[PHP] time... - RaGEZONE

passie · 08-18-2009

Hello,
i am working with time system for my new game if they make a building they get a time how long it takes before its done but now am getting confused
how i can make it show like this ;

HOURS : MINUTES : SECONDS

i got this ..

PHP Code:

  $now = date("Y-n-d H:i:s");

    

$to_time=strtotime($logs['time']); //time when it should be done 2009-08-18 23:15:59

$from_time=strtotime($now); // time now 2009-08-18 21:22:59

 
$timeleft = round(abs($to_time - $from_time) / 60);

 
$timeleft = str_replace(".", ":", $timeleft);

 
echo "<tr><td>".$logs['name']."</td><td>".round($timeleft)." minute('s)</td></tr>";

then it shows 114 minute('s)

so how i can make it like :
HOURS : MINUTES : SECONDS

thanks alot so far,
passie

foxx · 08-18-2009

PHP Code:

  <?php
function timeBetween($start_date,$end_date)
{
    $diff = $end_date-$start_date;
    $seconds = 0;
    $hours   = 0;
    $minutes = 0;

    if($diff % 86400 <= 0){$days = $diff / 86400;}
    if($diff % 86400 > 0)
    {
        $rest = ($diff % 86400);
        $days = ($diff - $rest) / 86400;
        if($rest % 3600 > 0)
        {
            $rest1 = ($rest % 3600);
            $hours = ($rest - $rest1) / 3600;
            if($rest1 % 60 > 0)
            {
                $rest2 = ($rest1 % 60);
            $minutes = ($rest1 - $rest2) / 60;
            $seconds = $rest2;
            }
            else{$minutes = $rest1 / 60;}
        }
        else{$hours = $rest / 3600;}
    }

    if($days > 0){$days = $days.':';}
    else{$days = false;}
    if($hours > 0){$hours = $hours.':';}
    else{$hours = false;}
    if($minutes > 0){$minutes = $minutes.':';}
    else{$minutes = false;}
    $seconds = $seconds.'';

    return $days.''.$hours.''.$minutes.''.$seconds;
}

$one_date = date("U");  
$two_date = strtotime($logs['time']); //"2009-08-18 23:15:59"
$result = timeBetween($one_date, $two_date);

if(strtotime($result)) {
echo date('H:i:s', strtotime($result));
} else {
echo "<strong>error:</strong> time already passed.";
}  
?>

this rewritten a little
http://www.linein.org/blog/2008/04/0...-dates-in-php/

passie · 08-18-2009

Originally Posted by foxx

PHP Code:

  <?php
function timeBetween($start_date,$end_date)
{
    $diff = $end_date-$start_date;
    $seconds = 0;
    $hours   = 0;
    $minutes = 0;

    if($diff % 86400 <= 0){$days = $diff / 86400;}
    if($diff % 86400 > 0)
    {
        $rest = ($diff % 86400);
        $days = ($diff - $rest) / 86400;
        if($rest % 3600 > 0)
        {
            $rest1 = ($rest % 3600);
            $hours = ($rest - $rest1) / 3600;
            if($rest1 % 60 > 0)
            {
                $rest2 = ($rest1 % 60);
            $minutes = ($rest1 - $rest2) / 60;
            $seconds = $rest2;
            }
            else{$minutes = $rest1 / 60;}
        }
        else{$hours = $rest / 3600;}
    }

    if($days > 0){$days = $days.':';}
    else{$days = false;}
    if($hours > 0){$hours = $hours.':';}
    else{$hours = false;}
    if($minutes > 0){$minutes = $minutes.':';}
    else{$minutes = false;}
    $seconds = $seconds.'';

    return $days.''.$hours.''.$minutes.''.$seconds;
}

$one_date = date("U");  
$two_date = strtotime($logs['time']); //"2009-08-18 23:15:59"
$result = timeBetween($one_date, $two_date);

if(strtotime($result)) {
echo date('H:i:s', strtotime($result));
} else {
echo "<strong>error:</strong> time already passed.";
}  
?>

this rewritten a little
http://www.linein.org/blog/2008/04/0...-dates-in-php/

how i supose to use it?
perhaps it sound as noob way but i use it like this

$now : 2009-08-18 22:36:00
$logs['time'] 2009-09-18 01:30:25

here's my full script how i use it

PHP Code:

