php – 类没有正确继承DB

发布时间：2020-12-13 16:51:32 所属栏目：PHP教程来源：网络整理

导读：所以我有一个看起来像这样的DB类 class db{ private $hostname = 'localhost'; private $username = 'root'; private $password = 'root'; private $con; public function db(){ try { $dbh = new PDO("mysql:host=$this-hostname;dbname=myDB",$this-userna

所以我有一个看起来像这样的DB类

class db{

    private $hostname = 'localhost';
    private $username = 'root';
    private $password = 'root';
    private $con;

    public function db(){
        try {
            $dbh = new PDO("mysql:host=$this->hostname;dbname=myDB",$this->username,$this->password);
        }catch(PDOException $e){
            echo $e->getMessage();
            exit();
        }
        $this->con = $dbh;
        echo 'Connected to database<br />';
    }

还有我的index.php

include('db.class.php');
include('todo.class.php');
include('dressTemplate.inc.php');

$db = new db;

$todo = new todo($db);

而我的todo.class.php就是这样开始的

class todo{

function todo(db $db){
    $this->db = $db;
} 

public function render($post) {

    $db &= $this->db;

但后来我得到了这个通知

Notice: Undefined variable: db in todo.class.php on line 11

Notice: Object of class db could not be converted to int in todo.class.php on line 11

如何在todo.class.php中正确定义db？

解决方法

试试这个：

class todo {

  var $db;

  __construct(&$db) {
    $this->db = $db;
  }

  public function render($post) {
    $db = &$this->db;
  }
}

（编辑：李大同）

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