Very simple request, not even a hack :P

[Mijae]

I have this on my page:

PHP Code:

$sinceopening = floor((time() - strtotime('June 24, 2002')) / 86400); 


Which shows total days since my forums went offline, and I can use the variable in the header. What I want is something similar, but to count days until a specified date (XX days remaining until our second anniversary type of thing).

Is there any way to make it a small variable to use in the header as well?

Thanks.

[Boofo]

There's a java script hack on here for vB2 called the countdown hack (or something similar) that will do what you want. Just do a search for countdown.

[Mijae]

Originally Posted by Boofo

There's a java script hack on here for vB2 called the countdown hack (or something similar) that will do what you want. Just do a search for countdown.

Ok.

I just read what I wrote, and I ment to say that the code above shows days since the forums went online :P

I'll go search now.

[Mijae]

Originally Posted by Mijae

Ok.

I just read what I wrote, and I ment to say that the code above shows days since the forums went online :P

I'll go search now.

Found some hacks but they are too much for what I want.

I just want a simple php command that will print how many days until a specified date. That simple.

[NTLDR]

PHP Code:

// ###################### Start fetch_countdown_timer ######################
// returns the time left untill $dateline
function fetch_countdown_timer($dateline) {

    global $vboptions;

    $diff = $dateline - (time() - $vboptions['hourdiff']);

    $days = ($diff - ($diff % 86400)) / 86400;
    $diff = $diff - ($days * 86400);
    $hours = ($diff - ($diff % 3600)) / 3600;
    $diff = $diff - ($hours * 3600);
    $minutes = ($diff - ($diff % 60)) / 60;
    $diff = $diff - ($minutes * 60);
    $seconds = ($diff - ($diff % 1)) / 1;

    return iif($days > 0, $days.'d ', '').iif("$hours > 0 && $days != 0", $hours.'h ', '').$minutes.'m';

} 


Add that to functions.php somewhere sensible, like before the end "downloaded" comment then do something like:

PHP Code:

$futureevent = fetch_countdown_timer(strtotime('June 24, 2002')); 


Then $futureevent will contain a string like '5d 21h 32m'.

[Boofo]

How would you do a function for time since like this?

[Mijae]

Nice code.

But could you rewrite that so it will only say how many days are left?

I dont care about hours, minutes, seconds...I just want days

[Boofo]

Just comment this out:

PHP Code:

   $hours = ($diff - ($diff % 3600)) / 3600; 
    $diff = $diff - ($hours * 3600); 
    $minutes = ($diff - ($diff % 60)) / 60; 
    $diff = $diff - ($minutes * 60); 
    $seconds = ($diff - ($diff % 1)) / 1; 


and change this:

PHP Code:

   return iif($days > 0, $days.'d ', '').iif("$hours > 0 && $days != 0", $hours.'h ', '').$minutes.'m'; 


to this:

PHP Code:

   return iif($days > 0, $days.'d ', ''); 


[Mijae]

You know what,

PHP Code:

$anniversary = floor(365 - (time() - strtotime('June 24, 2003')) / 86400); 


Works like a charm. The only thing is that it tells me there are 79 days left, so does this year have 366 days?

[NTLDR]

Yes, 2004 is a leap year