未定义变量，调用非对象上的成员函数 - Undefined variable, Call to a member function on a non-object

Undefined variable, Call to a member function on a non-object

我试图回显已传入类的变量。我有下面的代码(见下文)

engine.php

<?php
define('access', TRUE); //define to true so that we can access core.php
// engine
include_once('core.php'); //include core.php
$core = new core();
j_interpreter("login"); //call j_interpreter function
function j_interpreter($key){
    //switch request base on the request key
    switch($key) {
        case "login" :
            extract($_POST);
            $core->j_login($username, $password);
        break;
    default :
        echo "default";
    } //end of switch
}
?>

core.php

<?php
if(!defined('access')) :
//foreign access then dont allow!
die('Direct access not permitted');
endif;
class core{
public function ___construct(){
    //macro stuff here
}
public function j_login($username, $password){
    echo $username . " " . $password;
}
}

我试图获得用户名和密码的帖子数据已经从引擎。php传递到core.php，但遗憾的是它给了我错误

注意:未定义变量:core在C:'wamp'www'proj'core'engine.php
致命错误:在C:'wamp'www'proj'core'engine.php中调用非对象的成员函数j_login()

任何想法?

函数有它的作用域，在您的例子中，$core是在全局作用域中定义的。要在函数内部访问它，你需要像global $core;

那样提到它

function j_interpreter($key){
    global $core;
    //switch request base on the request key
    switch($key) {
        case "login" :
            extract($_POST);
            $core->j_login($username, $password);
        break;
    default :
        echo "default";
    } //end of switch
}

参见PHP手册中的变量作用域