date of birth to days old in php

[kevinkhan]

i have the following date 1990-08-27 and i wanna find out how many days old the person is.

 

Any ideas how to do it?

[phpSensei]

You can do, not the best solution, but it works.

 

return floor( (time() – strtotime($dateofbirth ) ) / 31556926);

 

Also I tried to see what Google would return, you should try it more often

 

http://stackoverflow.com/questions/3380990/php-calculate-persons-current-age

 

[TeNDoLLA]

$bday = new DateTime('1980-11-27');
$today = new DateTime('now');
$diff = $bday->diff($today);
echo $diff->format('%y years, %m months, %d days old.');

[kevinkhan]

$bday = new DateTime('1980-11-27');
$today = new DateTime('now');
$diff = $bday->diff($today);
echo $diff->format('%y years, %m months, %d days old.');

 

im getting the following error when i run this

 

Call to undefined method DateTime::diff()

[TeNDoLLA]

You need (PHP 5 >= 5.3.0) for DateTime functions to work. Otherwise you will have to stick with phpSensei's method.

[kevinkhan]

awh i only have 5.2.9 :(

 

any other method of calculated number of days?

 

and can you suggest any good hosting company's?

[Muddy_Funster]

On the hosting company front I would recomend 34sp.com

 

There is no way to easily compare dates in 5.2 that I know of, you would need to get/write a function to do it manualy.

[TeNDoLLA]

Yeah and writing such a function will probably never be 100% accurate since there is not same amount of days in every year, not same amount of hours in every day (swaps between summer and winter time) etc. Either way writing the function would probably be huge pain in the ass.

[ignace]

function calculate_age($birthday) {
    $time = strtotime($birthday);
    if ($time === FALSE || $time === -1) return FALSE;
    
    $a = idate('Y') - idate('Y', $time);
    $b = mktime(0, 0, 0, idate('n', $time), idate('j', $time), idate('Y'));
    if (time() < $b) --$a;
    return $a;
}

print calculate_age('1986-09-05').PHP_EOL; // 24
print calculate_age('1986-08-05').PHP_EOL; // 24
print calculate_age('1986-07-05').PHP_EOL; // 25

 

Just an idea. I prefer phpSensei's solution.

[gizmola]

I am in the process of writing up a small article on how to do this directly with mysql SQL in a query.  So if you have the birthdate in a mysql DATE type column, this query will get you a person's age, based on their birthdate.

 

This query assumes a table named "bday", with the birthdate column named "bdate".  You would need to adjust for your table and structure accordingly.  You'd want to add an appropriate WHERE clause to get only one row as well.

 

SELECT YEAR(CURDATE()) - YEAR(bdate) - 
IF(STR_TO_DATE(CONCAT(YEAR(CURDATE()), '-', MONTH(bdate), '-', DAY(bdate)) ,'%Y-%c-%e') >= CURDATE(), 1, 0) 
AS age FROM bday;

 

 

[ignace]

SELECT YEAR(CURDATE()) - YEAR(bdate) - 
IF(STR_TO_DATE(CONCAT(YEAR(CURDATE()), '-', MONTH(bdate), '-', DAY(bdate)) ,'%Y-%c-%e') >= CURDATE(), 1, 0) 
AS age FROM bday;

 

We think alike! Is it intended to calculate this for everyone in the bday table?

[phpSensei]

Gizmola, Cool stuff!