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PHP: Calculate Age from DOB

Posted: Sat Mar 17, 2007 6:08 am
by ner0
Hey guys, since I've spent a fair few hours working on this little snippet, I thought it'd be worth sharing.. as I had to search around for some ideas before I settled on my code:

All dates are in YYYY-MM-DD format, $u['DOB'] and &u['tzOffset'] are values taken from a database containing user data, tzOffset is stored as the offset in hours (i.e. -5 / +2 / -3.5)

The final $m['age'] is the value of the age, also included a check to see whether it's a person's actual birthday, incase you'd like to do anything with that.

Ignore the insane comments... they were for me.

Code: Select all

			//-- Calculate Age from DOB, times in YYYY-MM-DD FORMAT! --//
				$dob = $u['DOB'];
	
				# get user timezone... i.e. what we can take away from server time (GMT/UTC)
				$tz = $u['tzOffset'];
	
				# Get date of NOW, unfortunately, server TZ isn't GMT, to make it easier later, convert date INTO GMT here!
				$t = time();
	
					# How far in front/behind (of) GMT are we? (i.e. -0500)
					$gmtDiff = date("O");
	
						# make this figure usable (we need seconds)
						$gmtDiff = ($gmtDiff / 100) * 3600; # here we just turn the -0500 into -5 (-5 hours) then multiply by 3600 (secs per hour =D)
	
				# TRUE GMT TIME NOW?
				$t = $t - $gmtDiff;
	
					# User's Timezone offset? (in seconds)
					$tz = $tz * 60 * 60;
	
				# User's LOCAL TIME?.. hmm, this could be useful globally?.. yup yup.
				$uTime = $t + $tz;
				$m['uTime'] = $uTime;		// worth noting this is global'ed as a unix timestamp

				# Grab both dates as $n and $b (now and birthdate) arrays
				$n = explode("-",date("Y-m-d",$uTime));
				$b = explode("-",$dob);

					# Age step 1, check years..
					$age = $n[0] - $b[0];

					# Are we BEFORE their birth month?
					if ($n[1] < $b[1])
					{
						$age--;
					}

					# Are we IN their birth month?
					else if ($n[1] == $b[1])
					{
						# current month / birth month.. are we before, or on their birthday?
						if ($n[2] < $b[2])
						{
							$age--;
						}
						else if ($n[2] == $b[2])
						{
							$m['bday'] = "y";
						}
					}
					# Otherwise, we've passed.. or are on their birthday. so the original calc of this year minus birth year is correctimondo =]!

			$m['age'] = $age;
There will be much more elegant ways of doing this, I tried using timestamp difference to begin, though it was late and I couldn't think of the impact leap years would make.

Hope this is useful for someone!

PHP: Calculate Age from DOB

Posted: Sat Mar 17, 2007 10:34 am
by Josh
ROFL! You use comments like I do hahaha. They do work too /biggrin.gif' class='bbc_emoticon' alt=':D' />

PHP: Calculate Age from DOB

Posted: Sat Mar 17, 2007 12:32 pm
by ner0
I like to use comments to 'vent'... especially when I've been stuck coding strictly to php for so long, it's nice to go insane for a while lol

and for some reason using the 'explode()' command always makes me chuckle /blink.gif' class='bbc_emoticon' alt=':blink:' />