PHP值不會對數據庫

Question

session_start(); 
include('dbconn.php'); 
    if(isset($_POST['registerform']) || isset($_POST['loginform'])) { 

    if(isset($_POST['registerform']) && $_POST['registerform']){ 
     //register 

     if(isset($_POST['username']) && !empty($_POST['username']) 
     && isset($_POST['password']) && !empty($_POST['password']) 
     && isset($_POST['fname']) && !empty($_POST['fname']) 
     && isset($_POST['lname']) && !empty($_POST['lname']) 
     && isset($_POST['email']) && !empty($_POST['email'])){ 
     //ean einai gemata me kati ta paidia 

     $myName=trim($_POST['username']); 
     $myPass= md5(trim($_POST['password'])); 
     $fName=trim($_POST['fname']); 
     $lName=trim($_POST['lname']); 
     $eMail=trim($_POST['email']); 


     if($stmt = $mysqli->prepare("INSTERT INTO users(username,password,fname,lname,email) VALUES (?, ?, ?, ?, ?)")) { 


      $stmt->bind_param('sssss',$name,$pass,$fname,$lname,$email); 

      $name = $myName; 
      $pass = $myPass; 
      $fname = $fName; 
      $lname = $lName; 
      $email = $eMail; 

      $stmt-> execute(); 
      $stmt-> close(); 



     } 

     } 
    } 


     }else{ 
     echo "no form submited"; 
     } 

，這是數據庫連接

<?php 

$host = "localhost"; 
$username = "root"; 
$password = ""; 
$db_name = "phpwebsite"; 


$mysqli = new mysqli("$host","$username","$password","$db_name"); 
// Check connection 
if ($mysqli->connect_error) { 
    die("Connection failed: " . $mysqli->connect_error); 
} 

?> 

我認爲這個問題是插入statment因爲一切似乎是工作的罰款。當我按下注冊按鈕時，不會出現錯誤。唯一的一點是，該值不會對數據庫

Answer 1

的錯誤是在您的SQL查詢，我認爲它是一個類型INSTERT

if($stmt = $mysqli->prepare("INSTERT INTO users(username,password,fname,lname,email) VALUES (?, ?, ?, ?, ?)")) 

嘗試修正至

if($stmt = $mysqli->prepare("INSERT INTO users(username,password,fname,lname,email) VALUES (?, ?, ?, ?, ?)"))  

編輯：

對於您的第二個問題：

try { $stmt = $mysqli->prepare("INSERT INTO users(username,password,fname,lname,email) VALUES (?, ?, ?, ?, ?)") 

    $stmt->bind_param('sssss',$name, 
         $pass,$fname, 
         $lname,$email); 

    $name = $myName; 
    $pass = $myPass; 
    $fname = $fName; 
    $lname = $lName; 
    $email = $eMail; 

    $stmt-> execute(); 
    $stmt-> close(); 

} 
catch (PDOException $ex) { 
    // For testing, you could use a die and message. 
    die("Failed to run query: " . $ex->getMessage()); 



    }