SRL Archive - Web Development

Ok… pretend the account timer was made…. (m-d-Y format)
03-20-2009
if the account has been made for more then 30 days… then have a function return the word "done" … if its not been 30 days… have it return the remaining days till 30 days is up.. is that too hard to ask? :redface:
and can it read the dates from a array please? like
"Timer" => "03-20-2009"
"Hexinist" => "03-21-2009"

Code pl0x? I've tried and failed.. thanks in advance!

<?php
  $arr["Timer"] = "03-20-2009";
  $arr["Hexinist"] = "03-21-2009";
  $arr["nielsie95"] = "02-21-2009";

  foreach ($arr as $name => $d) {
    $ardate = explode("-", $d);
    $tdate = mktime(0, 0, 0, $ardate[0], $ardate[1], $ardate[2]);
    $togo = 30 - floor((time() - $tdate) / (60 * 60 * 24));
  
    echo "$name => ".(($togo <= 0) ? "done" : "$togo days to go")."<br \>\n";
  }
?>

wow looks like PHP tags need some work.

From: Bobarkinator
wow looks like PHP tags need some work.

Blame GeSHi, but I recon I can fix it myself.

From: Wizzup?
Blame GeSHi, but I recon I can fix it myself.

Weird, I thought GeShi was pretty stable, no?

From: nielsie95

<?php
  $arr["Timer"] = "03-20-2009";
  $arr["Hexinist"] = "03-21-2009";
  $arr["nielsie95"] = "02-21-2009";

  foreach ($arr as $name => $d) {
    $ardate = explode("-", $d);
    $tdate = mktime(0, 0, 0, $ardate[0], $ardate[1], $ardate[2]);
    $togo = 30 - floor((time() - $tdate) / (60 * 60 * 24));
  
    echo "$name => ".(($togo <= 0) ? "done" : "$togo days to go")."<br \>\n";
  }
?>

Thanks! :D

Edit..
I modified it a bit..
Now I'm getting..

Warning: mktime() expects parameter 4 to be long, string given in /home/mgate3/public_html/timer/bypass/includes.php on line 7

from this code

<?php
$arr["Timer"] = "03-20-2009";
$arr["Hexinist"] = "03-21-2009";

function DaysLeft($name) {
$ardate = explode("-", $arr[$name]);
$tdate = mktime(0, 0, 0, $ardate[0], $ardate[1], $ardate[2]);
$togo = 30 - floor((time() - $tdate) / (60 * 60 * 24));

if (togo <= 0) {
return "done";
exit;
} else {
return $togo;
}
}
echo DaysLeft("Timer");
?>

You can't access the array inside the function..

function DaysLeft($name, $arr) {
$ardate = explode("-", $arr[$name]);
$tdate = mktime(0, 0, 0, $ardate[0], $ardate[1], $ardate[2]);
$togo = 30 - floor((time() - $tdate) / (60 * 60 * 24));

if (togo <= 0) {
return "done";
exit;
} else {
return $togo;
}
}
echo DaysLeft("Timer", $arr);

oh, lol, thanks! :D
EDIT -
Now.. its really messed up.. no matter what, it returns done..

[PHP]<?php
$arr["Timer"] = "03-10-2009";
$arr["Hexinist"] = "03-21-2009";

function DaysLeft($name, $arr) {

$ardate = explode("-", $arr[$name]);
$tdate = mktime(0, 0, 0, $ardate[0], $ardate[1], $ardate[2]);
$togo = 30 - floor((time() - $tdate) / (60 * 60 * 24));

if (togo <= 0) {
return "done";
} else {
return $togo;
}
}

echo DaysLeft("Timer", $arr);
?>[/PHP]

How long did you try it for yourself..? :\

if ($togo <= 0) {

lol, wow.. i totally didn't notice that… :o