恼人的PHP错误：“严格的标准：只有变量应参考在传递”

Question

我有这个小脚本量身定做的，我不能让这个错误：

Strict Standards: Only variables should be passed by reference in C:\xampp\htdocs\includes\class.IncludeFile.php on line 34" off!

这里是页：

namespace CustoMS; 

if (!defined('BASE')) 
{ 
    exit; 
} 

class IncludeFile 
{ 
    private $file; 
    private $rule; 

    function __Construct($file) 
    { 
     $this->file = $file; 

     $ext = $this->Extention(); 
     switch ($ext) 
     { 
      case 'js': 
       $this->rule = '<script type="text/javascript" src="'.$this->file.'"></script>'; 
       break; 

      case 'css': 
       $this->rule = '<link type="text/css" rel="stylesheet" href="'.$this->file.'">'; 
       break; 
     } 
    } 

    private function Extention() 
    { 
     return end(explode('.', $this->file)); 
    } 

    function __Tostring() 
    { 
     return $this->rule; 
    } 
} 

请帮帮我。

Answer 1

功能end具有以下原型end(&$array)。

您可以通过创建变量并将其传递给函数来避免此警告。

private function Extention() 
{ 
    $arr = explode('.', $this->file); 
    return end($arr); 
} 

从文档：

The following things can be passed by reference:

• Variables, i.e. foo($a)
• New statements, i.e. foo(new foobar())
• References returned from functions, i.e.:

explode返回一个数组不数组的引用。

例如：

function foo(&$array){ 
} 

function &bar(){ 
    $myArray = array(); 
    return $myArray; 
} 

function test(){ 
    return array(); 
} 

foo(bar()); //will produce no warning because bar() returns reference to $myArray. 
foo(test()); //will arise the same warning as your example. 

Answer 2

private function Extention() 
{ 
    return end(explode('.', $this->file)); 
} 

端（）设置指针数组的最后一个元素。在这里，您将函数的结果提供给end而不是变量。

private function Extention() 
{ 
    $array = explode('.', $this->file); 
    return end($array); 
}