username and password passed as integers rather than strings - php

Question

I'm creating a login script and its failing each time even when I input the correct credentials. I firstly thought this was to do with the crypt function which has been explained in depth and have done further research upon to find a good method.
Anyway I have remove the crypt function in both the login and register to test the login script. Upon removing this I still got an error which stated Invalid username or password.
I wanted to see what was actually been passed in my POST variables, so I echoed these variables as they are defined e.g.
$username = isset($_POST['username']);
echo $username;
outputs: 1
however
echo $_POST['username'];
outputs admin (which is the username I have been using)
If I remove the isset from this POST variable then I am returned with the true value but this still fails my login script.
I don't understand whats going wrong as I've used this same statement in a different function to test it and it works, instead of using POST variables I am already setting the username and password and then wrapping in a foreach statement and this works.
It must be something to do with my POST variables and the way I am handling it but I don't have the experience/knowledge to solve the problem. I can't understand why my POST data is returned as a integer but then even when I remove the isset tag my statement is still false?
Any input and help is greatly welcomed and appreciated.
Below is my code, if there's anything else required please ask:
index.php
<form id="loginForm" method="POST" action="classes/class.Login.php">
<input type="text" id="username" name="username" placeholder="Username"/>
<input type="password" id="password" name="password" placeholder="Password" class="showpassword"/>
<input type="submit" name="submit" value="Log in"/>
classes/class.Login.php
public function loginuser() {
$username = isset($_POST['username']);
$password = isset($_POST['password']);
//$salt = "boo";
//$pw = crypt($password, $salt);
$stmt = $this->pdo->prepare("SELECT * FROM users WHERE username = :username AND password = :password LIMIT 1");
$stmt->bindValue(":username", $_POST['username'], PDO::PARAM_STR);
$stmt->bindValue(":password", $_POST['password'], PDO::PARAM_STR);
if ($stmt->fetch(PDO::FETCH_ASSOC))
{
$_SESSION['authorized'] = true;
$_SESSION['username'] = $username;
header('Location: testloginrequired.php');
} else {
echo "Invaild username or password. Try again";
}
}// end loginuser

Answer 1

Change:
$username = isset($_POST['username']);
to :
$username = ($_POST['username']);
isset function returns a boolean value (true,false) , so your $username would be boolean (true or 1) not a string!

Answer 2

$username = isset($_POST['username']);
Is just checking if the variable is set. Get rid of the isset and change it simply to
$username = $_POST['username'];
If that doesn't work, then do this
$username = settype($username, "string");

Answer 3

Use construction:
if (isset($_POST['username'])) $username = $_POST['username'];
If you will remove "isset" check, you will get warning, if key 'username' will not exists in POST.
Improvement of your code:
$username = (isset($_POST['username'])) ? $_POST['username'] : '';
This is ternary operator and it's suitable in your case. If key 'username' exists in POST, then $username will equals $_POST['username'], otherwise $username will empty. Enjoy :)