Question

// connection.php file 

<?php


class Connection{

   private $host;
   private $username;
   private $password;
   private $database;

   public function __construct($host,$username,$password,$database)
   {
     try{
     $conn = new PDO("mysql:host=".$host.";dbname=".$database, $username, $password);
     $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
     return $conn;
     }catch(PDOException $e)
     {
        return $e->getMessage();
     }
   }

}

?>

// user.php

<?php

class Users extends Connection{

  private $username;
  private $email;
  private $password;

  public function __construct($username,$email,$password){

    $this->username=$username;
    $this->email=$email;
    $this->password=$password;

  }

  public function addUser(){
    $conn = new Connection("localhost","root","","userdata");
    echo "database connection established";
    die();
    $sql = $conn->prepare("INSERT INTO information (UserName, Email, Password) VALUES (?, ?, ?)");
    $conn->bind_param("sss", $username, $email, $password);
    $username=this->username;
    $password=this->password;
    $email=this->email;
    $sql->execute();

    echo "New records created successfully";

    $sql->close();
    $conn->close();

  }
}

?>

process.php

<?php

include('lib/connection.php');

if(isset($_POST['username']) && isset($_POST['email']) && isset($_POST['password']))
{
  $user = new User("Gautam","gautam@mail.com","123456");
  $user->addUser();
}

?>

所以问题，当我尝试执行process.php文件时，记录没有添加到数据库中。所以我已经分别创建了连接类，我已经编写了数据库连接的逻辑，然后我继承了该类给用户我将用户添加到数据库然后我在process.php中调用了用户类的对象，但是没有将记录添加到数据库中。请说明与之相关的问题。

Answer 1

您的代码存在的问题是您在执行之前没有绑定任何参数。

检查这些行：

$sql = $conn->prepare()
$conn->bind_param("sss", $username, $email, $password);

此行应为$sql->bind_param("sss", $username, $email, $password);

并注意到PDO没有bind_param()函数：bindParam()

因此你需要

    $sql->bindParam(1, $this->username);
    $sql->bindParam(2, $this->email);
    $sql->bindParam(3, $this->password);

以下代码应该有效：

<?php

class Users extends Connection
{

    private $username;
    private $email;
    private $password;

    public function __construct($username, $email, $password)
    {
        $this->username = $username;
        $this->email    = $email;
        $this->password = $password;

    }

    public function addUser()
    {
        $conn = new Connection("localhost", "root", "", "userdata");
        echo "database connection established";

        $username = $this->username;
        $password = $this->password;
        $email    = $this->email;



        $sql = $conn->prepare("INSERT INTO information (UserName, Email, Password) VALUES (?, ?, ?)")->execute(array($username,$email,$password));

        if (!$sql) {

            print_r("error : " . $conn->errorInfo());
        } else {

            echo "New records created successfully";
        }
        $sql->close();
        $conn->close();
    }
}

?>

PS：不要以纯文本形式存储密码，使用php内置函数等用password_hash()和password_verify()来准备密码。

Answer 2

首先在你的班级中使用$ this代替这个，然后你拨打错误的班级名称User，这是用户，请修改并尝试，应该工作。

$user = new Users("Gautam","gautam@mail.com","123456");

在用户类

的adduser函数中

$username=$this->username;
$password=$this->password;
$email=$this->email;

如上所述改变它应该有效..

面向对象的PHP代码错误

2 个答案: