[英]why i can't reach my database?
這是我的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;
?>
這是主要的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 $header->sayfaHeader();
?>
但是,當我運行此代碼時,我看到此錯誤:
警告:mysql_query():提供的參數在第16行的C:\\ AppServ \\ www \\ ilk \\ class \\ header.class.php中不是有效的MySQL-Link資源
問題是什么?
變量$connect
不在類的范圍內。 要么省掉它,然后mysql_query
選擇最后一個可用資源。 或者,您可以將$connect
變量傳遞給該類:
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 $header->sayfaHeader();
我還想提一提,您應該研究Design Patterns ,因為它只是您正在創建的一些偽OOP。
您的$connect
變量是父類。 您需要指向它包含的$link
變量。
更改:
$query = mysql_query($sql, $connect) or mysql_error();
至
$query = mysql_query($sql,$link) or mysql_error();
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