Dynamic Dog Age

Discussion in 'PHP' started by johneva, Sep 15, 2011.

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  1. #1
    Hi

    I need a little help with some PHP to work out a dogs age and display it in years and months old so its always upto date.

    
$molly_dob = "16/5/2010" ;
$todays_date = date('d/m/y') ;

$molly_age = ???????
    PHP:
     
    Last edited: Sep 15, 2011
    johneva, Sep 15, 2011 IP
  2. JohnnySchultz

    JohnnySchultz Peon

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    #2
    you can try this.. didn't have time to make it in accurate years but its a good start

    <?php

$dog_bday = "2011-02-08";
$today = date("Y-m-d");

$datetime_bday  = strtotime($dog_bday);
$datetime_today = strtotime($today);
$diff = $datetime_today - $datetime_bday;

$minutes_diff =  $diff / 60;
$hours_diff =  $minutes_diff / 60;
$days_diff =  $hours_diff / 24;
$years_diff =  $days_diff / 365.242199;

// dog years converstion, re: http://www.dogyears.com/
if(($years_diff >= .17) && ($years_diff < .5))
	echo '<= 14 months';
else if(($years_diff >= .5) && ($years_diff < .67))
	echo '<= 5 years';
else if(($years_diff >= .67) && ($years_diff < 1))
	echo '<= 9 years';
else if(($years_diff >= 1) && ($years_diff < 2))
	echo '<= 24 years';
else if(($years_diff >= 2) && ($years_diff < 3))
	echo '<= 28 years';
else if(($years_diff >= 3) && ($years_diff < 4))
	echo '<= 32 years';
else if(($years_diff >= 4) && ($years_diff < 5))
	echo '<= 37 years';


?>
    PHP:
     
    JohnnySchultz, Sep 15, 2011 IP
  3. johneva

    johneva Well-Known Member

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    #3
    Sorry should have been more clear dont need it in dog years, just in human years. Although I do like the idea of doing both now you have come up with that.

    So I could say Molly is 1yr 5 months old and X old in doggy years.
     
    johneva, Sep 15, 2011 IP
  4. MyVodaFone

    MyVodaFone Well-Known Member

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    #4 
    
<?php 

date_default_timezone_set('America/Denver'); //  set to your server time if necessary 

$molly_dob = "2007-03-24";
$todays_date = date('y/d/m') ;

$diff = abs(strtotime($todays_date) - strtotime($molly_dob));

$years = floor($diff / (365*60*60*24));
$months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
$days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));

$molly_age = "$years Years, $months Months, $days Days"; 

echo $molly_age;

?>
    PHP:
    At time of post: 2 years, 8 months, 1 days

    You might like to add an if statement if its: year or years, month or months, day or days, something to remove the letter s
     
    Last edited: Sep 19, 2011
    MyVodaFone, Sep 19, 2011 IP