PHPはユーザーが存在するかどうかチェックする

Question

私は、データベースにユーザーが既に存在するかどうかを確認する機能を実装しようとしていますが、すべての登録データを挿入しますが、誰かが本当に事前にすべての答えを感謝しています。エラーがどこにあるか私が識別しやすくしてください。

<?php 

require '../ppuyakul/php/db_conn.php'; 


$message = ''; 

//Prepare date 
$DOB = date("Y-m-d", strtotime($_POST['year'].'-'. $_POST['month'].'-'. $_POST['day'])); 
$accessType = "0"; 

//Check enpty field 
if(!empty($_POST['email']) && !empty($_POST['password']) && !empty($_POST['fullname']) && !empty($_POST['username']) && !empty($_POST['password_confirmation']) && !empty($_POST['gender']) && !empty($_POST['country']) && !empty($_POST['state']) && !empty($_POST['city']) && !empty($_POST['day']) && !empty($_POST['month']) && !empty($_POST['year'])): 

    // Enter the new user in the database 
    $sql = "INSERT INTO assignment2 (fullname, username, email, password, gender, country, state, city, DOB, type) VALUES (:fullname, :username, :email, :password, :gender, :country, :state, :city, :DOB, :type)"; 
    $stmt = $conn->prepare($sql); 

    $stmt->bindParam(':fullname', $_POST['fullname']); 
    $stmt->bindParam(':username', $_POST['username']); 
    $stmt->bindParam(':email', $_POST['email']); 
    $stmt->bindParam(':password', password_hash($_POST['password'], PASSWORD_BCRYPT)); 
    $stmt->bindParam(':gender', $_POST['gender']); 
    $stmt->bindParam(':country', $_POST['country']); 
    $stmt->bindParam(':state', $_POST['state']); 
    $stmt->bindParam(':city', $_POST['city']); 
    $stmt->bindParam(':DOB', $DOB); 
    $stmt->bindParam(':type', $accessType); 

    $chk = $conn->prepare("SELECT username FROM assignment2 WHERE username = :name"); 
    $chk->bindParam(':name', $username); 
    $chk->execute(); 

    if($chk->rowCount() > 0): 
    $message = 'Error ! ! User already exists'; 
    else: 
     if($stmt->execute()): 
     $message = 'Successfully created new user'; 
      else: 
      $message = 'Sorry there must have been an issue creating your account'; 
      endif; 
     endif; 
    endif; 
?> 

Answer 1

@PaulのT.によると、私は最終的にここに解決策は、あなたの助けのおかげでそんなに再び、最終的なコードで見つかりました@ Paul T.

$username = $_POST['username']; 
    $chk = $conn->prepare("SELECT username FROM assignment2 WHERE username = :name"); 
    $chk->bindParam(':name', $username); 
    $chk->execute(); 


    if($chk->rowCount() > 0): 
    $message = 'Error ! ! User already exists'; 
    else: 
     if($stmt->execute()): 
     $message = 'Successfully created new user'; 
      else: 
      $message = 'Sorry there must have been an issue creating your account'; 
      endif; 
     endif; 
    endif;