Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given i

[rabesh]

<?php    
$limit=12;
if(isset( $_GET['page']))
      $page=$_GET['page'];
      if($page<=0) $page = 1;
      
      else {$start=0;}
   $sql=mysql_query('select * from tbl_gallery where status=1 AND category_name="0"');
   $count=mysql_num_rows($sql);
   $totalcount=ceil($count/$limit);
   $start=($page-1)*$limit;
   $s=mysql_query('select * from tbl_gallery where status=1 AND category_name="0" limit $start, $limit');
   while($result=mysql_fetch_array($s)){
      $start++;
      
?>

 

MOD EDIT: code tags added.

[Pikachu2000]

When posting code, enclose it within the forum's

 . . . 

BBCode tags.

[rabesh]

I am new user of this forum so i dont know...please solve my problem......

 

Error in    $s=mysql_query('select * from tbl_gallery where status=1 AND category_name="0" limit $start, $limit');

above line

[Pikachu2000]

That error indicates that a query failed. I don't see anywhere in that code where you make a connection to the database. Is that done somewhere else in the script?

 

One problem you have is that variables are not interpolated in single quoted strings, so the query should be rewritten using double quotes to enclose it.

 

$query = "select * from tbl_gallery where status=1 AND category_name='0' limit $start, $limit";
$s=mysql_query($query);

[rabesh]

Thanks alot Pikachu2000...my problem is solved

Error in qotation mark....

Any way thanks alot to Pikachu2000 specially and other who help me indirectly as well as the owner of this site...