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Month & Year Countdown (Date Time Comparison)

Posted on 2003-11-13
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Last Modified: 2008-02-01
Dear all Experts,

I can compare the hour by using the following line,
$hourDiff = (mktime(0,0,0,$month,$day,$year) - time())/3600;

in order to get day different,
$dayDiff = (mktime(0,0,0,$month,$day,$year) - time())/86400;

BUT,
      How do i get Month and Year Different? For example i need to know how many months from 01-01-2003 to 31-05-2003?
      $monthDiff = (mktime(0,0,0,$month,$day,$year) - time())/2678400; <---- This will only works for month that having 31 days (60sec*60mins*24hour*31days), what about this, (int)$endMonth - $startMonth?


Pls Advice... Thanks in Advance! :)



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Question by:paulsiew
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6 Comments
 

Author Comment

by:paulsiew
ID: 9737927
$endMonth = date("n",time());
$startMonth = (int)$month;

$monthDiff = $endMonth - $startMonth


how bout this? Pls Advice...
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Accepted Solution

by:
lozloz earned 100 total points
ID: 9738032
hi,

for years you could just do:

$endyear = date("Y", time());
$startyear = $year;
if($endyear > $startyear) {
  $difference = $endyear - $startyear;
}

where both are 4 digit numbers

for months:

$endmultiple = ($endyear - 1970) * 12;
$startmultiple = ($startyear - 1970) * 12;
$endmonth = date("n", time()) + $endmultiple;
$startmonth = $month * ($year + 1) + startmultiple;
$monthdifference = $endmonth - $startmonth;

cheers,

loz
0
 

Author Comment

by:paulsiew
ID: 9738095
what about this?

                        $yearNow           = date("Y",time());
                        $yearCreated = (int)$year;
                        $yearDiff           = $yearNow - $yearCreated;
                                                
                        if ($yearDiff > 0) {
                              $xtraMonth = $yearDiff * 12;
                        }
                                                
                        $monthNow      = date("n",time());
                        $monthCreated = (int)$month;
                        $monthDiff       = ($monthNow - $monthCreated) + $xtraMonth;
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LVL 13

Expert Comment

by:lozloz
ID: 9738277
hi,

i don't get what this int thing is at the front of month and year?

yes that code should work (better than mine)

loz
0
 

Author Comment

by:paulsiew
ID: 9744434
i'm casting the value into integer type

ref --> http://www.phpnoise.com/tutorials/22/5
0
 
LVL 13

Expert Comment

by:lozloz
ID: 9744827
ok, i don't think they're particularly necessary but there's no real downside to them - have you tested the code? did it work?

loz
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