working out age from DOB

[fireice87]

Hey iv got a script that works out the users age from there dob

$ageTime = mktime(0, 0, 0, 9, 8, 1987); // Get the person's birthday timestamp
$t = time(); // Store current time for consistency
$age = ($ageTime < 0) ? ( $t + ($ageTime * -1) ) : $t - $ageTime;
$year = 60 * 60 * 24 * 365;
$ageYears = $age / $year;

echo 'You are ' . floor($ageYears) . ' years old.';

 

Im trying to expand this so that i can pull the DOB from the database and use variables to insert the DOB into the mktime( ). below is the code i have written to pull the dob from the database and add it too the calculation for the users age problem im having is that the age doesnt make it to the database.

the error i get is when it should be inserted into the database

 

iv used the code to get it into the database multiple times in diffrent scripts so i know thats fine so im thinking the error is something to do with the way im tring to use the variables in the calculation.

 

any help would be great as i cant work out the issue here... thanks 

$conn = @mysql_connect( "localhost", "username", "password")
or die ("could not connect to mysql");  #Connect to mysql
$rs = @mysql_select_db( "site23_thewu", $conn )
or die ("Could not select database"); #select database 

$sql0 = "DELETE 'age' FROM profile_quickfacts WHERE user= '$user'";

$empty=  $run= @mysql_query($sql0, $conn )
or die(" Could not add quick facts");

  $sql = "Select `day`,`month`,`year` FROM profile_quickfacts WHERE user= '$user' "; 

   $result= @mysql_query($sql, $conn )
or die(" Could not select data");

while($row = mysql_fetch_row($result))
{
$Day = $row[1];
$Month = $row[2];
$Year = $row[3];
} 

$ageTime = mktime(0, 0, 0, ($Day),  ($Month), ($Year)); // Get the person's birthday timestamp
$t = time(); // Store current time for consistency
$age = ($ageTime < 0) ? ( $t + ($ageTime * -1) ) : $t - $ageTime;
$year = 60 * 60 * 24 * 365;
$ageYears = $age / $year;

// echo 'You are ' . floor($ageYears) . ' years old.';

$sql3 = "INSERT INTO profile_quickfacts(`age`)
VALUES (\"$ageYears\");";

    $run= @mysql_query($sql3, $conn )
or die(" Could not add quick facts");

[Caesar]

What is the error you're actually getting/seeing?

[sasa]

try

...
//$ageTime = mktime(0, 0, 0, ($Month),  ($Day), ($Year)); // Get the person's birthday timestamp
$ageYears = date('Y') - $Year;
$ageYears -= (date('m') * 100 + date('d')) < ($Month * 100 + $Day) ? 1 : 0;
//$t = time(); // Store current time for consistency
//$age = ($ageTime < 0) ? ( $t + ($ageTime * -1) ) : $t - $ageTime;
//$year = 60 * 60 * 24 * 365;
echo $ageYears;
...

[Barand]

it doesn't have to be so complex

<?php
function calcAge($dob)
{
    $tob = strtotime($dob);
    $age = date('Y') - date('Y', $tob);    // age in years
    return (date('md') < date('md', $tob)) ? $age - 1 : $age;    // but if we haven't reached birthday this year, sub 1
}

echo calcAge ('1987-09-08');
?>