PHP - subtracting today's year with an integer or string

Question

I have the following columns in my table: birth_year, birth_month, birth_day

example data from the table is

birth_year         birth_month        birth_day
1996               April              1

I only need to get the age of a user.

It doesn't need to be accurate. I just need to subtract the year today and his/her birth_year

example:

his birth_year is 1996

and the date today is 2015

if($row = mysqli_fetch_assoc($query_run)
{
       $birth_day = $row['birth_year'];    //data from the database which equals to 1996
       $today = date("Y");                 //year today
       $age = $today - $birth_day;        //age of this user
        echo $age;                        //this should output 19
}

My problem is the $age variable from the above echoes a blank or null value.

here are the sources that might be similar to my question:

source 1 source 2

The difference between my question and those is that I can't follow the format of DateTime because my registration form looks like this.

You should not store date data in three separate columns in a database. All DBMS have dedicated types for that, like `datetime` or `timestamp`. — Anders, Aug 29 '15 at 12:29
Instead of `echo`ing your variables, use `var_dump` so you can see it's type. You're probably using arithmetic with a string.. — Joaquín O, Aug 29 '15 at 12:35
I used die($age) instead of echo, I thought it will give me the same output. — KennethC, Aug 29 '15 at 12:40

Niki van Stein · Accepted Answer · 2015-08-29T12:27:16.820

2

Try to use

$age = intval( $today ) - intval( $birth_day );

Also check if your mysql query returns what you expect (you did not provide it so it is hard to check).

edited Aug 29 '15 at 12:27

answered Aug 29 '15 at 12:23

Niki van Stein

10,564
3
29
62

your code outputs nothing, yes my query worked. i tried echoing it just now. and it gives me a value of 1996 – KennethC Aug 29 '15 at 12:25
Sorry I wrote it with capital but it should be intval without capital. Can you try again cause it should work. – Niki van Stein Aug 29 '15 at 12:27
It's still the same, sir. – KennethC Aug 29 '15 at 12:30
I also tried echoing those 2 variables individually, both of them have values, but when i tried to subtract $today - $birth_year it returns nothing – KennethC Aug 29 '15 at 12:31
I tried the suggested code, and it does indeed work. – Anders Aug 29 '15 at 12:32
How is that possible? I even tried to substitute $birth_year manually into a '1996' and 1996 but it still returns nothing – KennethC Aug 29 '15 at 12:35
Turn on your error messages, perhaps something else in the code goes wrong. – Niki van Stein Aug 29 '15 at 12:37
Oh it works now, Im using die($age) a couple of times to see the output clearer, but i think it's the main reason why it gives me nothing – KennethC Aug 29 '15 at 12:39
1

No problem. I advise you to add error_reporting(E_ALL); ini_set('display_errors', 1); so you see all errors happening. – Niki van Stein Aug 29 '15 at 12:42

score 0 · Answer 2 · answered Aug 29 '15 at 12:46

0

try this with the datetime:

if($row = mysqli_fetch_assoc($query_run){
    $birth_day = $row['birth_year'];
    $bday = new DateTime($birth_day);
    $today = new DateTime(date("Y"));
    $age = $today->diff($bday);
    echo $age->y;
}

answered Aug 29 '15 at 12:46

Dimak

13
3

PHP - subtracting today's year with an integer or string

2 Answers2