I have this error appearing
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /websites/123reg/LinuxPackage21/th/er/ag/theragereport.co.uk/public_html/feed/feed_view.php on line 148
from this code
$sql3="SELECT view FROM $tbl_name2 WHERE id='40'";
$result3=mysql_query($sql3);
$rows=mysql_fetch_array($sql3);
$view=$rows['view'];
// if have no counter value set counter = 1
if(empty($view)){
$view=1;
$sql4="INSERT INTO $tbl_name(view) VALUES('$view') WHERE id='40'";
$result4=mysql_query($sql4);
}
// count more value
$addview=$view+1;
$sql5="update $tbl_name set view='$addview' WHERE id='40'";
$result5=mysql_query($sql5);
any help would be hot
Regards Ryan Cosans