How to get previous year using PHP

Php

Php Problem Overview

How am I able to get the value of the previous year using PHP. Are there any predefined functions for it?

Php Solutions

Solution 1 - Php

try

echo date("Y",strtotime("-1 year"));

Solution 2 - Php

$year = date("Y");
$previousyear = $year -1;

http://php.net/manual/de/function.date.php

Solution 3 - Php

There are many ways, you can either take away the amount of seconds in a year from time() like so:

$prevYear = date('Y', time() - 60*60*24*365 );

Or if you'd prefer, use the PHPs clever strtotime() function:

$prevYear = date('Y', strtotime('-1 year'));

Or even like others have said, if it's from todays year just do date('Y') -1

Solution 4 - Php

Shortest approach:

$lastYear = (int)date("Y") - 1;

Solution 5 - Php

function adddate($vardate,$added)
{
$data = explode("-", $vardate);
$date = new DateTime();            
$date->setDate($data[0], $data[1], $data[2]);
$date->modify("".$added."");
$day= $date->format("Y-m-d");
return $day;    
}

echo "Example : " . adddate("2010-08-01","-1 year");

Solution 6 - Php

If you want to display the entire date exactly 1 year ago, including the month and year:

 <?php echo date("M d Y", strtotime("-1 year")); ?>

Solution 7 - Php

you can give a value of the input tag as :

<?php echo date("Y-m-d",strtotime("-1 year"));?>

Solution 8 - Php

Try This

date('Y', strtotime('last year'));

Solution 9 - Php

This thread needs an update.

Nowadays you would use the PHP date-time object and do something like this:

$lastYear = new DateTime();
$lastYear->sub(new DateInterval('P1Y'));
echo $lastYear->format('Y');

Solution 10 - Php

<?php
    echo date("Y",strtotime("-1 year"));  //last year "2021"
    echo date("Y");                       //current year "2022"
    echo date("Y-m-d",strtotime("-1 year"));  //one year ago "2021-03-31"
    echo date("Y");                      //current date "2022-03-31"
?>
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