How do I calculate one's retirement date? - php

Question

I'm trying to use one's date of birth to calculate when he'll be 50, if he's not 50 already.
If person is not 50, add a year to his age then check if it'll be 50. If not, iterate until it's true. Then get the date he turned 50. in PHP
Here's the code, not complete.
$rAge = 50;
$retir = date('j F Y ', strtotime("+30 days"));
$oneMonthAdded = strtotime(date("d-m-Y", strtotime($DOB)). "+1 year");
$re = date("d-m-Y", $oneMonthAdded);
$futDate = date("d-m-Y", strtotime(date("d-m-Y", strtotime($re))));
$date_diff = strtotime($futDate)-strtotime($DOB);
$future_age = floor(($date_diff)/(60*60*24*365));
Help please.

Answer 1

try this code bro!
// your date of birth
$dateOfBirth = '1950-11-26';
// date when he'll turn 50
$dateToFifty = date('Y-m-d', strtotime($dateOfBirth . '+50 Years'));
// current date
$currentDate = date('Y-m-d');
$result = 'retired';
// checks if already fifty
if($currentDate <= $dateToFifty) {
$result = $dateToFifty;
}
echo $result;

Answer 2

I use simplest php code to find out retire date. y
<?php
$dob = '1970-02-01';
$dob_ex = explode("-",$dob);
$age_diff = date_diff(date_create($dob), date_create('today'))->y;
$year_of_retire = 50 - $age_diff;
$end = date('Y', strtotime('+'.$year_of_retire.'years'));
$date_of_retire = $end."-".$dob_ex[1]."-".$dob_ex[2];
echo $date_of_retire;
?>
you can use if...else condition to echo values according to you.
like
if($year_of_retire > 0){
echo $date_of_retire;
} else if($year_of_retire < 0){
echo "retired";
}