Davie33 Posted June 22, 2012 Share Posted June 22, 2012 Hi am having a problem with this error i am seeing and dont know why. This is my error...Warning: mysql_fetch_row() expects parameter 1 to be resource, null given in C:\wamp\www\ydiycomputers\inc\includes.php on line 8 <?php if (isset($_GET['page'])) { $pageno = ($_GET['page']); } else { $pageno = 1; } $query = "SELECT count(id) FROM items WHERE category = 'id'"; $query_data = mysql_fetch_row($result); $numrows = $query_data[0]; $rows_per_page = 5; $lastpage = CEIL($numrows/$rows_per_page); $pageno = (int)$pageno; IF ($pageno < 1) { $pageno = 1; } ELSEIF ($pageno > $lastpage) { $pageno = $lastpage; } $limit = 'LIMIT ' .($pageno - 1) * $rows_per_page .',' .$rows_per_page; ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <link rel="stylesheet" type="text/css" href="<?php ECHO $urlpath;?>template/style/style.css" /> <link rel="stylesheet" type="text/css" href="<?php ECHO $urlpath;?>template/style/home.css" /> <link rel="stylesheet" type="text/css" href="<?php ECHO $urlpath;?>template/style/cats.css" /> <?php SWITCH($_GET['act']){ CASE 'cats': $query = mysql_query("SELECT * FROM `cats` WHERE id = 'id'"); $row = mysql_fetch_array($query); ?> <title><?php ECHO $sitedomain;?> - <?php ECHO $row['catname']; ?></title> <?php BREAK; DEFAULT:?> <title><?php ECHO $sitedomain;?> - <?php ECHO $slogan;?></title> <?php BREAK; } ?> </head> Quote Link to comment Share on other sites More sharing options...
mikosiko Posted June 22, 2012 Share Posted June 22, 2012 you are not executing the query ($query), hence $result doesn't exist Quote Link to comment Share on other sites More sharing options...
The Little Guy Posted June 22, 2012 Share Posted June 22, 2012 I usually get that when my query is invalid. I also see you don't have "mysql_query" in the code either Quote Link to comment Share on other sites More sharing options...
Davie33 Posted June 22, 2012 Author Share Posted June 22, 2012 you are not executing the query ($query), hence $result doesn't exist so it sould be like this ? $result= "SELECT count(id) FROM items WHERE category = 'id'"; $query_data = mysql_fetch_row($result); Quote Link to comment Share on other sites More sharing options...
Davie33 Posted June 22, 2012 Author Share Posted June 22, 2012 I usually get that when my query is invalid. I also see you don't have "mysql_query" in the code either I didnt even notice that thanks for that Quote Link to comment Share on other sites More sharing options...
Davie33 Posted June 22, 2012 Author Share Posted June 22, 2012 Thank you got it sorted thanks guys. Quote Link to comment Share on other sites More sharing options...
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