Top Banotāji

[razery]

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.

[tk1]

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.

[razery]

paldies par nespamu.

Edited January 20, 2008 by razery

[FuNk1Y]

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

[razery]

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

[eM`]

To, kā viņu labot neprasi mums, jo mēs nezinam kas tev ir iekš tās 36 rindas.

[lazda2]

Lūk mana pamācība!! Varbuut noderēs

[razery]

<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?

[lazda2]

taks taisi visu peec manas pamaaciibas!!!!

[razery]

es jau taissiju pec tavas pamacibas.

[lazda2]

ko tu izmanto app , wamp ..... ??

[Drumais]

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

[razery]

easyPHP1-7