Calculate time difference with formatted time - php

Question

I have two times: a starting time and the duration. I want to subtract the duration from the starting time. The time I read from an mysql db and is already formatted. My code:
$start= $row["start"]; //output is for e.g. 08:00:00
$dur = $row["duration"]; //output is for e.g. 01:00:00
$sub = $start - $dur;
// I want the output to be 07:00:00
// the result now is 7 and I got an error (non well formed numeric value)
Can someone help me?

Answer 1

Alternatively you can achieved like this
$date = "1970-01-01";
$start = $date." ".$row["start"];
$dur = $date ." ".$row["duration"];
$date1=date_create($start);
$date2=date_create($dur);
$diff=date_diff($date1,$date2);
echo $diff->format("%H:%I:%S");

Answer 2

$start= "08:00:00";
$dur = "01:00:00";
$diff = differenceInHours($start, $dur);
echo convertTime($diff);
function differenceInHours($startdate,$enddate){
$starttimestamp = strtotime($startdate);
$endtimestamp = strtotime($enddate);
$difference = abs($endtimestamp - $starttimestamp)/3600;
return $difference;
}
function convertTime($dec)
{
// start by converting to seconds
$seconds = ($dec * 3600);
// we're given hours, so let's get those the easy way
$hours = floor($dec);
// since we've "calculated" hours, let's remove them from the seconds variable
$seconds -= $hours * 3600;
// calculate minutes left
$minutes = floor($seconds / 60);
// remove those from seconds as well
$seconds -= $minutes * 60;
// return the time formatted HH:MM:SS
return lz($hours).":".lz($minutes).":".lz($seconds);
}
// lz = leading zero
function lz($num)
{
return (strlen($num) < 2) ? "0{$num}" : $num;
}

Answer 3

You should always save date in your database in the appropriate format. And not as a "already formated" value. Except in very specific case.
Anyway, to solve your problem you can do soemthing like this
$start = new DateTime('08:00:00');
$duration = new DateTime('01:00:00');
$interval = date_diff($start, $duration);
echo $interval->format('%H:%I:%S'); //ouput will be 07:00:00