cs1h Posted April 14, 2008 Share Posted April 14, 2008 Hi, I am trying to get a star rating script working but I get the following error. Fatal error: Call to a member function on a non-object in D:\Inetpub\vhosts\asfdhjfk.com\httpdocs\tester\rate.php on line 11 the code is, <?php include("DB.php"); mysql_connect($server, $db_user, $db_pass) or die ("Could not connect to mysql because ".mysql_error()); $id = $_GET['id']; $title = ''; $sql = "SELECT name FROM movies WHERE movie_id=$id"; while( $sql->fetchInto( $row ) ) { $title = $row[0]; } ?> <html> <head> <title><?php echo($title); ?></title> <script src="prototype.js"></script> <script> function rate( value ) { new Ajax.Updater( 'rating', 'ratemovie.php?id=<?php echo($id)?>&v='+value ); } </script> </head> <body> <h1><?php echo($title); ?></h1> <div id="rating"> <img src="star_off.gif" onClick="rate(1)"></img> <img src="star_off.gif" onClick="rate(2)"></img> <img src="star_off.gif" onClick="rate(3)"></img> <img src="star_off.gif" onClick="rate(4)"></img> <img src="star_off.gif" onClick="rate(5)"></img> <br/><br/> <?php $res2 = $db->query( 'SELECT count( rating ), sum(rating ) FROM ratings WHERE movie_id=?', $id ); while( $res2->fetchInto( $row ) ) { ?> Votes: <?php echo($row[0]); ?><br/> Average Rating: <?php echo($row[1]/$row[0]); ?> <?php } ?> </div> </body> </html> I don't know what to do about it. Can anyone help? Thanks, Colin Quote Link to comment Share on other sites More sharing options...
trq Posted April 14, 2008 Share Posted April 14, 2008 $sql is a string, not an object. You likely need to call $db query as you do further down, however, I don't see where you define $db either. Quote Link to comment Share on other sites More sharing options...
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