Малко помощ с Warning: mysql_num_rows(): supplied argument..

[youlive]

Та имам следната грешка
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/radiocho/public_html/dating/main.php on line 119
Error displaying info

<?php 
$qa = mysql_query("SELECT COUNT(*) AS br FROM `users`");
$aa = mysql_fetch_array($qa);
$a = $aa['br'] - 4;
$q = "SELECT * FROM `users` LIMIT $a, 4";
$result=mysql_query($q);
$num_rows = mysql_num_rows($result);
if(!$result || ($num_rows < 0)){
   echo "Error displaying info";
}
else if($num_rows > 0){
   for($i=0; $i<$num_rows; $i++){
      $uname = mysql_result($result,$i,"username");
      $pic = mysql_result($result,$i,"pic");
      $sex = mysql_result($result,$i,"sex");
      $points = mysql_result($result,$i,"points");
      if($sex=='m') {
	  ?>

Решение ?

 

[gamena007]

 <?php
$qa = mysql_query("SELECT COUNT(*) AS br FROM `users`") or die(mysql_error());
$aa = mysql_fetch_array($qa);
$a = $aa['br'] - 4;
$q = "SELECT * FROM `users` LIMIT $a, 4";
$result=mysql_query($q);
$num_rows = mysql_num_rows($result);
if(!$result || ($num_rows < 0)){
   echo "Error displaying info";
}
else if($num_rows > 0){
   for($i=0; $i<$num_rows; $i++){
      $uname = mysql_result($result,$i,"username");
      $pic = mysql_result($result,$i,"pic");
      $sex = mysql_result($result,$i,"sex");
      $points = mysql_result($result,$i,"points");
      if($sex=='m') {
     ?>

Какво ти показва?

 

[lortnoc]

Та имам следната грешка
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/radiocho/public_html/dating/main.php on line 119
Error displaying info

<?php 
$qa = mysql_query("SELECT COUNT(*) AS br FROM `users`");
$aa = mysql_fetch_array($qa);
$a = $aa['br'] - 4;
$q = "SELECT * FROM `users` LIMIT $a, 4";
$result=mysql_query($q);
$num_rows = mysql_num_rows($result);
if(!$result || ($num_rows < 0)){
   echo "Error displaying info";
}
else if($num_rows > 0){
   for($i=0; $i<$num_rows; $i++){
      $uname = mysql_result($result,$i,"username");
      $pic = mysql_result($result,$i,"pic");
      $sex = mysql_result($result,$i,"sex");
      $points = mysql_result($result,$i,"points");
      if($sex=='m') {
	  ?>

Решение ?

$a е отрицателно число