Very simple request, not even a hack :P

[Mijae]

I have this on my page:

$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? :D

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]

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]

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]

// ###################### 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:


$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 :D

[Boofo]

Just comment this out:

$hours = ($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:

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

to this:

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

[Mijae]

You know what,


$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 ;)

[Boofo]

I would like to have the code like above that countsdown the days hours and miutes, too. Can a time since be done like it is above?

[Boofo]

Lee, will your code account for leap years?

[NTLDR]

It should work fine as it doesn't deal with years, just days, hours and minutes :)

[Boofo]

But wouldn't it have to take Feb, 29th into account as a day on a leap year?

Which I supoose is a moot point as Feb 29th is gone and won't be here for another 4 yeras. ;)

[NTLDR]

PHP should be clever enough to know if I give it a dateline for say 1 March 2004 that the day before it is 29 Feb 2004 ;)

[Boofo]

Oh, ok, I didn't know PHP was able to figure that out. I guess I'm still used to the older versions of Turbo Pascal's way of doing things. ;)