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Solved

calculate age from dob??

Posted on 2009-04-10
6
338 Views
Last Modified: 2013-12-12
Below is how i store a dob... by using echo $getuserprofile['dob']; i can echo a users dob...but who can i work out there age?
<?php  
	$months = array (1 => 'January', 'February', 'March', 'April', 'May', 'June','July', 'August', 'September', 'October', 'November', 'December');
$weekday = array('Sunday','Monday','Tuesday','Wednesday','Thursday','Friday','Saturday');
$days = range (1, 31);
$years = range (1910, 2015);
 
//**********************************************
 
echo "Day: <select name='day'>";
foreach ($days as $value) {
   echo '<option ';
   if($splitDate[0] == $value)
      echo 'selected="selected"';
   echo ' value="'.$value.'">'.$value.'</option>\n';
} echo '</select>';
 
echo "Month: <select name='month'>";
foreach ($months as $value) {
 
   echo '<option ';
   if($splitDate[1]==$value) 
      echo "selected='selected'";
   echo 'value="'.$value.'">'.$value.'</option>\n';
} echo '</select>';
 
echo "Year: <select name='year'>";
foreach ($years as $value) {
   echo '<option ';
   if($splitDate[2]==$value)
      echo 'selected="selected"';
   echo " value='".$value."'>".$value."</option>\n";
} 
$dob = $day.'-'.$month.'-'.$year; 
 ?>

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Comment
Question by:runnerjp
  • 3
  • 2
6 Comments
 
LVL 19

Expert Comment

by:Michael701
ID: 24116607
try this
$today=getdate();
$age=$today['year']-$dob_year-1;
// now have they had a birthday this year?
if ($today['month']>$dob_month)
  $age++;
else
  if ($today['month']==$dob_month)
    if ($today['mday']>=$dob_day)
      $age++;
 
echo $age;

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LVL 109

Accepted Solution

by:
Ray Paseur earned 500 total points
ID: 24116680

<?php // RAY compute_age_from_dob.php
 
// TEST DATA
$dob = '15'.'-'.'09'.'-'.'1950';
 
// GET TIMESTAMPS
$ts_dob = strtotime($dob);
$ts_now = time();
 
// DIFFERENCE IN YEARS
$years = date('Y', $ts_now) - date('Y', $ts_dob);
 
// IF BEFORE THE BIRTHDAY
$mdnow = date('m-d', $ts_now);
$mddob = date('m-d', $ts_dob);
if ($mdnow < $mddob) $years--;
 
// SHOW THE ANSWER
echo $years;

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Author Comment

by:runnerjp
ID: 24116846
soryy i should have said i display dob as 19-December-1987   so would i need to somehow convert the month into numeric?
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LVL 109

Expert Comment

by:Ray Paseur
ID: 24116858
That should work OK with strtotime()

http://www.laprbass.com/RAY_strtotime.php?s=19-December-1987

HTH, ~Ray
0
 
LVL 19

Expert Comment

by:Michael701
ID: 24117010

// since you already $months, this should get the month number
$dob_month = array_keys ( $months, $dob_month_text );
 
 
// but really you should
 
echo "Month: <select name='month'>";
foreach ($months as $month_number=>$month_name) {
 
   echo '<option ';
   if($splitDate[1]==$month_name) 
      echo "selected='selected' ";
   echo 'value="'.$month_number.'">'.$month_name.'</option>\n';
} echo '</select>';

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0
 
LVL 109

Expert Comment

by:Ray Paseur
ID: 24117044
One note, not directly at issue here, but useful as a "best practices" approach to handling dates.  Make all internal representations into ISO8601 date/time strings.  php date('c') gives you the traditional format

YYYY-MM-DD HH:MM:SS

This is useful because you can use it with strtotime and date() to format dates and times for output, and you can see the columns line up in reports.  Also, it collates and compares correctly whether in a timestamp or a character string.
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