PHP Access 对象包含在一个类中。

Question

我在 Windows 7 中使用带有 php 和 mysql 的 wamp 服务器。有一个连接到 mysql 数据库的类，一个 db__connect 文件用于建立连接。

下面是 db_connect 的代码；

<?php


include_once 'psl-config.php';   // Needed because functions.php is not included
$mysqli = new mysqli(HOST, USER, PASSWORD, DATABASE);
if ($mysqli->connect_error) {
    header("Location: ../error.php?err=Unable to connect to MySQL");
    exit();
}

下面是类代码:

<?php

include_once $_SERVER['DOCUMENT_ROOT'].'/datacentre/admin/includes/db_connect.php';

class User {
	
	function getPermission( $email, $perms) {
	
		   if ($stmt = $mysqli->prepare("SELECT members.email , groups.permission FROM members 
		                                  JOIN groups ON members.group_id = groups.group_id 
										  WHERE email = ? LIMIT 1")) {
        $stmt->bind_param('s', $email);  // Bind "$email" to parameter.
        $stmt->execute();    // Execute the prepared query.
        $stmt->store_result();
        // get variables from result.
        $stmt->bind_result($user_email, $user_perms);
		print "yes we can ".$user_perms;
		exit();
		
		if($perms & $user_perms)
		{return true;
	      exit();}
	
	return false;
     exit();  
}
		
	}


?>

错误信息如下

Answer 1

您需要在您的用户类中声明 $mysqli，尽管我敢打赌您会希望在其他类中重用该连接。您最好将 $mysqli 传递给用户类

    new User($email, $perms, $mysqli);

这当然假设您在全局实例化 User 类而不是在另一个类中。