Date minus 1 year?

I've got a date in this format:

2009-01-01

How do I return the same date but 1 year earlier?

You can use strtotime:

$date = strtotime('2010-01-01 -1 year');

The strtotime function returns a unix timestamp, to get a formatted string you can use date:

echo date('Y-m-d', $date); // echoes '2009-01-01'

Use strtotime() function:

  $time = strtotime("-1 year", time());
  $date = date("Y-m-d", $time);

Using the DateTime object...

$time = new DateTime('2099-01-01');
$newtime = $time->modify('-1 year')->format('Y-m-d');

Or using now for today

$time = new DateTime('now');
$newtime = $time->modify('-1 year')->format('Y-m-d');

an easiest way which i used and worked well

date('Y-m-d', strtotime('-1 year'));

this worked perfect.. hope this will help someone else too.. :)

// set your date here
$mydate = "2009-01-01";

/* strtotime accepts two parameters.
The first parameter tells what it should compute.
The second parameter defines what source date it should use. */
$lastyear = strtotime("-1 year", strtotime($mydate));

// format and display the computed date
echo date("Y-m-d", $lastyear);

On my website, to check if registering people is 18 years old, I simply used the following :

$legalAge = date('Y-m-d', strtotime('-18 year'));

After, only compare the the two dates.

Hope it could help someone.

Although there are many acceptable answers in response to this question, I don't see any examples of the sub method using the \Datetime object: https://www.php.net/manual/en/datetime.sub.php

So, for reference, you can also use a \DateInterval to modify a \Datetime object:

$date = new \DateTime('2009-01-01');
$date->sub(new \DateInterval('P1Y'));

echo $date->format('Y-m-d');

Which returns:

2008-01-01

For more information about \DateInterval, refer to the documentation: https://www.php.net/manual/en/class.dateinterval.php

You can use the following function to subtract 1 or any years from a date.

 function yearstodate($years) {

        $now = date("Y-m-d");
        $now = explode('-', $now);
        $year = $now[0];
        $month   = $now[1];
        $day  = $now[2];
        $converted_year = $year - $years;
        echo $now = $converted_year."-".$month."-".$day;

    }

$number_to_subtract = "1";
echo yearstodate($number_to_subtract);

And looking at above examples you can also use the following

$user_age_min = "-"."1";
echo date('Y-m-d', strtotime($user_age_min.'year'));