CodingForums.com - Subtracting Time from a Time

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- - Subtracting Time from a Time - Help (http://www.codingforums.com/showthread.php?t=275472)

Subtracting Time from a Time - Help

Code:

<?php

$channel = "mattbenson";

$json_file = @file_get_contents("http://api.justin.tv/api/stream/list.json?channel={$channel}", 0, null, null);

$json_array = json_decode($json_file, true);



if( array_key_exists( '0',       $json_array )

 && array_key_exists( 'channel', $json_array[0] )

 && $json_array[0]['name'] == "live_user_{$channel}" )

{

    $channelTitle = $json_array[0]['channel']['title'];

    $title        = $json_array[0]['channel']['status'];

    $viewers      = $json_array[0]['channel_count'];

    $uptime       = $json_array[0]['up_time'];

$today = date("D M j d:i:s Y");   

$uptime1 = ($today - $uptime);



    printf('Online');



printf ($uptime1);



print "<br>";

print "date <--- ";

printf ($today);

print "<br>";

print "uptime <--- ";

printf ($uptime);

}

else

{

    printf('Offline');

}



?>

It displays like this
http://gyazo.com/fb3c95029c79da2a1a0...png?1349530088

Problem is it doesn't subtract is just displays 0

Anyhelp?

You cannot subtract strings.
$today should be time(), and $uptime MUST be an integer. Then you can subtract and create $uptime1. That will give you the number of seconds $uptime is which you can format into whatever display.
If you have 5.3+, you can use the dateinterval.

PHP Code:


		
			
$dtUptime = new DateTime($json_array[0]['up_time']); // this now assumes a string GNU valid format
$dtNow = new DateTime();
$diDiff = $dtNow->diff($dtUptime, true);
print $diDiff->format("Uptime: %a days");