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Unlocking character error

Discussion on Unlocking character error within the Metin2 Private Server forum part of the Metin2 category.

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Old   #1
 
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Unlocking character error

Hy everyone,I have a problem with my server,with some characters,some characters work,others don't.At the login the character stays with his hand up.I have resolved it changing the map where the character is found out.But i can't always complete them manually.So I found a script:



<?php
$pwd = Array();
$query = Array();
$db = Array();
$results = Array();

$db['host'] = "host";
$db['user'] = "user";
$db['password'] = "password";
$db['database'] = Array();
$db['database']['player'] = "player";
$db['database']['account'] = "account";

$db['conn'] = mysql_connect($db['host'], $db['user'], $db['password']) or die('Could not connect: '.mysql_error());
?>

<form method="post" action="">
<label><input type="text" name="pg" /> Introdu numele caracterului </label><br />
<label><input type="text" name="pw" /> Introdu parola contului </label><br />
<input type="hidden" name="sbug" value="1" />
<input type="submit" value="Sbugga personaggio" />
</form>
<br />

<?php
if($_POST['sbug']) {
mysql_select_db($db['database']['account']) or die('Could not select database: '.mysql_error());
echo '$_POST[\'sbug\'] trovato e connesso al database account'."<br />\n";
$pg = mysql_real_escape_string($_POST['pg']);
$pwd['user'] = mysql_real_escape_string($_POST['pw']);
$query['pwd'] = "SELECT password FROM account WHERE login = '$pg'";
$pwd['fromdb'] = mysql_query($query['pwd']);
$menge = mysql_num_rows($pwd['fromdb']);
$results['pwd'] = mysql_fetch_row($pwd['fromdb']);
echo $pwd['fromdb']."<br />\n";
echo $results['pwd'][0]."<br />\n";

if($pwd['user'] == $results['pwd'][0]) {#$pwd['fromdb']) { #&& $menge == "1") {
echo 'Le due password coincidono';
$query['id'] = "SELECT id FROM account WHERE login = '$pg' AND password = '".$pwd['user']."'";
$id = mysql_query($query['id']);
if ($id != "") {
mysql_select_db($db['database']['player']) or die('Could not select database: '.mysql_error());
echo '$id diverso da "" e connesso al database player';
$query['sbug'] = "UPDATE `player`.`player` SET `x`=436377, `y`=215769, `map_index`=61, `exit_x`=436378, `exit_y`=215769, `exit_map_index`=61 WHERE (`name`='$pg') AND (`account_id`='$id')";
$success = mysql_query($query['sbug']);
if($success != "") {
echo "PG Sbuggato";
} else {
echo "PG NON Sbuggato";
}
}
}
}
?>




And I implemented it in my site,but whatever I write in it it shows
$_POST['sbug'] trovato e connesso al database account
Resource id #4
What can I modify in it?Can you please tell me? Thank you very much.
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Old 10/06/2011, 00:55   #2


 
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EDIT###

Sry fail...

I watch the script...


Mfg. Benhero

Have you try this...
1. Charakter Name!
2. Account Password from your Charakter.

Mfg
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Old 10/06/2011, 01:01   #3
 
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Still $_POST['sbug'] trovato e connesso al database account
Resource id #4
Any other ideas? Thank you anyway for helping.
EDIT: I've tried anything,even correct and false names and passwords,nothing,at everyone of them it appears the same.Obviously it shoul appear PG Sbuggato or PG NON Sbuggato isn't that right?
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Old 10/06/2011, 01:06   #4


 
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okok^^ Try this one:
PHP Code:
<?php
$pwd = Array();
$query = Array();
$db = Array();
$results = Array();

$db['host'] = "host";
$db['user'] = "user";
$db['password'] = "password";
$db['database'] = Array();
$db['database']['player'] = "player";
$db['database']['account'] = "account";

$db['conn'] = mysql_connect($db['host'], $db['user'], $db['password']) or die('Could not connect: '.mysql_error());
?>

