Website user password verification

I have a login web page where a user enters their email and password and I need to check if:

• the entered email exists in the DB and

• the entered password matches the hashed one from the db.

The password in the DB was hashed with

$pw = password_hash($_POST["pw"], PASSWORD_BCRYPT);

and is stored in a VARCHAR(255) column.
Email is set to primary and unique in the DB so there cannot be a duplicate record for this.

To achieve the above I came to the below which works as intended for me.
However, since I am new to this and because of the double DB connection I was wondering if there is a way to shorten or otherwise improve this code.

$email = $_POST["email"];
$pw = $_POST["pw"]; 

$stmt = $conn->prepare("SELECT email FROM Users WHERE email = ?");
$stmt->bind_param('s', $email);
$stmt->execute();
$result = $stmt->get_result();
if(mysqli_num_rows($result) == 0){
    echo "Email has not been registered yet";
}else{
    $stmt = $conn->prepare("SELECT pw FROM Users WHERE email = ? LIMIT 1");
    $stmt->bind_param('s', $email);
    $stmt->execute();
    $result = $stmt->get_result();
    $pwHashed = $result->fetch_assoc();
    if(password_verify($pw, $pwHashed["pw"])){
        echo "Password correct";
    }else{
        echo "Password incorrect";
    }   
}

You really don't need two queries to do this. You can just use your second query, and if it doesn't return any results, report that. Eg:

$stmt = $conn->prepare("SELECT pw FROM Users WHERE email = ? LIMIT 1");
$stmt->bind_param('s', $email);
$stmt->execute();
$result = $stmt->get_result();

if(mysqli_num_rows($result) == 0){
    echo "Email has not been registered yet";
}

$pwHashed = $result->fetch_assoc();
if(password_verify($pw, $pwHashed["pw"])){
    echo "Password correct";
}else{
    echo "Password incorrect";
}

Other than this, your code looks good. You use prepared statements, so it's secure, and your style is consistent and easy to read. I would probably write pw as password as it's more readable, but otherwise everything looks good.

StackExchange Code Review Q#94528, answer score: 3