PHP calculating hours

Question

Something is not working properly with the following code and I am going mad trying to understand why it is not calculating correctly:

$hours = date('H:i' , strtotime('03:00') - strtotime('02:00'));

echo $hours;

Result: 02:00

Expected Result: 01:00

Could anyone help me guessing what is going wrong?

02:00 as the result

@Albzi thank you for your answer. When I use only one strtotime, it calculated correctly. — Alexandre Ribeiro
– Alexandre Ribeiro, Commented Oct 7, 2016 at 11:48

Mihai Matei · Accepted Answer · 2016-10-07 11:29:23Z

3

You can use the DateTime object to accomplish it:

$date1 = new DateTime('03:00');
$date2 = new DateTime('02:00');
$dateInterval = $date1->diff($date2);

echo $dateInterval->format('%H:%S'); // result would be 01:00

answered Oct 7, 2016 at 11:29

Mihai Matei

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Comments

Albzi · Accepted Answer · 2016-10-07 11:31:27Z

2

You probably want something like this instead:

$hours = date('H:i' , strtotime('03:00 - 02:00'));

Or even:

$hours = date('H:i' , strtotime('03:00 - 2 hours'));

This will print 01:00.

https://3v4l.org/on4JH

edited Oct 7, 2016 at 11:31

answered Oct 7, 2016 at 11:28

Albzi

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6 Comments

Albzi Over a year ago

@LifeTimeProgrammer Look at this link: 3v4l.org/EIZ4s It shows OP's behavior he doesn't want.

Karthi Over a year ago

yes i looked up. but when my local server its working as well as your code also .

Albzi Over a year ago

That's PHP for you haha @LifeTimeProgrammer

Alexandre Ribeiro Over a year ago

Thanks for your answer!

Albzi Over a year ago

@rjdown Nice catch. Time/date is a pain!

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