只有变量应该通过引用传递 [英] Only variables should be passed by reference

问题描述

// Other variables
$MAX_FILENAME_LENGTH = 260;
$file_name = $_FILES[$upload_name]['name'];
//echo "testing-".$file_name."<br>";
//$file_name = strtolower($file_name);
$file_extension = end(explode('.', $file_name)); //ERROR ON THIS LINE
$uploadErrors = array(
    0=>'There is no error, the file uploaded with success',
    1=>'The uploaded file exceeds the upload max filesize allowed.',
    2=>'The uploaded file exceeds the MAX_FILE_SIZE directive that was specified in the HTML form',
    3=>'The uploaded file was only partially uploaded',
    4=>'No file was uploaded',
    6=>'Missing a temporary folder'
);

有什么想法吗?2天后仍然卡住.

Any ideas? After 2 days still stuck.

推荐答案

将explode的结果赋值给一个变量，并将该变量传递给end:

Assign the result of explode to a variable and pass that variable to end:

$tmp = explode('.', $file_name);
$file_extension = end($tmp);

问题是，end 需要一个引用，因为它修改了数组的内部表示(即它使当前元素指针指向最后一个元素).

The problem is, that end requires a reference, because it modifies the internal representation of the array (i.e. it makes the current element pointer point to the last element).

explode('.', $file_name) 的结果不能转化为引用.这是 PHP 语言中的一个限制，可能出于简单原因而存在.

The result of explode('.', $file_name) cannot be turned into a reference. This is a restriction in the PHP language, that probably exists for simplicity reasons.

这篇关于只有变量应该通过引用传递的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！