Udregne alder ved hjælp af fødselsdato?

http://www.eksperten.dk/spm/658942

du kan bare vende datoen om:

idag: 20051025
føds: 19870915

og så trække tallene fra hindanden:

(20051025-19870915)/10000

= 18

jeg er ikke helt sikker med prøv dette:

<?php

function age($BirthDate, $do)
{

if ($do == "") { $do = "1"; } //Checks if do is set, default set is number of days

list($month1, $day1, $year1) = split('[-.]', $BirthDate); //Splits date apart

$year = date("Y"); //Gets Current Year
$month = date("m"); //Gets Current Month
$day = date("d"); //Gets Current Day

$date = mktime(0,0,0,$month,$day,$year); //Gets Unix timestamp for current date
$date1 = mktime(0,0,0,$month1,$day1,$year1); //Gets Unix timestamp for date set

$difference = $date-$date1; //Calcuates Difference
$days2 = floor($difference /60/60/24); //Calculates Days Old

//If you choice "2" which is status unstead of days you can modify it here!
if ($days2 <= "20") { $days1 = "2 Weeks"; }
if ($days2 <= "13") { $days1 = "1 Week"; }
if ($days2 <= "7") { $days1 = "New"; }
if ($days2 >= "21") { $days1 = "Old"; }
//End Status Modify

if ($do == "1") { echo $days1; } //Echo Status
if ($do == "2") { echo $days2; } //Echo Days
}
?>

Then just put this where ever you want the days or status to show
age('MM-DD-YYYY', 'S')

Replace MM with the month it was made or born
DD with the day
YYYY with the year

S is what format you want it to show, if you want the days of something such as how many days a post has been posted enter "2".
If you want status like "New, 1 week, 2 weeks, Old" then enter "1".

Hope you enjoy.

Gordon
FreePgs.com Tech

hentet fra http://dk.php.net/date

her har du svaret ;)

http://www.pcdummy.nl/~myphp/snippets/1

Har prøvet:

  function birthday ($birthday){
    list($bruger[birthyear],$bruger[birthmon],$bruger[birthday]) = explode("-",$birthday);
    $year_diff  = date("Y") - $bruger[birthyear];
    $month_diff = date("m") - $bruger[birthmon];
    $day_diff  = date("d") - $bruger[birthday];
    if ($day_diff < 0 || $month_diff < 0)
      $year_diff--;
    return $year_diff;
  }

Synes bare ikke det virker?

Når jeg skriver print"birthday"; skriver den ikke noget.

Prøv dette...

<?php
echo "Din alder er : " . birthday ("1987-09-15");
  //calculate years of age (input string: YYYY-MM-DD)
  function birthday ($birthday){
    list($year,$month,$day) = explode("-",$birthday);
    $year_diff  = date("Y") - $year;
    $month_diff = date("m") - $month;
    $day_diff  = date("d") - $day;
    if ($day_diff < 0 || $month_diff < 0)
      $year_diff--;
    return $year_diff;
  }
?>

thx:)
svar?

Mig???

ja