Проблем с логина

C

chepa

Registered
Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\login.php on line 84

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\xampp\htdocs\login.php on line 84

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\login.php on line 87

Ето от 84 и 87 ред кода в червено:




<?php

$result = mysql_query("SELECT * FROM register ORDER BY id LIMIT 1")


while($row = mysql_fetch_array($result))
{

$uid = htmlspecialchars($_GET['id']);
$result = mysql_query("SELECT * FROM register WHERE Username = '$username'") or die(mysql_error());
$br = mysql_num_rows($result);
if ($br == "1")
{
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$username = $row['Username'];


}

}
}
 
Оправи си config файла, явно не си го настроил както трябва
и какви са тея bb кодове които си слагал, прой таака

$result = mysql_query("SELECT * FROM register ORDER BY id LIMIT 1") or die (mysql_error());


while($row = mysql_fetch_array($result))
{

$uid = htmlspecialchars($_GET['id']);
$result = mysql_query("SELECT * FROM register WHERE Username = '$username'") or die(mysql_error());
$br = mysql_num_rows($result);
if ($br == "1")
{
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$username = $row['Username'];


}

}
}
Увеличете...
 
пробвай така
Код: 
$result = mysql_query("SELECT * FROM register ORDER BY id LIMIT 1") or die (mysql_error());


while($row = mysql_fetch_array($result))
{

$uid = htmlspecialchars($_GET['id']);
$result = mysql_query("SELECT * FROM register WHERE Username = '$username'") or die(mysql_error());
$br = mysql_num_rows($result);
if ($br == "1")
{
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$username = $row['Username'];


}

}
}
 
lub4o каза:
пробвай така
Код: 
$result = mysql_query("SELECT * FROM register ORDER BY id LIMIT 1") or die (mysql_error());


while($row = mysql_fetch_array($result))
{

$uid = htmlspecialchars($_GET['id']);
$result = mysql_query("SELECT * FROM register WHERE Username = '$username'") or die(mysql_error());
$br = mysql_num_rows($result);
if ($br == "1")
{
while($row = mysql_fetch_array($result))
{
$id = $row['id'];
$username = $row['Username'];


}

}
}
Увеличете...
Нестава...

Това дава

Warning: mysql_query() [function.mysql-query]: Access denied for user 'ODBC'@'localhost' (using password: NO) in C:\xampp\htdocs\login.php on line 84

Warning: mysql_query() [function.mysql-query]: A link to the server could not be established in C:\xampp\htdocs\login.php on line 84
Access denied for user 'ODBC'@'localhost' (using password: NO)
 
Код: 
$result = mysql_query("SELECT * FROM register ORDER BY id LIMIT 1") or die (mysql_error());


while($row = mysql_fetch_array($result))
{

$uid = htmlspecialchars($_GET['id']);
$result1 = mysql_query("SELECT * FROM register WHERE username = '$username'") or die(mysql_error());
$br = mysql_num_rows($result);
if ($br == "1")
{
$rows = mysql_fetch_array($result1);
$id = $rows['id'];
$username = $rows['Username'];

}
}
 
lub4o каза:
Код: 
$result = mysql_query("SELECT * FROM register ORDER BY id LIMIT 1") or die (mysql_error());


while($row = mysql_fetch_array($result))
{

$uid = htmlspecialchars($_GET['id']);
$result1 = mysql_query("SELECT * FROM register WHERE username = '$username'") or die(mysql_error());
$br = mysql_num_rows($result);
if ($br == "1")
{
$rows = mysql_fetch_array($result1);
$id = $rows['id'];
$username = $rows['Username'];

}
}
Увеличете...


Няма разлика явно кода е верен от Phpmyadmin е ясно аз се логвам там ама ..

Ето кода на connect2.php ..


Код: 
<?php
$dbhost = "localhost";
$dbuser = "root";
$dbpass = "Паролата..";
$dbname = "subs";
$conn = mysql_connect($dbhost, $dbuser, $dbpass)or die(mysql_error());
mysql_select_db($dbname, $conn)or die(mysql_error());
mysql_query("SET CHARACTER SET cp1251");
mysql_query("SET NAMES cp1251");

?>
 
Аз ползвам този код за логин

<?php
session_start();
include 'config.php';

if(isset($_POST['submit']))
{
$username = $_POST['username'];
$password = md5($_POST['password']);
$sql = mysql_query("SELECT * FROM users WHERE username = '$username'") or die (mysql_error());
$row = mysql_fetch_assoc($sql);
$broi = mysql_num_rows($sql);
if($broi == 0)
{
echo 'Няма такъв потребител';
}else{
if($password != $row[password])
{
echo 'Грешна парола';
}else{
$_SESSION['username'] = $row['username'];

$_SESSION['user_id'] = $row['id'];

echo 'Добре дошъл '."$_SESSION[username]";
echo 'Добре дdddшъл '."$_SESSION[user_id]";
}
}
} ?>

<form action="" method="post">
Потребител: <input type="text" name="username" /> <br />
Парола: <input type="password" name="password" /><br />
<input type="submit" name="submit" value="submit" />
</form>
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С него няма да имаш проблем само си промени имената на таблиците
 
Ам ще ми кажеш ли как да направя така че да ми запазва логина 60 минути след което да го дропва COOKIE мисля че беше
 
Трябва да си влезнал или регистриран, за да отговориш тук.

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