eaglelegend Posted June 14, 2009 Share Posted June 14, 2009 Hi, I am getting rather annoyed with this little bit of code as it is giving me an error, on my application on facebook... any ideas? The code: it is meant to get the users that have recently viewed my application from the database then display their pictures on the app. $sql = "SELECT * FROM usertable order by uid DESC limit 7"; $result = mysql_query($sql, $connection); $num = @mysql_num_rows($result); $i = 0; print"<div align=center><b>Recent Visitors</b><br />"; while ($num > $i) { $uids = mysql_result($result, $i, "uid"); $scores = mysql_result($result, $i, "point"); $num_games = mysql_result($result, $i, "gameplayed"); s ?> <fb:profile-pic uid="<?php echo $uids; ?>" size="square" linked="true" /> echo '<fb:tabs>'; echo '<fb:tab-item href='.$appCanvasUrl.' title="Arcade Site List" selected="true" />'; echo '<fb:tab-item href='.$appCanvasUrl.'invite.php title="Invite Friends" />'; echo '</fb:tabs>'; <?php //$game = mysql_result($result,$i, "game"); $i++; } print"</div>"; ?> This is the error im getting Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in THE_CODE on line 18 Quote Link to comment Share on other sites More sharing options...
papaface Posted June 14, 2009 Share Posted June 14, 2009 $connection is not defined anywhere. Quote Link to comment Share on other sites More sharing options...
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