Login page unsuccessful

LiquidFusi0n · September 25, 2011

I always feel bad for my first post on a new forum being a question. But this has me stuck

I am writing a very simple log-in script. To check I am using 'mysql_num_rows' and seeing that only one row is returned. I have tried debugging this, and using Google but to no avail.

The page is saying 'Login unsuccessful' even when the values are correct. Any help is greatly appreciated.

<html>
<head>
<title> Log-in </title>
</head>
<body>

<form action="login.php" method="POST"> 
Username: <input type="text" name=username">
Password: <input type="password" name="password">
<input type="submit" name="login" value="Login">
</form>

<?php 

if(isset($_POST['username'])){
        $un = $_POST['username'];
}

if(isset($_POST['password'])){
        $pass = $_POST['password'];
}

$db = mysql_connect("127.0.0.1:3306", "root" , "password"); //Not actual password btw 

if(!$db) {
        die("Error connecting to database: " . mysql_error());
}

$db_used = mysql_select_db("login", $db);

if(!$db_used){
        die("Could not select database: " . mysql_error());
}

$user_name = mysql_real_escape_string($un);
$password = mysql_real_escape_string($pass);

$query = "SELECT * FROM details WHERE un=" . $user_name . " AND pwd=" . $password. ""; 

$query2 = "SELECT * FROM details WHERE un='$user_name' AND pwd='$password'"; //This was for debugging, seeing if it was the passing that was failing 

$result = mysql_query($query);

if(!$result){
        echo "There was an error processing your request: " . mysql_error() . ""; 
}

$check = mysql_num_rows($result);

if($check == 1){ 
        echo "Login successful, welcome back " . $user_name . ""; 
}
else{
        echo "Login unsuccessful, please ensure you are using the correct details";
}

?>

</body>
</html>

Thanks in advance guys....

--LiquidFusi0n

WebStyles · September 25, 2011

here's a simplified version that should work: (not the best way to do it though...)

<html>
<head>
<title> Log-in </title>
</head>
<body>

<?php 
if($_SERVER['REQUEST_METHOD']=="POST"){
$db = mysql_connect("127.0.0.1:3306", "root" , "password")or die("Error connecting to database: " . mysql_error());
$db_used = mysql_select_db("login", $db)or die("Could not select database: " . mysql_error());

$user_name = mysql_real_escape_string($_POST['username'],$db);
$password = mysql_real_escape_string($_POST['password'],$db);

$query = mysql_query("SELECT * FROM `details` WHERE `un` = '$user_name' AND `pwd` = '$password'",$db) or die(mysql_error());
if(mysql_num_rows($query) == 1){ 
        echo "Login successful, welcome back " . $user_name . ""; 
}else{
        echo "Login unsuccessful, please ensure you are using the correct details";
}
}else{
?>
<form action="login.php" method="POST"> 
Username: <input type="text" name=username">
Password: <input type="password" name="password">
<input type="submit" name="login" value="Login">
</form>
<?php
}
?>

</body>
</html>

LiquidFusi0n · September 25, 2011

Thank you, I will use that if I can't get my own code fixed

I am still failing to see why my code is not working... the syntax/logic looks sound to me?

--LiquidFusi0n

LiquidFusi0n · September 26, 2011

Apologies for this 'self-bump' but this is getting beyond a joke :wtf:

I have now added error checking at every stage and no errors are given. This to means it MUST be an error with 'mysql_num_rows'.

Anyways I have re-done the code for you guys and I will post the database structure so you can see for yourselves what is looks like

MySQL Query With Result

mysql> SELECT * FROM details WHERE un='test' AND pwd='pass';
+----+------+------+
| id | un   | pwd  |
+----+------+------+
|  2 | test | pass |
+----+------+------+
1 row in set (0.00 sec)

Code:

<html>
<head>
<title> Log-in </title>
</head>
<body>

<form action="login.php" method="POST"> 
Username: <input type="text" name=username">
Password: <input type="password" name="password">
<input type="submit" name="login" value="Login">
</form>

<?php 

$un = ''; 

if(isset($_POST['username'])){
        $un = $_POST['username'];
}

$pass = ''; 

if(isset($_POST['password'])){
        $pass = $_POST['password'];
}

$db = mysql_connect("127.0.0.1:3306", "root" , "password"); //Not actual password btw 

if(!$db) {
        die("Error connecting to database: " . mysql_error());
}

$db_used = mysql_select_db("login", $db);

if(!$db_used){
        die("Could not select database: " . mysql_error());
}

$user_name = mysql_real_escape_string($un);
$password = mysql_real_escape_string($pass);

$query = "SELECT * FROM details WHERE un='$user_name' AND pwd='$password'"; //Exact same command works when given the MySQL directly 

$result = mysql_query($query);

if(!$result){
        die("SQL query failed: " . mysql_error()); //Query is not returning an error 
}

if(mysql_num_rows($result) == 1) {
        echo "Login successful welcome back, " . $user_name . ""; 
}
else{
        echo "Login unsuccessful";
}

?>

</body>
</html>

This is now past how to do it I really want to know WHY it isn't working

--LiquidFusi0n

LiquidFusi0n · September 26, 2011

OK Thread can be closed. After some more de-bugging it was clear that username wasn't being passed. Want to see the culprit?

Username: <input type="text" name=username">

Typo's will be the end of us >:(

--LiquidFusi0n

Pikachu2000 · September 26, 2011

Echo, echo, echo and echo some more when debgging!

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