无法访问传递的实例对象类

When i catch class in an method i am getting error. When i call anotherMethod i got error like this.

See the bellow code. How can i solve this please help.

class Java{

    function anotherMethod(Php $phpAccess){
        $phpAccess->framework();
        $phpAccess->cms();
    }

}


class Php{

    public function framework()
    {
        echo "Laravel is a popular php framework. </br>";
    }

    public function cms()
    {
        echo "WordPress is popular php cms. </br>";
    }

}

$php = new Php();
$java = new Java($php);
echo $java->anotherMethod();

But when i catch class in an constructor then it's giving right output.See this code bellow.

class Java{
  function __construct(Php $phpAccess){
    $phpAccess->framework();
    $phpAccess->cms();
  }
}

You are getting errors because you are passing the PHP object to the Java constructor but the PHP object is a dependency for the anotherMethod instead. Change the Java class as below to work as you want:

class Java
{
    private $phpAccess;

    public function __construct(Php $phpAccess)
    {
        $this->phpAccess = $phpAccess;
    }

    public function anotherMethod()
    {
        $this->phpAccess->framework();
        $this->phpAccess->cms();
    }
}

class Php
{
    public function framework()
    {
        echo "Laravel is a popular php framework. </br>";
    }

    public function cms()
    {
        echo "WordPress is popular php cms. </br>";
    }
}

$php = new Php();
$java = new Java($php);
$java->anotherMethod();

You call function anotherMethod without any parameter. In your class the method is defined WITH paramater. So you have to call anotherMethod WITH parameter $php

echo $java->anotherMethod($php);