Prepared statement returning nothing - php

Question

I'm trying to confirm if a user submitted valid credentials.
HTML
<form action="assets/php/account/login_user.php" method="post" id="login" class="center">
<div id="register_errorloc"></div>
<div class="form-group">
<input type="text" name="login_username" id="login_username" class="form-control" placeholder="Username" />
</div>
<div class="form-group">
<input type="password" name="login_password" id="login_password" class="form-control" placeholder="Password" />
</div>
<div class="form-group">
<input type="submit" value="login" id="submit_login" class="btn btn-success form-control"/>
</div>
</form>
PHP
$username = trim($_POST['login_username']);
$password = hash('sha512',trim($_POST['login_password']));
//to check if values are correct by myself
echo '<br>' . $username . '<br>' . $password . '<br>';
include_once '../../../assets/php/DB/DBConnect.php';
//check if valid credentials are entered
$stmt = $conn->prepare("SELECT * FROM tbl_users WHERE username = ?");
echo $stmt->bind_param('s', $username);
echo $stmt->execute();
echo $stmt->store_result();
if($stmt->num_rows == 1){
echo 'Good Creds';
}else{
echo 'Bad Creds';
}
When I'm doing this it always shows 0 results. When I change the querystring to
$stmt = $conn->prepare('SELECT * FROM tbl_users WHERE username = "abc"');
I get the result I'm looking for. I'm confused because I'm using the same approach on a different page and it's working like a charm.
Screenshots
Login form
Submitted form
Row I'm looking for
Edit: Added all code.
Edit2: Added screenshots
Edit3: Added Form HTML
Edit4:
Nothing is wrong with the form submit. When I'm hardcoding the variable like this:
$username ='abc';
It still doesn't work

Answer 1

I'm sorry but you posted screenshots that don't help us.
In order to avoid further commenting, am submitting the following.
Your form and its related elements may be failing and require a POST method and that name attributes exist for each of them.
I.e.:
<form method="post" action="???">
<input type="text" name="login_username">
...
Edit:
Try the following, comments are getting too extensive:
$check = $conn->prepare("SELECT * FROM tbl_users WHERE username=?");
$check->bind_param("s", $username);
$check->execute();
$check->store_result();
$row_chqry = $check->num_rows;
if($row_chqry > 0 ) {
echo "Exists";
$check->close();
$conn->close();
}
Edit #2:
OP posted an answer stating that they were using the same variable for their db connection.
Something I would have spotted right away and causing a conflict, had full code been posted. I've seen that happen quite a few times before, and remember posting answers for them too. As to where they are exactly, it would take some time for me to find them; but I do remember posting answers for related questions.
Oh well, problem solved.
Solutions:
Use a different variable for your username(s).
Use unique variables.

Answer 2

I finally found the error:
When connecting to my DB I use:
$servername = "";
$username = "";
$password = "";
$db = "";
// Create connection
$conn = mysqli_connect($servername, $username, $password, $db);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}
In my submission script I also use $username. Including the DBconnection file caused my local variable ($username that I get from the form) to be overwritten with the username found in the DBconnect.