  <?php
include "config.php";

$query = mysql_query("select * from `logs` where `username`='{$_SESSION['username']}'");



?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<table>
<tr>
<td>Name</td> <td>Timeleft</td>
</tr>
<?php


while($logs = mysql_fetch_assoc($query)){
    
function timeBetween($start_date,$end_date)
{
    $diff = $end_date-$start_date;
    $seconds = 0;
    $hours   = 0;
    $minutes = 0;

    if($diff % 86400 <= 0){$days = $diff / 86400;}
    if($diff % 86400 > 0)
    {
        $rest = ($diff % 86400);
        $days = ($diff - $rest) / 86400;
        if($rest % 3600 > 0)
        {
            $rest1 = ($rest % 3600);
            $hours = ($rest - $rest1) / 3600;
            if($rest1 % 60 > 0)
            {
                $rest2 = ($rest1 % 60);
            $minutes = ($rest1 - $rest2) / 60;
            $seconds = $rest2;
            }
            else{$minutes = $rest1 / 60;}
        }
        else{$hours = $rest / 3600;}
    }

    if($days > 0){$days = $days.':';}
    else{$days = false;}
    if($hours > 0){$hours = $hours.':';}
    else{$hours = false;}
    if($minutes > 0){$minutes = $minutes.':';}
    else{$minutes = false;}
    $seconds = $seconds.'';

    return $days.''.$hours.''.$minutes.''.$seconds;
}

$one_date = date("U");  
$two_date = strtotime($logs['time']); //"2009-08-18 23:15:59"
$result = timeBetween($one_date, $two_date).'';

if(strtotime($result)) {
echo date('H:i:s', strtotime($result));
} else {
echo "<strong>error:</strong> time already passed.";
}  

$now = date("Y-n-d H:i:s");
$timeleft = timeBetween($now, $logs['time']);

echo "<tr><td>".$logs['name']."</td><td>".$timeleft." minute('s)</td></tr>";
}
?>
</table>
</body>
</html>

i get this : error: time already passed.33:29

hopefull you can help me again.

foxx · 08-18-2009

Well try this.

PHP Code:

  <?php
include "config.php";

$query = mysql_query("select * from `logs` where `username`='{$_SESSION['username']}'");

function timeBetween($start_date,$end_date)
{
    $diff = $end_date-$start_date;
    $seconds = 0;
    $hours   = 0;
    $minutes = 0;

    if($diff % 86400 <= 0){$days = $diff / 86400;}
    if($diff % 86400 > 0)
    {
        $rest = ($diff % 86400);
        $days = ($diff - $rest) / 86400;
        if($rest % 3600 > 0)
        {
            $rest1 = ($rest % 3600);
            $hours = ($rest - $rest1) / 3600;
            if($rest1 % 60 > 0)
            {
                $rest2 = ($rest1 % 60);
            $minutes = ($rest1 - $rest2) / 60;
            $seconds = $rest2;
            }
            else{$minutes = $rest1 / 60;}
        }
        else{$hours = $rest / 3600;}
    }

    if($days > 0){$days = $days.':';}
    else{$days = false;}
    if($hours > 0){$hours = $hours.':';}
    else{$hours = false;}
    if($minutes > 0){$minutes = $minutes.':';}
    else{$minutes = false;}
    $seconds = $seconds.'';

    return $days.''.$hours.''.$minutes.''.$seconds;
}


?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<table>
<tr>
<td>Name</td> <td>Timeleft</td>
</tr>
<?php
while($logs = mysql_fetch_assoc($query)){
    
    $one_date = date("U");  
    $two_date = strtotime($logs['time']); //"2009-08-18 23:15:59"
    $result = timeBetween($one_date, $two_date);

    if(strtotime($result)) {
        $message = date('H:i:s', strtotime($result));
    } else {
        $message = "time passed.";
    }  
    echo "<tr><td>".$logs['name']."</td><td>".$message."</td></tr>";
}
?>
</table>
</body>
</html>

There might be some problems when the difference is longer than 24 hours though. You'd need to rewrite that original script bit more, in that case.