Only variables should be passed by reference

Question

// 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'
);

Any ideas? After 2 days still stuck.

Answer 1

Just as you can't index the array immediately, you can't call end on it either. Assign it to a variable first, then call end.

$basenameAndExtension = explode('.', $file_name);
$ext = end($basenameAndExtension);

Answer 2

Php 7 compatible proper usage:

$fileName      = 'long.file.name.jpg';
$tmp           = explode('.', $fileName);
$fileExtension = end($tmp);

echo $fileExtension;
// jpg

Answer 3

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

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

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).

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.

Answer 4

Try this:

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

The reason is that the argument for end is passed by reference, since end modifies the array by advancing its internal pointer to the final element. If you're not passing a variable in, there's nothing for a reference to point to.

See end in the PHP manual for more info.

Answer 5

PHP offical Manual : end()

Parameters

array

The array. This array is passed by reference because it is modified by the function. This means you must pass it a real variable and not a function returning an array because only actual variables may be passed by reference.

Answer 6

PHP complains because end() expects a reference to something that it wants to change (which can be a variable only). You however pass the result of explode() directly to end() without saving it to a variable first. At the moment when explode() returns your value, it exists only in memory and no variable points to it. You cannot create a reference to something (or to something unknown in the memory), that does not exists.

Or in other words: PHP does not know, if the value you give him is the direct value or just a pointer to the value (a pointer is also a variable (integer), which stores the offset of the memory, where the actual value resides). So PHP expects here a pointer (reference) always.

But since this is still just a notice (not even deprecated) in PHP 7, you can savely ignore notices and use the ignore-operator instead of completely deactivating error reporting for notices:

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