2017-02-27 112 views
-1

为什么下面的代码不工作?变量和函数不能被“$ objkt”访问。为什么不调用PHP类的变量和函数?

<?php 
     print "This file displays function info."; 
     echo "<br/>"; 
     class User 
     { 
      public $name="MyName"; 
      public $pwd="PaSsWoRd"; 
      function info() 
      { 
       print_r(ucfirst(strtolower($name))); 
       echo "<br/>"; 
       print_r(ucfirst(strtolower($pwd))); 
       echo "<br/>"; 
      } 
     } 
     $objkt = new User; 
     $objkt->name; 
     $objkt->pwd; 
     $objkt->info(); 
    ?> 

。输出:

This file displays function info. 

    Notice: Undefined variable: name in C:\xampp\htdocs\project2\infotest.php on line 17 

    Fatal error: Cannot access empty property in C:\xampp\htdocs\project2\infotest.php on line 17 

回答

0

$名字是里面的 “信息” 方法未定义的变量

function info() 
      { 

      print_r(ucfirst(strtolower($this->name))); 
      echo "<br/>"; 
      print_r(ucfirst(strtolower($this->pwd))); 
      echo "<br/>"; 
      } 

阅读PHP OOP

+0

也'$ objkt->名;'不没有。 – AbraCadaver

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