<form method="post" action="">
<label><input type="text" name="pg" /> Introdu numele caracterului </label><br />
<label><input type="text" name="pw" /> Introdu parola contului </label><br />
<input type="hidden" name="sbug" value="1" />
<input type="submit" value="Sbugga personaggio" />
</form>
<br />

<?php
if($_POST['sbug']) {
mysql_select_db($db['database']['account']) or die('Could not select database: '.mysql_error());
echo "$_POST[\'sbug\'] trovato e connesso al database account"."<br />\n";
$pg = mysql_real_escape_string($_POST['pg']);
$pwd['user'] = mysql_real_escape_string($_POST['pw']);
$query['pwd'] = "SELECT password FROM account WHERE login = '$pg'";
$pwd['fromdb'] = mysql_query($query['pwd']); 
$menge = mysql_num_rows($pwd['fromdb']);
$results['pwd'] = mysql_fetch_row($pwd['fromdb']);
echo $pwd['fromdb']."<br />\n";
echo $results['pwd'][0]."<br />\n";

if($pwd['user'] == $results['pwd'][0]) {#$pwd['fromdb']) { #&& $menge == "1") {
echo 'Le due password coincidono';
$query['id'] = "SELECT id FROM account WHERE login = '$pg' AND password = '".$pwd['user']."'";
$id = mysql_query($query['id']); 
if ($id != "") {
mysql_select_db($db['database']['player']) or die('Could not select database: '.mysql_error());
echo '$id diverso da "" e connesso al database player';
$query['sbug'] = "UPDATE `player`.`player` SET `x`=436377, `y`=215769, `map_index`=61, `exit_x`=436378, `exit_y`=215769, `exit_map_index`=61 WHERE (`name`='$pg') AND (`account_id`='$id')";
$success = mysql_query($query['sbug']);
if($success != "") {
echo "PG Sbuggato";
} else {
echo "PG NON Sbuggato";
}
}
}
}
?>
Mfg. Benhero
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Old 10/06/2011, 01:14   #5
 
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Parse error: syntax error, unexpected T_ENCAPSED_AND_WHITESPACE, expecting T_STRING or T_VARIABLE or T_NUM_STRING in /home/advance7/public_html/deblocare.php on line 28
I've tried to look out to the line 28..but what to modify? :-??
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Old 10/06/2011, 01:17   #6


 
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Hmm... Delete line 28 ... and try again O.o

Soo... i think... here is the fix.. try it x,x sry when its fail..

PHP Code:
<?php
$pwd = Array();
$query = Array();
$db = Array();
$results = Array();

$db['host'] = "localhost";
$db['user'] = "benhero";
$db['password'] = "made2507";
$db['database'] = Array();
$db['database']['player'] = "player";
$db['database']['account'] = "benhero";

$db['conn'] = mysql_connect($db['host'], $db['user'], $db['password']) or die('Could not connect: '.mysql_error());
?>

<form method="post" action="">
<label><input type="text" name="pg" /> Introdu numele caracterului </label><br />
<label><input type="text" name="pw" /> Introdu parola contului </label><br />
<input type="hidden" name="sbug" value="1" />
<input type="submit" value="Sbugga personaggio" />
</form>
<br />

<?php
if($_POST['sbug']) {
mysql_select_db($db['database']['account']) or die('Could not select database: '.mysql_error());
echo "\$_POST['sbug'] trovato e connesso al database account"."<br />\n";
$pg = mysql_real_escape_string($_POST['pg']);
$pwd['user'] = mysql_real_escape_string($_POST['pw']);
$query['pwd'] = "SELECT password FROM account WHERE login = '$pg'";
$pwd['fromdb'] = mysql_query($query['pwd']); 
$menge = mysql_num_rows($pwd['fromdb']);
$results['pwd'] = mysql_fetch_row($pwd['fromdb']);
echo $pwd['fromdb']."<br />\n";
echo $results['pwd'][0]."<br />\n";

