How to handle the dates greater than "2057-12-31"

[jeeva]

Hi,

i want find the [b]number of days[/b] between the dates "2045-12-31 and 2047-12-31".

i have tried with so many date functions.  but i am not able to find the solution.

suggestions please................


jeeva

[Hypnos]

The built in PHP date functions won't go past 2038.

http://phplens.com/phpeverywhere/adodb_date_library

[sasa]

try[code]<?php
function in_day($a){
$monts=array(0,31,28,31,30,31,30,31,31,30,31,30,31);
$a = explode('-',$a);
if (($a[0] % 400 == 0) or ($a[0] % 4 == 0 and $a[0] % 100 != 0))  $monts[2] = 29;
$y = $a[0] -1;
$out = $y *365 + (int) ($y / 4) - (int) ($y / 100) + (int) ($y / 400);
for ($i=0;$i<$a[1];$i++) $out += $monts[$i];
$out += $a[2] - 1;
return $out;
}

$date1='2045-12-31';
$date2='2047-12-31';
echo in_day($date2) - in_day($date1);

?>[/code]

[Jessica]

sasa: Does that check for leap years?

[kenrbnsn]

Yes it does, in this line:
[code]<?php
if (($a[0] % 400 == 0) or ($a[0] % 4 == 0 and $a[0] % 100 != 0))  $monts[2] = 29;
?>[/code]

Ken

[Ninjakreborn]

Nice function.

[Jessica]

I wasn't sure if that's what that was - I missed the 29 part, that makes it obvious ;) thanks Ken

[subbukumararaja]

Hi jeeva try this one,

it will help u


<?php
function in_day($a){
$monts=array(0,31,28,31,30,31,30,31,31,30,31,30,31);
$a = explode('-',$a);
if (($a[0] % 400 == 0) or ($a[0] % 4 == 0 and $a[0] % 100 != 0))  $monts[2] = 29;
$y = $a[0] -1;
$noOfDays = $y *365 + (int) ($y / 4) - (int) ($y / 100) + (int) ($y / 400);
for ($i=0;$i<$a[1];$i++) $noOfDays += $monts[$i];
$noOfDays += $a[2] - 1;
return $noOfDays;
}

$date1='2045-12-31';
$date2='2047-12-31';
echo in_day($date2) - in_day($date1);

?>