[SOLVED] calculate age

[Dragen]

Hi,

I'm trying to write a code to calculate my age. I've got a fully working script below:

<?php
$crntdate = date('d/m/Y');
$birth = "22/05/1988";

$crntsplit = explode("/", $crntdate);
$birthsplit = explode("/", $birth);

$crntdatestamp = mktime(0, 0, 0, $crntsplit[1], $crntsplit[0], $crntsplit[2]);
$birthstamp = mktime(0, 0, 0, $birthsplit[1], $birthsplit[0], $birthsplit[2]);

$seconds = $crntdatestamp - $birthstamp;

$years = floor($seconds / 31536000);

echo $years . " years";
?>

Just wondering if there was a better way of doing it as finding the seconds and dividing it all seems a bit of an odd way round..

Thanks

[Anzeo]

I don't know if my solution is better, but I'll post it right away. (I used my code from a tutorial and augmented it to use it for age calculations)

[Dragen]

? where is it lol

[Dragen]

I've found a major problem with this method.. It works okay if I want to display my age, but I'm now trying to use a slightly modified version to say how long my site has been running for..

what I was wanting is something like

this site has been running for about 1 year and 3 months

the problem with my code is it can't define months. It will successfully tell me how many months between todays date and the start date, but not for examply

1 year and 3 months

I end up with something ridiculous like

1 year and 43 months

[Anzeo]

Like I said I've got a code which display's the amount of days between two given date's

 

<?php
function DateDiff($dformat, $endDate, $beginDate)
{
$date_parts1=explode($dformat, $beginDate);
$date_parts2=explode($dformat, $endDate);
$start_date=gregoriantojd($date_parts1[1], $date_parts1[2], $date_parts1[0]);
$end_date=gregoriantojd($date_parts2[1], $date_parts2[2], $date_parts2[0]);
return $end_date - $start_date;
}
echo "".DateDiff("-",date("Y-m-d"),$RegDatum)."";
?>

 

If you would divide this result by 365 you'd get an approximat age

you could add another clause checking wether your birthdate has been passed or not.

[Dragen]

ah, thanks.

I was trying to stay away from datediff though, because I heard it was unreliable..

[Anzeo]

I don't think it's unreliable, as far as I've worked with it it seems to work fine.

Only thing is, it's probably not the best method for age calculating.

[sasa]

your 1st code do noi work

try

<?php
$crntdate = "18/05/2007";
$birth = "22/05/1988";

$crntsplit = explode("/", $crntdate);
$birthsplit = explode("/", $birth);

$crntdatestamp = mktime(0, 0, 0, $crntsplit[1], $crntsplit[0], $crntsplit[2]);
$birthstamp = mktime(0, 0, 0, $birthsplit[1], $birthsplit[0], $birthsplit[2]);

$seconds = $crntdatestamp - $birthstamp;

$years = floor($seconds / 31536000);

echo $years . " years";
?>

try

<?php
function my_old($birth, $curent = '') {
$curent = $curent ? $curent : date('d/m/Y');
$birth = explode('/',$birth);
$curent = explode('/', $curent);
$days = array(31,31,28,31,30,31,30,31,31,30,31,30,31);
if (($birth[2] % 4 == 0 and $birth[2] % 100 != 0) or $birth[2] % 400 == 0) $days[2] = 29;
if ($birth[0] > $curent[0]) $curent[0] += $days[--$curent[1]];
if ($birth[1] > $curent[1]) {
	$curent[1] += 12;
	$curent[2] --;
}
return array('year' => $curent[2] - $birth[2], 'month' => $curent[1] - $birth[1], 'day' => $curent[0] - $birth[0]);
}

print_r(my_old('22/05/1988'));
?>

[Dragen]

it does work.. I'm using on my site at the moment.

 

What do you think is wrong?

[sasa]

if today is $crntdate = "18/05/2007"; and i'm born $birth = "22/05/1988"; i'm not 19 years old

[Dragen]

here's my code:

function birthdday($crntdate, $birth){
$crntsplit = explode("/", $crntdate);
$birthsplit = explode("/", $birth);

$crntdatestamp = mktime(0, 0, 0, $crntsplit[1], $crntsplit[0], $crntsplit[2]);
$birthstamp = mktime(0, 0, 0, $birthsplit[1], $birthsplit[0], $birthsplit[2]);

$seconds = $crntdatestamp - $birthstamp;

$years = floor($seconds / 31536000);

echo "<strong>Age:</strong> " . $years . " years";
}

then I'm calling the function using:

<?php birthdday(date('d/m/Y'), "22/05/1988"); ?>

It says that I am 18.

I'm not finding any problem with the age...

[sasa]

try

<?php birthdday(date('d/m/Y'), "21/04/1988"); ?>

[Dragen]

ah, it's not checking the days for some reason.. just months and years.

Okay I understand now.. thanks!

I'll have a look at your code when I have time... I'll let you know how it works for me.

[Dragen]

okay, I've finnally tested the code you gave me sasa.

With a couple of modifications I've got it working perfectly to calculate both my age and how long my site's been running for!

Thanks.

 

here's my code:

<?php
function birthday($view, $birth, $curent = '') {
$crnt = $crnt ? $crnt : date('d/m/Y');
$birth = explode('/',$birth);
$crnt = explode('/', $crnt);
$days = array(31,31,28,31,30,31,30,31,31,30,31,30,31);
if (($birth[2] % 4 == 0 and $birth[2] % 100 != 0) or $birth[2] % 400 == 0) $days[2] = 29;
if ($birth[0] > $crnt[0]) $crnt[0] += $days[--$crnt[1]];
if ($birth[1] > $crnt[1]) {
	$crnt[1] += 12;
	$crnt[2] --;
}
$age = array('year' => $crnt[2] - $birth[2], 'month' => $crnt[1] - $birth[1], 'day' => $crnt[0] - $birth[0]);
if($view == age){
	echo "<strong>Age:</strong> " . $age['year'] . " years";
}elseif($view == site){
	echo $age['year'] . " ";
	if($age['year'] == 1){
		echo "year ";
	}else{
		echo "years ";
	}
	if($age['month'] > 0){
	echo "and "  . $age['month'] . " ";
		if($age['month'] == 1){
			echo "month ";
		}else{
			echo "months ";
		}
	}
}
}
?>

If added some if statements to check if year and month is plural or not and also wether months should be included at all.

 

Thanks!