警告：mysql_num_rows（）：提供的参数不是有效的MySQL结果资源

Question

attempting to implement a facebook style pagination and i've been getting this error? Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource ,

I'm not entirely sure what this really means, but I can't find a fix for it from googling, anything silly mistakes that stand out? here's the code :

    <?php
include('functions.php');


if(isset($_POST['lastmsg']) &&is_numeric($_POST['lastmsg']))
{
$lastmsg=$_POST['lastmsg'];
$query = mysql_query("SELECT * FROM links WHERE id>'$lastmsg' ORDER BY id ASC LIMIT 9 ");

while($result = mysql_fetch_array($query))
{
$msg_id=$result['id'];
$message=$result['link'];
?>

<li> <?php echo $message; ?> </li>
<?php
}


?>
<?php

if(mysql_num_rows($result)==9){
   ?>
<div class="facebook_style" id="facebook_style"> 
    <a id="<?php echo $message; ?>" href="#" class="load_more" >
        Show Older Posts <img src="arrow1.png" /></a> </div>
<?php
 }else {

    echo '  <div  id="facebook_style">
  <a  id="end" href="#" class="load_more" >No More Posts</a>
  </div>';

 }
}
?>

Answer 1

You want to pass mysql_num_rows() your $query value rather than your $results value. (Or you want to refactor your code to name $query to something else like $query_result and pass that to mysql_num_rows() .)

mysql_num_rows() expects a MySQL result resource, which in your case is returned by mysql_query() , which you've assigned to $query .

$result , in contrast, is just an array representing a single row of the result. (So a better refactoring might be to rename $result to $row and rename $query to $result . Then you can leave your call to mysql_num_rows() as is.)

Answer 2

mysql_num_rows() expects a result resource (in your case $query ) rather than a row pulled from mysql_fetch_array() .

if(mysql_num_rows($query )==9) {
  // etc...
}