php - 如何访问另一个页面上另一个类中的mysqli连接？

Question

如何在用户类中打开数据库连接，在那里我可以进行数据库操作？为什么需要在 DBConnection 类中定义内置创建的函数..？？？？

我创造了

1. db.php
2. 用户.php
3. 结果.php

在 db.php 中

class DBConnection  
{       
  protected $mysqli;
  private  $db_host="127.0.0.1";
  private  $db_name="test";
  private  $db_username="root";
  private  $db_password="";

  public function __construct()
    {
        $this->mysqli=new mysqli($this->db_host,$this->db_username,
                $this->  db_password,$this->db_name) or die($this->mysqli->error);

         return $this->mysqli;
    }

public function query($query)  // why i need to creat this function
     {
    return $this->mysqli->query($query);
     }

public function real_escape_string($str)
 {
    return $this->mysqli->real_escape_string();
 }


   function __destruct(){
     //Close the Connection
     $this->mysqli->close();
    }
}
?>

在 User.php 中

<?php
  require "db.php"; 

  class User {  

  public function __construct($conn)
{
    $this->mysqli=$conn;

}

   public function addUser($name,$username,$email,$pwd)
   {
     $query="  ";

     $res=$this->mysqli->query($query);

  //pls chek the query function in DBConnection,what is the need to define query               >       function in DBConnection ?

    return $res;
    }
   }    
?>

在result.php

    <?php

      require "user.php";

      $conn=new DBConnection();

      if(isset($_POST['submit']))
  {

 $name = $conn->real_escape_string(trim(strip_tags($_POST['name'])));
   $username =  $conn->real_escape_string(trim(strip_tags($_POST['username'])));
   $email =     $conn->real_escape_string(trim(strip_tags($_POST['email'])));
   $password    = $conn->real_escape_string(trim(strip_tags($_POST['pass'])));

//echo $name."".$username."".$email."".$password;
  $uObj=new User($conn);
  $uObj->addUser($name,$username,$email,$password);


echo " hi your name is <I>".$name."</I> and your email ID is <I>".$email."</I>";
        }


?>

Answer 1

您的 DBConnection 类需要一个额外的方法:

public function getLink()
{
    return $this->mysqli;
}

您原来的 User 类似乎是 DBConnection 的子类，因为 DBConnection 上的 mysqli 属性是protected 和 User 类有一个 parent::__construct() 调用。

最好使用依赖注入(inject)，这样您的 User 类将通过构造函数接收其数据库连接:

public function __construct(DBConnection $db)
{
    $this->mysqli = $db->getLink();
}

然后您可以从您的代码运行:

$db = new DBConnection;
$uObj = new User($db);