PHP Converting DateInterval to int

Question

I'm using this code:

$due_date = new DateTime($_POST['due_date']);

$today = new DateTime();
$months = $due_date->diff($today);

$months->format("%m");

$fine = 0.02 * $price * $months; // i got error in this line

$bill = $price + $fine;

I want to calculate, if someone is late to pay then they will be fined per month. And the error message is:

Object of class DateInterval could not be converted to int

You never do something with the return value of [`$months->format("%m");`](http://php.net/manual/en/dateinterval.format.php). — PeeHaa, Jun 25 '16 at 13:45
`$months` is a Datetime object here is a sample one... `DateInterval Object ( [y] => 0 [m] => 4 [d] => 12 [h] => 6 [i] => 56 [s] => 9 [weekday] => 0 [weekday_behavior] => 0 [first_last_day_of] => 0 [invert] => 0 [days] => 133 [special_type] => 0 [special_amount] => 0 [have_weekday_relative] => 0 [have_special_relative] => 0 )` — , Jun 25 '16 at 13:56

score 7 · Accepted Answer · 2016-06-25T14:00:14.913

The error message appears because $months is not an int, but a Datetime object like this one:

DateInterval Object
(
    [y] => 0
    [m] => 4
    [d] => 12
    [h] => 6
    [i] => 56
    [s] => 9
    [weekday] => 0
    [weekday_behavior] => 0
    [first_last_day_of] => 0
    [invert] => 0
    [days] => 133
    [special_type] => 0
    [special_amount] => 0
    [have_weekday_relative] => 0
    [have_special_relative] => 0
)

You can get the integer value of months like this

$due_date = new DateTime('13-02-2016');

$today = new DateTime();
$months = $due_date->diff($today);

echo $months->m;

Check the above result in PHP Sandbox

So basically your code will look like

$due_date = new DateTime($_POST['due_date']);

$today = new DateTime();
$months = $due_date->diff($today);

$fine = 0.02 * $price * $months->m; // i got no error in this line

$bill = $price + $fine;

@Pevara thanks, `format` it was not needed in this answer – Jun 25 '16 at 14:02 — , Jun 25 '16 at 14:02

score 2 · Answer 2 · answered Jun 25 '16 at 13:48

2

You calculate the difference in months, but you never actually use that value. The format method outputs something, but doesn't change the actual DateInterval.

Try it like this:

$due_date = new DateTime($_POST['due_date']);

$today = new DateTime();
$diff = $due_date->diff($today);

$months = $diff->format("%m");

$fine = 0.02 * $price * $months;

$bill = $price + $fine;

answered Jun 25 '16 at 13:48

Pevara

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`$months = $due_date->diff($today)->format("%m"));` – splash58 Jun 25 '16 at 13:49
@splash58 Sure, or `$months = (new DateTime($_POST['due_date']))->diff(new DateTime())->format("%m");`... I was just trying to point out the `diff`erence ;) – Pevara Jun 25 '16 at 13:52
If it doesn't work in your version of php (< 5.4), you probably shouldn't be using it anymore.. – Pevara Jun 25 '16 at 14:17

score 1 · Answer 3 · answered Sep 13 '19 at 14:35

Peter Darmis's answer is wrong. I can't downvote nor comment it so I will add this new answer.

DateInterval represents a period of time by deconstructing the interval in different parts, years, months, days etc... So "m" in the proposed solution will never be bigger than 12.

So you should probably do something like this:

$due_date = new DateTime('13-02-2016');

$today = new DateTime();
$diff = $due_date->diff($today);

$months = ($diff->y * 12) + $diff->m;

echo $months;

Check it in the sandbox: http://sandbox.onlinephpfunctions.com/code/907b39ffee4586c4c9481ff3e0ea1aeb2d27c7b8

PHP Converting DateInterval to int

3 Answers3