenda_id=' . $info['agenda_id'] . '">View Agenda</a> '." </td>";
        echo "<hr>";
        }
    }
echo "</tr>";
echo "</table>";


?>

viewagenda.php 如下:

 <?php
$id = $_GET['agenda_id'];

$query = mysql_query("SELECT * FROM Agendas INNER JOIN Meetings ON Meetings.meeting_id = Agendas.meeting_id WHERE agenda_id = '$id'")
or die(mysql_error()); 

if (mysql_num_rows($result) == 0) {
       echo '<hr><h2>There Arent Any Agendas For This Meeting Yet</h2>';
    } else {

while($info = mysql_fetch_array($result))
{

        echo "<tr>";
        echo "<td><br/>" .'Title: '. $info['title']." </td>";
        echo "<td><br/>" .'Subject: '. $info['subject']. "</td>";
        echo "<td><br/>" .'Duration: '. $info['duration']. "</td>";

        echo "<hr>";
        }
    }
echo "</tr>";
echo "</table>";
?>

我得到的错误信息是:

'警告:mysql_num_rows():提供的参数不是第 70 行 E:\webareas\hj942\conference\Chairperson\viewagenda.php 中的有效 MySQL 结果资源'

我的表格如下:

session :meeting_id、头衔、主席、秘书、事件

议程:议程 ID、主题、持续时间、 session ID

Answer 1

我猜你的意思是:

$result = mysql_query("SELECT * FROM Agendas INNER JOIN Meetings ON  ...

因为您使用的是 $result mysql_num_rows 中的变量:

if (mysql_num_rows($result) == 0) {