Determine Time Difference

[treeleaf20]

I have two timestamps, they are:

17:20:34 and 19:40:20

I would like to take the greater one minus the smaller to get the total time in between them. When I do this by doing:

$totaltime = $resultset['time_in'] - $resultset['time_out'];

It just returns 0, can someone please help me out here.

Thanks.

[treeleaf20]

Also would it be possible to break them out like you worked 2 hours 20 minutes and 46 seconds based on the above time stamp without breaking the timestamp out into hours, minutes and seconds?

Thanks

[_theworks]

read this
thread

[treeleaf20]

I reviewed that thread but that deals with if you have the date. I already have the timestamps so I would just like to determine the difference in them instead of trying to convert a date to a timestamp.

[bpat1434]

Perhaps something like....

<?php

function timeDiff($timebegin, $timefinish, $rtn=TRUE)
{
	if(!is_numeric($timebegin))
	{
		// $timebegin is NOT a UNIX timesamp, so we need to make one
  		list($s_hr, $s_min, $s_sec) = explode(':', $timebegin);
  		$start = mktime($s_hr, $s_min, $s_sec, 1,1,1999);
	}
	else
	{
		// $timebegin is numerical, assume it's a timestamp:
		$start = $timebegin;
	}
	if(!is_numeric($timefinish))
	{
		// $timefinish is NOT a UNIX timestamp, so we need to make one
		list($f_hr, $f_min, $f_sec) = explode(':', $timefinish);
		$end = mktime($f_hr, $f_min, $f_sec, 1,1,1999);
	}
	else
	{
		// $timefinish is numerical, assume it's a timestamp:
		$end = $timefinish;
	}
  $diff = $end-$start;

  $dif['hour'] = floor( $diff/(60*60) );
  $diff = $diff-($dif['hour']*(60*60));

  $dif['min'] = floor( $diff/60 );
  $dif['sec'] = $diff-($dif['min']*60);

  $difference = $dif['hour'].' hours '.$dif['min'].' minutes '.$dif['sec'].' seconds';

  if($rtn == TRUE)
  {  return $difference; }
  else
  {  echo $difference; }
}

$difference = timeDiff(mktime(12,54,00,1,1,1999), mktime(18,54,00,1,1,1999));
echo 'The difference between '.date('H:i:s', mktime(12,54,00,1,1,1999)).' and '.date('H:i:s', mktime(18,54,00,1,1,1999)).' is:<br>'.$difference;

$difference = timeDiff('12:54:00', mktime(18,54,00,1,1,1999));
echo '<hr>The difference between 12:54:00 and '.date('H:i:s', mktime(18,54,00,1,1,1999)).' is:<br>'.$difference;

?>

Working Example (It is running the exact code above.)

Something like that?

~Brett

[laserlight]

I already have the timestamps so I would just like to determine the difference in them instead of trying to convert a date to a timestamp.

The problem is that your timestamps cannot be used directly to calculate time difference (aside from being unable to difference days).

//convert from timestamp to time in seconds, on the same day
function timeInSec($timestamp) {
	$temp = explode(':', $timestamp);
	return ($temp[0] * 3600) + ($temp[1] * 60) + $temp[2];
}

$totaltime = timeInSec($resultset['time_in']) - timeInSec($resultset['time_out']);

[treeleaf20]

Yeah that looks great, thanks for the help. I have a follow up question now. Users could be in any timezone, how would I get their current time zone and use that time zone in the date() instead of the servers time? Also what would happen if a user would log their time at 22:00:00 (10pm) and then they log their time out at 01:00:00 (1 am) the next day. I would like for that to still calculate the time difference so 3 hours in that case. Any ideas how to do that?

Thanks for the help.

[laserlight]

Users could be in any timezone, how would I get their current time zone and use that time zone in the date() instead of the servers time?

One way would just be to ask your users, e.g. via a control panel. Other ways include using clientside scripting that passes the data to the server.

Also what would happen if a user would log their time at 22:00:00 (10pm) and then they log their time out at 01:00:00 (1 am) the next day. I would like for that to still calculate the time difference so 3 hours in that case.

Cannot be done (without guessing) unless the time logged includes the date.

[bpat1434]

well, are you using time-stamps? If so, it's possible by comparing the dates to see if it's within the same day. Then, if not, calculate to 24:00:00, then from 00:00:00 to the end time. Will take extra steps, but if you think through it logically, I think you can get it.

~Brett

[bpat1434]

Hmm... worthy of a double post... here's the updated code to give you the hour difference if the time falls over a day:

<?php

function timeDiff($timebegin, $timefinish, $rtn=TRUE)
{
	if(!is_numeric($timebegin))
	{
		// $timebegin is NOT a UNIX timesamp, so we need to make one
  		list($s_hr, $s_min, $s_sec) = explode(':', $timebegin);
  		$start = mktime($s_hr, $s_min, $s_sec, 1,1,1999);
	}
	else
	{
		// $timebegin is numerical, assume it's a timestamp:
		$start = $timebegin;
	}
	if(!is_numeric($timefinish))
	{
		// $timefinish is NOT a UNIX timestamp, so we need to make one
		list($f_hr, $f_min, $f_sec) = explode(':', $timefinish);
		$end = mktime($f_hr, $f_min, $f_sec, 1,1,1999);
	}
	else
	{
		// $timefinish is numerical, assume it's a timestamp:
		$end = $timefinish;
	}
  if($start <= $end)
  {
    $diff = $end-$start;

$dif['hour'] = floor( $diff/(60*60) );
$diff = $diff-($dif['hour']*(60*60));

