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laksh_khamesra · 12-09-2008, 02:08

This code i have made to take out age of the user(the date in the mktime is just for testing). But many problems are occurring like if the user was born on 2nd of a month today is 1st of a month then the date comes in minus (-) which makes it too confusing... After all, i m completely stuck & confused. Please tell how to take out users age.

PHP Code:


		
			
$td = mktime(0,0,0,05,24,1991);

$td1 = time();



$aa = date("Y", $td);

$bb = date("Y", $td1);

$age = $bb - $aa;



$cc = date("m", $td);

$dd = date("m", $td1);

$month = $dd - $cc;



$ee = date("d", $td);

$ff = date("d", $td1);

$days = $ff - $ee;



echo "$age years, $month months and $days days";

Cborrow · 12-09-2008, 07:45

Try this.

[Please sign in to see the solution]

datahound · 12-09-2008, 07:57

Did not work when I tried it, sorry.

datahound · 12-09-2008, 07:58

11 months out.

Cborrow · 12-09-2008, 08:21

Yeah, I'm not sure what it is but mktime returns the month as 12. If I change the day to 09 the month changes to 11. Anthing over Aug 9th renders the month as 12.

Edit: Find a function online that seems to do what you want.

time since [date] [php] [time] [since]

feha · 12-09-2008, 17:19

try this
PHP Get Age Function | Trevor Davis

My-PHP.tk ••• Your Free PHP Resource! •••

or google for "age php function"

combatsheep · 15-09-2008, 09:03

PHP Code:


		
			
<?php

$date = strtotime("11/04/1984"); // could be your sql result or timestamp



$day = date('jS',$date);

$month = date('F',$date);

$year = date('Y',$date);





echo "My birthdate is the $day of $month , $year";

?>

Remember date() accepts US dates by default so 4/11/1984 in US is 11/04/1984.

look at php.net/date for formatting options.

feha · 15-09-2008, 09:12

@combatsheep
I think he wanted to calculate the age on given date ...
not date conversion

12-09-2008, 02:08	#1 (permalink)
laksh_khamesra Registered User Join Date: Sep 2008 Posts: 13	How to take out user's AGE in PHP? This code i have made to take out age of the user(the date in the mktime is just for testing). But many problems are occurring like if the user was born on 2nd of a month today is 1st of a month then the date comes in minus (-) which makes it too confusing... After all, i m completely stuck & confused. Please tell how to take out users age. PHP Code: `$td = mktime(0,0,0,05,24,1991); $td1 = time(); $aa = date("Y", $td); $bb = date("Y", $td1); $age = $bb - $aa; $cc = date("m", $td); $dd = date("m", $td1); $month = $dd - $cc; $ee = date("d", $td); $ff = date("d", $td1); $days = $ff - $ee; echo "$age years, $month months and $days days";`

12-09-2008, 07:45	#2 (permalink)
Cborrow I like code. Join Date: Dec 2004 Location: Chesapeake, VA Posts: 297	Try this. [Please sign in to see the solution] __________________ My SourceForge page \| Cory_Borrow.Blog(); Last edited by Cborrow : 12-09-2008 at 14:27.

12-09-2008, 07:57	#3 (permalink)
datahound Spare Parts Join Date: Jan 2005 Location: Bracknell Forest Posts: 6,875	Did not work when I tried it, sorry. __________________ poochmedia \| Computer Spare Parts \| Computer Disposals \| Telephone Disposals

12-09-2008, 07:58	#4 (permalink)
datahound Spare Parts Join Date: Jan 2005 Location: Bracknell Forest Posts: 6,875	11 months out. __________________ poochmedia \| Computer Spare Parts \| Computer Disposals \| Telephone Disposals

12-09-2008, 08:21	#5 (permalink)
Cborrow I like code. Join Date: Dec 2004 Location: Chesapeake, VA Posts: 297	Yeah, I'm not sure what it is but mktime returns the month as 12. If I change the day to 09 the month changes to 11. Anthing over Aug 9th renders the month as 12. Edit: Find a function online that seems to do what you want. time since [date] [php] [time] [since] __________________ My SourceForge page \| Cory_Borrow.Blog(); Last edited by Cborrow : 12-09-2008 at 14:24.

12-09-2008, 17:19	#6 (permalink)
feha Design Destroyer Join Date: Aug 2008 Location: Somewhere in Universe Posts: 198	try this PHP Get Age Function \| Trevor Davis My-PHP.tk ••• Your Free PHP Resource! ••• or google for "age php function" __________________ SIGNATURE ?

15-09-2008, 09:03	#7 (permalink)
combatsheep trained to kill Join Date: Apr 2008 Location: Liverpool, UK Posts: 133	PHP Code: `<?php $date = strtotime("11/04/1984"); // could be your sql result or timestamp $day = date('jS',$date); $month = date('F',$date); $year = date('Y',$date); echo "My birthdate is the $day of $month , $year"; ?>` Remember date() accepts US dates by default so 4/11/1984 in US is 11/04/1984. look at php.net/date for formatting options.

15-09-2008, 09:12	#8 (permalink)
feha Design Destroyer Join Date: Aug 2008 Location: Somewhere in Universe Posts: 198	@combatsheep I think he wanted to calculate the age on given date ... not date conversion __________________ SIGNATURE ?

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