dongzhanyan3667 2013-10-17 06:19
浏览 31

对PHP中的类变量的建议

I am getting an error displayed in calling my $dog2 method greet() that is undefined variable but I can't understand where I am making a mistake. Please help me :)

 <?php

        class Dog{

            public $numLegs=4;

            public $name;

            public function __construct($name){

                $this->name=$name;

            }

            public function bark(){

                return "woof";

            }

            public function greet(){

                return $name." is very beatifull dog hmmm";

            }

        }

        $dog1= new Dog("Barker");

        $dog2= new Dog("Amigo");

        echo $dog1->bark();

        echo $dog2->greet();


    ?>
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1条回答 默认 最新

  • douzhe9075 2013-10-17 06:20
    关注

    you cant direcly access name like $name you have to use $this->name

    change the following line

     return $this->name." is very beatifull dog hmmm";
    

    $this->name refers to the current class's variable name, When using $this->name, you are accessing the property with the name name of the current object.

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