$dif['min'] = floor( $diff/60 );
$dif['sec'] = $diff-($dif['min']*60);

$difference = $dif['hour'].' hours '.$dif['min'].' minutes '.$dif['sec'].' seconds';
  }
  else
  {
    // Dealing with an over-the day time period...
    $mid = mktime(0,0,0,1,2,1999);
    $dif1 = $mid-$start;
    $dif2 = $end-$mid;

// The first time-period
$dif[1]['hour'] = floor( $dif1/(60*60) );
$dif1 = $dif1-($dif[1]['hour']*(60*60));
$dif[1]['min'] = floor( $dif1/60 );
$dif[1]['sec'] = $$dif1-($dif[1]['min']*60);

// The second time-period
$dif[2]['hour'] = floor( $dif2/(60*60) );
$dif2 = $dif2-($dif[2]['hour']*(60*60));
$dif[2]['min'] = floor( $dif2/60 );
$dif[2]['sec'] = $dif2-($dif[2]['min']*60);

// Now combine them to get the total time
$dif['tot']['hour'] = $dif[1]['hour']+$dif[2]['hour'];
$dif['tot']['min'] = $dif[1]['min']+$dif[2]['min'];
$dif['tot']['sec'] = $dif[1]['sec']+$dif[2]['sec'];

// Set the difference variable:
$difference = $dif['tot']['hour'].' hours '.$dif['tot']['min'].' minutes '.$dif['tot']['sec'].' seconds';
  }

  if($rtn == TRUE)
  {  return $difference; }
  else
  {  echo $difference; }
}

$difference = timeDiff(mktime(12,54,00,1,1,1999), mktime(18,54,00,1,1,1999));
echo 'The difference between '.date('H:i:s', mktime(12,54,00,1,1,1999)).' and '.date('H:i:s', mktime(18,54,00,1,1,1999)).' is:<br>'.$difference;

$difference = timeDiff('22:00:00', mktime(01,00,00,1,2,1999));
echo '<hr>The difference between 22:00:00 and '.date('H:i:s', mktime(01,00,00,1,1,1999)).' is:<br>'.$difference;

?>

As always: Working Example

~Brett

[laserlight]

If so, it's possible by comparing the dates to see if it's within the same day. Then, if not, calculate to 24:00:00, then from 00:00:00 to the end time.

What if the actual period of time stretches across more than two days? This is why a date should be provided. hh:mm:ss is just not enough, unless there's a guarantee that the period of time will not be beyond two days.

[_theworks]

laserlight wrote:

One way would just be to ask your users, e.g. via a control panel. Other ways include using clientside scripting that passes the data to the server.

or you could simply grab the users IP, run a whois on the IP then parse the results to find out what country the proxy is in.

[treeleaf20]

Not neccessarily a different country. When I logged my time it was 3 hours late I'm assuming my hosting company's server are elsewhere then I am. I just want to be able to determine the timezone in the US and adjust the server time accordingly. I saw some examples on php.net when I searched on the date function but couldn't figure out how to apply it to my problem. Also in the the post by _theworks on this thread it had something similar to what I would want with the time zone just once again I couldn't figure out how to use it.

[bpat1434]

Easiest way is to have each user input their timezone via a control panel or a drop-down on the side of the site.

Running the IP WHOIS query could possibly take a while, but it's up to you.

~Brett

[treeleaf20]

OK new development. I would now like a user to be able to clock out if they go to lunch and then clock back in or something along those lines. What would be the best practice to store their times and then figure out the total time worked for the day. Would I need to put everything into an array and perform the functions on that? Any code examples of this would be greatly appreciated. Thanks in advance again.

[bpat1434]

You could just store their clock-in and clock-out times in a database or cookie. Then, just see if there is already a cookie there from previously that day, and then add the totals together (i.e. add the differences together and evaluate hours, minutes and seconds total).

~Brett

[treeleaf20]

What if I had something like everytime they clocked in or out it would create a new record in the database. I would also include the date. So say for the date 1/7/2006 a user has times of:

8:00:23-11:53:25 which equals 4 hours 53 minutes and 2 seconds and then another timestamp from 1:00:00 to 5:00:00 for another 4 hours. What would be the best way to loop through the database and perform the calculations to come up with the total hours worked for the day.

Thanks for the continued help.

[bpat1434]

SELECT *
  FROM table
  WHERE id='users_id'
    AND date=DATE_FORMAT(NOW(), '%m/%d/%Y')
  ORDER BY date ASC

Then just loop through each item, and add the differences....

<?php
$diffs = array();
$total = 0;
while($row = mysql_fetch_array($result))
{
  $start = strtotime($row['time_in']);
  $end = strtotime($row['time_out']);
  array_push($diffs, ($start-$end));
}

foreach($diffs as $span)
{
  $total = $total+$span;
}

$hours = floor($total/(60*60));
$total = $total-($hours*(60*60));
$minutes = floor($total/(60));
$seconds = $total-($minutes*60);

echo 'Total Time Logged In: '.$hours.' hours, '.$minutes.' minutes, '.$seconds.' seconds';
?>

Perhaps something like that?

~Brett

[treeleaf20]

Thanks Brett, just antoher quick question for you. What would the insert statement look like so tI can use your query that you provided?

Thanks again.

[bpat1434]

INSERT INTO table
  (time_in, date) 
  VALUES ('$time_in', DATE_FORMAT(NOW(), '%m/%d/%Y'))

UPDATE table 
  SET time_out=DATE_FORMAT(NOW(), '%h:%i%s')
  WHERE time_out = NULL
    AND date=DATE_FORMAT(NOW(), '%m/%d/%Y')
    AND id='$user_id'

~Brett