PHP - Handle Undefined Index Gracefully

Let's say I'm processing some input in the form of an array or object, and I can't predict which keys are defined.

// input
$item = [
    'name' => 'Item name',
];

$year = $item['year']; // Undefined index: year

If I try to reference a non-existent key and don't handle that properly, I'll get this nice little error, and my code will break:

PHP error:  Undefined index: year on line x

So let's assume that I want to handle this by assigning null to an undefined value or index. This can be done a few different ways.

Option 1

Long form. I've seen this a lot in older code bases (especially pre-7.0) and I just don't like it. It's too lengthy and awkward.

$year = isset($item['year']) ? $item['year'] : null; // null

Option 2

Null coalescing operator (??). Way cleaner and much more elegant. PHP 7.0+.

$year = $item['year'] ?? null; // null

Option 3

A more graceful approach with error reporting suppression. I haven't used PHP's @ (error control) operator in a long time and had almost forgot about it. Frustrated with the verbosity of error handling in an older PHP project that did not have access to null coalesce, I discovered this much shorter syntax and it does exactly what I need.

If you want to assign anything but null, this method won't work, of course.

$year = @$item['year']; // null

NB Test this well to ensure it works in your local and/or production environments. I haven't found any issues in any of mine, but caveat emptor. Also, if you have custom error handling/reporting in place, this might not work. Always test your code when trying this method!

Bonus

To expand on this, let's say I want to assign a default year, if the year in the input is not defined. Using the long form approach, I could do the following - and it's messy and hard to follow:

$current_year = 2019;

$year = isset($item['year']) ? $item['year'] : (isset($year) ? $year : $current_year); // 2019

Using the error control operator it can be simplified to:

$year = isset($item['year']) ? $item['year'] : @$current_year; // 2019

Or even further for PHP 7.0+:

$year = $item['year'] ?? @$current_year; // 2019