Get difference between time using strtotime returns different value

Question

I have two different time in a 24-hr format which is time in and timeout. Given the scenario below:

$timeIn = "19:00";
$timeOut = "09:00";
//calculated it using strtotime
$timeDiff = strtotime($timeIn) - strtotime($timeOut);
$timeDiff = ($timeDiff/60)/60;
//result is 10:00

With this one, instead of returning 14hrs, it returns 10hrs only. How can I catch this one out?

UPDATE

Thanks for the comments guys. Here is how I do it:

if($timeIn > $timeOut){
    $timeDiff = 24 - abs(strtotime($timeIn)/60)/60;
    $timeDiff += $timeOut;
}else{
    $timeDiff = strtotime($timeIn) - strtotime($timeOut);
    $timeDiff = ($timeDiff/60)/60;
}

you are doing 19 - 9 resulting in 10 if you want 14 you could change $timeIn to 23:00 — Collin ter Steege, Sep 14 '18 at 09:05
Also, that won't return "10:00", but some integer instead… — deceze, Sep 14 '18 at 09:06
maybe you want $timeOut-$timeIn? (that would need another calculation though) — Jeff, Sep 14 '18 at 09:07
19 - 9 equals 10, nothing unusual here. you have to use datetime if you want the time between 19:00 and 9:00 the next morning — ᴄʀᴏᴢᴇᴛ, Sep 14 '18 at 09:07
@Jeff If you will manually count it, 7PM (19:00) to 9AM is 14 hrs. — Eem Jee, Sep 14 '18 at 09:08
but in your question you have 9 to 19. the other way round. As I said, you want `$timeOut-$timeIn` — Jeff, Sep 14 '18 at 09:09
@Jeff Please check this line `$timeDiff = strtotime($timeIn) - strtotime($timeOut);` — Eem Jee, Sep 14 '18 at 09:10
How is PHP supposed to guess that those two times aren't on the same day? — deceze, Sep 14 '18 at 09:10
_after you UPDATE_: you shouldn't do it that way. Now you get negative values on a normal day (eg in: 9 out: 19). a hint: you want $timeOut - $timeIn — Jeff, Sep 14 '18 at 09:18
no. It's basic maths. If you want to know the difference of two values you subtract the smaller (earlier) one from the bigger (later) one. — Jeff, Sep 14 '18 at 09:23

score 4 · Answer 1 · answered Sep 14 '18 at 09:11

4

Oh.. you could use datetime, as this example:

<?php
$timeIn = new DateTime("2019-09-14 19:00:00");
$timeOut = new DateTime("2019-09-15 09:00:00");

$interval = $timeOut->diff($timeIn);
echo $interval->h;
?>

answered Sep 14 '18 at 09:11

AnTrakS

733
4
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Thanks fot this one. But my data has no date, so I have tried my approach and works well. – Eem Jee Sep 14 '18 at 09:15
3

@ramj It *doesn't* work well. And without a date, those times are pretty useless. In timezones which observe DST, you may even get wrong values depending on what day those times actually are. – deceze Sep 14 '18 at 09:16

Thiago Guelfi · Answer 2 · 2018-09-14T09:21:09.887

0

You need to calculate it in timestamp, but to make the complete time you need to use full date format

Example:

$timeIn = "2018-09-14 19:00:00";
$timeOut = "2018-09-15 09:00:00";
//calculated it using strtotime
$timeDiff = strtotime($timeIn) - strtotime($timeOut);

and it gonna work

edited Sep 14 '18 at 09:21

answered Sep 14 '18 at 09:09

Thiago Guelfi

26
3

2

That's still a difference of 10 hours… – deceze Sep 14 '18 at 09:10
Reading between the lines, I think $timeOut is intended to be the following day, 15th. – Tim Sep 14 '18 at 09:11

score 0 · Answer 3 · edited Sep 14 '18 at 10:04

0

The function returning 10 hours is correct and expectable behaviour.

19-9=10

PHP doesn't know that you want to count the hours from 19:00 to 9:00 the next day. Plase refer to D.Dimitrovs answer which show the correct way using DateTime

edited Sep 14 '18 at 10:04

Roy Bogado

4,299
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answered Sep 14 '18 at 09:13

dmuensterer

1,875
11
26

2

I don't understand how an answer stating that another answer is correct (and repeating some comments) could be useful. – Jeff Sep 14 '18 at 09:14

Eem Jee · Answer 4 · 2018-09-14T09:24:09.360

0

For this approach, I have resolved it using this formula and no need to use datetime for I dont have date on my data.

if($timeIn > $timeOut){
    $timeDiff = 24 - abs(strtotime($timeIn)/60)/60;
    $timeDiff += $timeOut;
}else{
    $timeDiff = strtotime($timeIn) - strtotime($timeOut);
    $timeDiff = ($timeDiff/60)/60;
}

Hope this well help the others.

edited Sep 14 '18 at 09:24

answered Sep 14 '18 at 09:16

Eem Jee

1,239
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Get difference between time using strtotime returns different value

4 Answers4