if($pwd['user'] == $results['pwd'][0]) {#$pwd['fromdb']) { #&& $menge == "1") {
echo 'Le due password coincidono';
$query['id'] = "SELECT id FROM account WHERE login = '$pg' AND password = '".$pwd['user']."'";
$id = mysql_query($query['id']); 
if ($id != "") {
mysql_select_db($db['database']['player']) or die('Could not select database: '.mysql_error());
echo '$id diverso da "" e connesso al database player';
$query['sbug'] = "UPDATE `player`.`player` SET `x`=436377, `y`=215769, `map_index`=61, `exit_x`=436378, `exit_y`=215769, `exit_map_index`=61 WHERE (`name`='$pg') AND (`account_id`='$id')";
$success = mysql_query($query['sbug']);
if($success != "") {
echo "PG Sbuggato";
} else {
echo "PG NON Sbuggato";
}
}
}
}
?>
Mfg
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Old 10/06/2011, 01:28   #7
 
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Still
$_POST['sbug'] trovato e connesso al database account
Resource id #4
Than you very much for your help,any other ideas are welcomed.Thank you.
EDIT
Without line 28
Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 4 in /home/advance7/public_html/deblocare.php on line 33
$pwd['fromdb'] : Resource id #4
$results['pwd']
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Old 10/06/2011, 01:43   #8


 
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Here... a other version XD iŽll try everything to fix it...
PHP Code:
<?php
$pwd = Array();
$query = Array();
$db = Array();
$results = Array();

$db['host'] = "host";
$db['user'] = "user";
$db['password'] = "password";
$db['database'] = Array();
$db['database']['player'] = "player";
$db['database']['account'] = "account";

$db['conn'] = mysql_connect($db['host'], $db['user'], $db['password']) or die('Could not connect: '.mysql_error());
?>

<form method="post" action="">
<label><input type="text" name="pg" /> Introdu numele caracterului </label><br/>
<label><input type="text" name="pw" /> Introdu parola contului </label><br/>
<input type="hidden" name="sbug" value="1" />
<input type="submit" value="Sbugga personaggio" />
</form>
<br/>

<?php
if($_POST['sbug']) {
mysql_select_db($db['database']['account']) or die('Could not select database: '.mysql_error());
echo '$_POST[\'sbug\'] trovato e connesso al database account'."<br/>\n";
$pg = mysql_real_escape_string($_POST['pg']);
$pwd['user'] = mysql_real_escape_string($_POST['pw']);
$query['pwd'] = "SELECT password FROM account WHERE login = '$pg'";
$pwd['fromdb'] = mysql_query($query['pwd']); 
$menge = mysql_num_rows($pwd['fromdb']);
$results['pwd'] = mysql_fetch_row($pwd['fromdb']);
echo $pwd['fromdb']."<br/>\n";
echo $results['pwd']."<br/>\n";

if($pwd['user'] == $results['pwd']) {#$pwd['fromdb']) { #&& $menge == "1") {
echo 'Le due password coincidono';
$query['id'] = "SELECT id FROM account WHERE login = '$pg' AND password = '".$pwd['user']."'";
$id = mysql_query($query['id']); 
if ($id != "") {
mysql_select_db($db['database']['player']) or die('Could not select database: '.mysql_error());
echo '$id diverso da "" e connesso al database player';
$query['sbug'] = "UPDATE `player`.`player` SET `x`=436377, `y`=215769, `map_index`=61, `exit_x`=436378, `exit_y`=215769, `exit_map_index`=61 WHERE (`name`='$pg') AND (`account_id`='$id')";
$success = mysql_query($query['sbug']);
if($success != "") {
echo "PG Sbuggato";
} else {
echo "PG NON Sbuggato";
}
}
}
}
?>
Mfg. Benhero
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Old 10/06/2011, 10:29   #9
 
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Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/advance7/public_html/deblocare.php on line 59 srry,it doesn't work either.Got any other idea? Thanks anyway.
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Old 10/06/2011, 18:00   #10


 
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The script dosenŽt have Line 59.


Mfg. Benhero
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