razery Posted January 20, 2008 Report Share Posted January 20, 2008 ja ja jaaa, zinu, ka juus man tagad bljausiet virsu par to ka shii teema jau ir bijusi izrunata... Bet man searchaa rada error: Sorry, an error occurred. If you are unsure on how to use a feature, or don't know why you got this error message, try looking through the help files for more information. Nu, luudzu kaads iemetiet to kodu par TOP BANOTAJIEM (admin) webaa. Tjipa rindinja ar tiem adminiem, kuri ir visvairaak nobanojushi un CIK. Link to comment Share on other sites More sharing options...
tk1 Posted January 20, 2008 Report Share Posted January 20, 2008 Sorry, an error occurred. If you are unsure on how to use a feature, or don't know why you got this error message, try looking through the help files for more information. Link to comment Share on other sites More sharing options...
razery Posted January 20, 2008 Author Report Share Posted January 20, 2008 (edited) paldies par nespamu. Edited January 20, 2008 by razery Link to comment Share on other sites More sharing options...
FuNk1Y Posted January 20, 2008 Report Share Posted January 20, 2008 amxbans web direktorijā izveido jaunu php dokumentu (topadmin.php) <style type="text/css"> #admintop { font-size: 15px; color: #B22222; } #admintop td{ border: dotted; border-width: 1 0 0 0; padding-right: 20px; } #admintop th{ text-align: left; } </style> <?php echo "<table id='admintop'> <tr> <th>Admins</th> <th>Bani</th> </tr>"; $con = mysql_connect("1.MySQL host","2.MySQL user","3.MySQL parole"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("amxbans", $con); $result = mysql_query("SELECT count(*), admin_nick FROM amx_bans GROUP BY admin_nick ORDER BY count(*) DESC LIMIT 5"); // DESC LIMIT 5 noziimee, ka vinjsh paraadiis 5 aktiivaakos while($row = mysql_fetch_array($result)) { $admins = $row['admin_nick']; $skaits = $row['count(*)']; echo "<tr> <td>$admins</td> <td>$skaits</td> </tr>"; } mysql_close($con); ?> </table> Ierakstam savus datus: 1.MySQL host 2.MySQL user 3.MySQL parole Avots Link to comment Share on other sites More sharing options...
razery Posted January 20, 2008 Author Report Share Posted January 20, 2008 Ko nozīmē un kā labot šo erroru? Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in d:\mans webs\www\amxbans\admintop.php on line 36 Admins Bani Link to comment Share on other sites More sharing options...
eM` Posted January 21, 2008 Report Share Posted January 21, 2008 To, kā viņu labot neprasi mums, jo mēs nezinam kas tev ir iekš tās 36 rindas. Link to comment Share on other sites More sharing options...
lazda2 Posted January 21, 2008 Report Share Posted January 21, 2008 Lūk mana pamācība!! Varbuut noderēs Link to comment Share on other sites More sharing options...
razery Posted January 21, 2008 Author Report Share Posted January 21, 2008 <style type="text/css"> /*-------------admintop---------------*/ #admintop { font-size: 15px; color: #B22222; } #admintop td{ border: dotted; border-width: 1 0 0 0; padding-right: 20px; } #admintop th{ text-align: left; } </style> <?php echo "<table id='admintop'> <tr> <th>Admins</th> <th>Bani</th> </tr>"; $con = mysql_connect("localhost","mysql lietotaajvards","parole"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("amxbans", $con); $result = mysql_query("SELECT count(*), admin_nick FROM amx_bans GROUP BY admin_nick ORDER BY count(*) DESC LIMIT 5"); // DESC LIMIT 5 noziimee, ka vinjsh paraadiis 5 aktiivaakos while($row = mysql_fetch_array($result)) { $admins = $row['admin_nick']; $skaits = $row['count(*)']; echo "<tr> <td>$admins</td> <td>$skaits</td> </tr>"; } mysql_close($con); ?> </table> Luuk viss fails. nevaru atrast probleemu... kads var palidzet? Link to comment Share on other sites More sharing options...
lazda2 Posted January 21, 2008 Report Share Posted January 21, 2008 taks taisi visu peec manas pamaaciibas!!!! Link to comment Share on other sites More sharing options...
razery Posted January 22, 2008 Author Report Share Posted January 22, 2008 es jau taissiju pec tavas pamacibas. Link to comment Share on other sites More sharing options...
lazda2 Posted January 22, 2008 Report Share Posted January 22, 2008 ko tu izmanto app , wamp ..... ?? Link to comment Share on other sites More sharing options...
Drumais Posted January 22, 2008 Report Share Posted January 22, 2008 (edited) iskatas sads apaksa pievienoju failus ieks config.inc.php pieliekam velvienu rindinu klat $config->top_admin_limit = "5"; tur kur ir 5 tur vari ievadis savu top garumu top_admin.rar Edited January 22, 2008 by Drumais Link to comment Share on other sites More sharing options...
razery Posted January 22, 2008 Author Report Share Posted January 22, 2008 easyPHP1-7 Link to comment Share on other sites More sharing options...
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