php - 烦人的 PHP 错误 : "Strict Standards: Only variables should be passed by reference in"

Question

我制作了这个小脚本，但我无法得到这个错误:

严格的标准:只有变量应该在 C:\xampp\htdocs\includes\class.IncludeFile.php 中第 34 行通过引用传递!

这是页面:

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

函数 end 具有以下原型(prototype) 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);
}

来自文档:

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 返回一个数组而不是对数组的引用。

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.