flemingmike Posted February 25, 2009 Share Posted February 25, 2009 ie says error on line 44. here are the lines surrounding line 44. (it was working before, and just stopped. no changes to the script, only changes to database was more entries added. $totalteams=mysql_query("SELECT COUNT(*) FROM ladder_$id "); $totalteams=mysql_fetch_array($totalteams); $totalteams="$totalteams[0]"; here is my whole code: <? include("./includes/incglobal.php"); global $config; $groups=mysql_query("SELECT id,name,logo,active,priority,created FROM groups WHERE active='1' ORDER BY priority"); while(list($gid,$gname,$glogo,$gactive,$gpriority,$gcreated)=mysql_fetch_row($groups)){ $totalla=mysql_query("SELECT COUNT(*) FROM ladders WHERE grid='$gid' ORDER BY id"); $totalla=mysql_fetch_array($totalla); $totalla="$totalla[0]"; if(!$glogo){ $headlogo=""; }else{ $headlogo="<img src='$glogo' border='0'><br />"; } $out[body].="<center><br />$headlogo <table width='96%' cellpadding='1' cellspacing='1' bgcolor='#000000' border='0'> <tr> <td width='37%' style='background: url($config[bg]) repeat-x;' colspan='7'><b>$gname</b></td> </tr> <tr> <td width='37%' style='background: url($config[bg2]) repeat-x;'> Ladder</td> <td width='30%' style='background: url($config[bg2]) repeat-x;'> Game</td> <td width='8%' style='background: url($config[bg2]) repeat-x;'>Join</td> <td width='8%' style='background: url($config[bg2]) repeat-x;'>Rules</td> <td width='10%' style='background: url($config[bg2]) repeat-x;'>Matches</td> <td width='12%' style='background: url($config[bg2]) repeat-x;'>Count</td> <td width='3%' style='background: url($config[bg2]) repeat-x;'></td> </tr>"; if($totalla == 0){ $out[body].=" <tr bgcolor='$config[altcolora]'><td width='100%' background='$config[cellbg]' colspan='7'> There are currently no ladders in this group</td></tr>"; } $ladders=mysql_query("SELECT id,name,game,gamelink,grid,isteam,active FROM ladders WHERE grid='$gid' ORDER BY id"); while(list($id,$name,$game,$gamelink,$grid,$isteam,$active)=mysql_fetch_row($ladders)){ $totalteams=mysql_query("SELECT COUNT(*) FROM ladder_$id "); $totalteams=mysql_fetch_array($totalteams); $totalteams="$totalteams[0]"; if ($active == 0) { $light = "<img src='./images/ineligible.png' border='0' alt='Ladder is inactive!'>"; $link = "$name"; }else{ $light = "<img src='./images/eligible.png' border='0' alt'Ladder is active!'>"; $link = "<a href='$config[scripturl]/standings.php?ladder[id]=$id' class='content'>$name</a>"; } if($gamelink){ $glink="<a href='$gamelink'>$game</a>"; }else{ $glink="$game"; } if($config[altcolorx]==$config[altcolora]){ $config[altcolorx]="$config[altcolorb]"; }else{ $config[altcolorx]="$config[altcolora]"; } if($config[cellbgx]==$config[cellbg]){$config[cellbgx]="$config[cellbg2]";}else{$config[cellbgx]="$config[cellbg]";} $out[body].=" <tr bgcolor='$config[altcolorx]'> <td width='30%' background='$config[cellbgx]'> $link </td> <td width='30%' background='$config[cellbgx]'> $glink</td> <td width='8%' background='$config[cellbgx]'><a href='./join.php?ladder=$id'>Join!</a></td> <td width='8%' background='$config[cellbgx]'><a href='./rules.php?rules[ladderid]=$id'>Rules</a></td> <td width='10%' background='$config[cellbgx]'><a href='./matchdb.php?ladder=$id'>All Match</a></td> <td width='12%' background='$config[cellbgx]'>$totalteams</td> <td width='3%' background='$config[cellbgx]'>$light</td> </tr>"; } $out[body].="</table></center>"; } $out[body].="<br /><br />"; include("$config[html]"); ?> here is the error im recieving: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mash905/public_html/betting/ladders.php on line 44 Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/mash905/public_html/betting/ladders.php on line 44 Link to comment https://forums.phpfreaks.com/topic/146858-i-might-just-be-stupid-phpmysql-problem-pleasethx-in-advance/ Share on other sites More sharing options...
limitphp Posted February 25, 2009 Share Posted February 25, 2009 I'm probably wrong here, but maybe because ladder_$id is not a valid table name? Link to comment https://forums.phpfreaks.com/topic/146858-i-might-just-be-stupid-phpmysql-problem-pleasethx-in-advance/#findComment-771024 Share on other sites More sharing options...
premiso Posted February 25, 2009 Share Posted February 25, 2009 For anyone helping on this topic, it is double posted. http://www.phpfreaks.com/forums/index.php/topic,240254.0.html Please use that one as it contains more updated information than this thread. Just an FYI. Link to comment https://forums.phpfreaks.com/topic/146858-i-might-just-be-stupid-phpmysql-problem-pleasethx-in-advance/#findComment-771026 Share on other sites More sharing options...
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