php - 为什么我无法访问我的数据库？

Question

这是我的 dbconnect.class.php

<?php 

class Connect
{
    //public $error;
    public $db;
    public function __construct()
    {
        $link = mysql_connect("localhost","root","1");
        $db = mysql_select_db("tarih",$link);
        $this->db = $db;
    }
}

$connect = new Connect();
$connect = $connect->db;
?>

这是主 php 文件 (header.class.php)

<?php
require_once ('dbconnect.class.php');
class Header extends Connect
{ 
    public $headers = array();
    public function __construct() 
    { 
        /* 
        * Bu sınıf sayfaların header bilgilerini işler. 
        */  
    } 

    public function sayfaHeader($sayfa = true) 
    { 
        $sql = "SELECT * FROM header WHERE id='" . $sayfa . "'"; 
        $query = mysql_query($sql,$connect) or mysql_error(); 
        echo $sql;
    } 


} 

$header = new Header();
echo $header->sayfaHeader();
?>

但是当我运行此代码时，我看到此错误:

Warning: mysql_query(): supplied argument is not a valid MySQL-Link resource in C:\AppServ\www\ilk\class\header.class.php on line 16

问题是什么？

Answer 1

变量$connect不在类的范围内。要么您将其忽略，然后 mysql_query 选择最后一个可用资源。或者您可以将 $connect 变量传递到类中:

require_once ('dbconnect.class.php');
class Header extends Connect
{ 
    public $headers = array();
    protected $database;

    public function __construct($database) 
    { 
        /* 
        * Bu sınıf sayfaların header bilgilerini işler. 
        */  
        $this->database = $database;
    } 

    public function sayfaHeader($sayfa = true) 
    { 
        $sql = "SELECT * FROM header WHERE id='" . $sayfa . "'"; 
        $query = mysql_query($sql,$this->database) or mysql_error(); 
        echo $sql;
    } 
} 

$header = new Header($connect); // here you pass-through your resource
echo $header->sayfaHeader();

---

我还想提一下，您应该查看 Design Patterns ，因为它只是您正在创建的一些伪 OOP。