Question 235372 in PHP Question posted in
Official manual can be found here.

Php – How to handle array_shift() on Null?

Please take a look at this code:

$end = isset($newvar) ? array($newvar) : null;
while($ends = array_shift($end)){
  ...

It was working well when I was using PHP 7.2, but after upgrading to 8.1, it throws:

PHP Fatal error: Uncaught TypeError: array_shift(): Argument #1 ($array) must be of type array, null given in /path/to/qanda.php:469

Any idea how can I fix it?

Tags:

2

Answers


  1. 0

    Just use an empty array instead:

    $end = isset($newvar) ? array($newvar) : [];

    array_shift will return null on the first call with an empty array as input, so the loop will not execute.

    Login or Signup to reply.
  2. 0

    The most basic solution would be to replace the null value with an empty array to comply with the type requirements:

    $end = isset($newvar) ? array($newvar) : [];
while($ends = array_shift($end)){

    You could also create the array and use the null coalescing operator on $newvar:

    $end = [$newvar ?? null];
while($ends = array_shift($end)){

    But I don’t understand why you would create an array with a single value and then create a loop using array_shift‘s return value. The loop body will only run once.
    Maybe just use a condition ?

    if (isset($newvar)) {
    Login or Signup to reply.
Please signup or login to give your own answer.

Back To Top
Search