duanpang5583 2014-07-09 15:34
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PHP:尝试获取非对象的属性或方法

I am trying to save an object as property in an another class, PHP throwing notices and fatal errors. Simplified version of my code:

<?php
    class A {
        public function a() {
            // do something
        }
    }
    $A = new A();

    class B {
        private $A;
        public function __constructor($A) {
            $this->A = $A;
        }
        private function b() {
            if($this->A->a()) { // This line is referred by PHP
                // do something
            }
        }
    }

    $B = new B($A);
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1条回答 默认 最新

  • douhe6181 2014-07-09 15:36
    关注

    You have a typo. Change __constructor to __construct and PHP will process your code correctly. Constructors in PHP are always named __construct. See the documentation for more details.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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