How do I use PHP to get the current year?

You can use either date or strftime. In this case I'd say it doesn't matter as a year is a year, no matter what (unless there's a locale that formats the year differently?)

For example:

<?php echo date("Y"); ?>

On a side note, when formatting dates in PHP it matters when you want to format your date in a different locale than your default. If so, you have to use setlocale and strftime. According to the php manual on date:

To format dates in other languages, you should use the setlocale() and strftime() functions instead of date().

From this point of view, I think it would be best to use strftime as much as possible, if you even have a remote possibility of having to localize your application. If that's not an issue, pick the one you like best.

<?php echo date("Y"); ?>

My super lazy version of showing a copyright line, that automatically stays updated:

&copy; <?php 
$copyYear = 2008; 
$curYear = date('Y'); 
echo $copyYear . (($copyYear != $curYear) ? '-' . $curYear : '');
?> Me, Inc.

This year (2008), it will say:

© 2008 Me, Inc.

Next year, it will say:

© 2008-2009 Me, Inc.

and forever stay updated with the current year.

Or (PHP 5.3.0+) a compact way to do it using an anonymous function so you don't have variables leaking out and don't repeat code/constants:

© 
<?php call_user_func(function($y){$c=date('Y');echo $y.(($y!=$c)?'-'.$c:'');}, 2008); ?> 
Me, Inc.

With PHP heading in a more object-oriented direction, I'm surprised nobody here has referenced the built-in DateTime class:

$now = new DateTime();
$year = $now->format("Y");

or one-liner with class member access on instantiation (php>=5.4):

$year = (new DateTime)->format("Y");

http://us2.php.net/date

echo date('Y');

strftime("%Y");

I love strftime. It's a great function for grabbing/recombining chunks of dates/times.

Plus it respects locale settings which the date function doesn't do.

This one gives you the local time:

$year = date('Y'); // 2008

And this one UTC:

$year = gmdate('Y'); // 2008

Here's what I do:

<?php echo date("d-m-Y") ?>

below is a bit of explanation of what it does:

d = day
m = month
Y = year

Y will gives you four digit (e.g. 1990) and y for two digit (e.g. 90)

For 4 digit representation:

<?php echo date('Y'); ?>

2 digit representation:

<?php echo date('y'); ?>

Check the php documentation for more info: https://secure.php.net/manual/en/function.date.php

echo date('Y') gives you current year, and this will update automatically since date() give us the current date.

print date('Y');

For more information, check date() function documentation: https://secure.php.net/manual/en/function.date.php

use a PHP function which is just called date().

It takes the current date and then you provide a format to it

and the format is just going to be Y. Capital Y is going to be a four digit year.

<?php echo date("Y"); ?>

<?php echo date("Y"); ?>

This code should do

Just write:

date("Y") // A full numeric representation of a year, 4 digits
          // Examples: 1999 or 2003

Or:

date("y"); // A two digit representation of a year     Examples: 99 or 03

And 'echo' this value...

use a PHP date() function.

and the format is just going to be Y. Capital Y is going to be a four digit year.

<?php echo date("Y"); ?>

If your server supports Short Tags, or you use PHP 5.4, you can use:

<?=date("Y")?>