在PHP中的另一个类中调用一个类 [英] Call a class inside another class in PHP

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本文介绍了在PHP中的另一个类中调用一个类的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

嘿,我想知道如何做这样做,当我尝试下面的代码在一个类的函数中,它产生一些我不能捕获的PHP错误

Hey there I'm wondering how this is done as when I try the following code inside a function of a class it produces some php error which I can't catch

public $tasks;
$this->tasks = new tasks($this);
$this->tasks->test();

我不知道为什么类的启动需要$ this作为参数:S

I don't know why the initiation of the class requires $this as a parameter either :S

感谢

class admin
{
    function validate()
    {
        if(!$_SESSION['level']==7){
            barMsg('YOU\'RE NOT ADMIN', 0);
            return FALSE;
        }else{
            **public $tasks;** // The line causing the problem
            $this->tasks = new tasks(); // Get rid of $this->
            $this->tasks->test(); // Get rid of $this->
            $this->showPanel();
        }
    }
}
class tasks
{
    function test()
    {
        echo 'test';
    }
}
$admin = new admin();
$admin->validate();


推荐答案

类的方法(函数。)如果你不需要使用该方法之外的任务对象,你可以这样做:

You can't declare the public $tasks inside your class's method (function.) If you don't need to use the tasks object outside of that method, you can just do:

$tasks = new Tasks($this);
$tasks->test();

当你使用一个变量时,你只需要使用$ this->可在整个课程。

You only need to use the "$this->" when your using a variable that you want to be available throughout the class.

您的两个选项:

class Foo
{
    public $tasks;

    function doStuff()
    {
        $this->tasks = new Tasks();
        $this->tasks->test();
    }

    function doSomethingElse()
    {
        // you'd have to check that the method above ran and instantiated this
        // and that $this->tasks is a tasks object
        $this->tasks->blah();
    }

}

class Foo
{
    function doStuff()
    {
        $tasks = new tasks();
        $tasks->test();
    }
}

您的代码:

class Admin
{
    function validate()
    {
        // added this so it will execute
        $_SESSION['level'] = 7;

        if (! $_SESSION['level'] == 7) {
            // barMsg('YOU\'RE NOT ADMIN', 0);
            return FALSE;
        } else {
            $tasks = new Tasks();
            $tasks->test();
            $this->showPanel();
        }
    }

    function showPanel()
    {
        // added this for test
    }
}
class Tasks
{
    function test()
    {
        echo 'test';
    }
}
$admin = new Admin();
$admin->validate();

这篇关于在PHP中的另一个类中调用一个类的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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