days between two dates

i need difference between two date in day format . i my excel month/day/year format for example i) 4/25/2011 in one column & in another column 5/30/2011 , how to calculate in excel format ONLY

function dateDiff($start, $end) {

  $start_ts = strtotime($start);

  $end_ts = strtotime($end);

  $diff = $end_ts - $start_ts;

  return round($diff / 86400);

}

echo dateDiff("2011-03-07", "2011-06-06").'days';

$days = (strtotime(date("Y-m-d")) - strtotime("2005-11-20")) / (60 * 60 * 24);

print $days;

either use abs(); or in the order shown above because the result is negative.

Slavi

i want simple function which return no of days between two given days

any help will appreciable

Thank u Alexandar. You saved me much time !

This is a great job Jackson . Thanks Jackson

i want to calculate for two different date.
the start_date i name it c_date(state for $commencement_date). the second one is $today. the result i name the variable is $service_year. but somebody told me, that my wrong is in i state that one year as 365 days, but not really are( for leap year is 366 right) so, how can i fix this problem?

$c_date = date('d-m-Y',strtotime($row1['commencement_date']));  



$today = date ("d-m-Y");



$service_year = round(dateDiff("-", date("d-m-Y",time()), $c_date)/365,2);







(this is the function for convert year - year, month and date)

function year2monthsNdays($years)

{

$array = explode(".",$years);

$year = $array[0];

$month = ($array[1]>9) ? $array[1]/100 : $array[1]/10;

if ($month) {

$days = round($month*365,2);

$daysArray = explode(".",$days);

$months = round($daysArray[0]/30,2);

$monthArray = explode(".",$months);

$monthInt = $monthArray[0];

$daysInt = round($monthArray[1]*30/100,1);

}

$a = "$year-$monthInt-$daysInt";



return $a;

}



$service_year_convert = year2monthsNdays($service_year);

my query gvs this ans

name date
sandy 01/jan/2010
avi 02/jan/2010
sandy 02/jan/2010
sats 03/jan/2010
nil 03/jan/2010
sandy 03/jan/2010
avi 03/jan/2010
avi 04/jan/2010
sandy 05/jan/2010

& i want ot print in this manner

name/date 01/jan/2010 02/jan/2010 03/jan/2010
avi 0 1 1
sandy 1 1 1
sats 0 0 1
nil 0 0 1

first of all thanks a lot.saves a lot of time

Oops. Should be <?php

nice work guys.. i found the solution.. thanks to all..

Thnx Friends this code works for me also im very happy do the best

thxn again

Hi! Can i ask for some help here....
I've tested the following script and it works.
However, when i use date in d/m/y format instead of m/d/y, thing goes wrong and the correct number of days can't be generated.
What can i do to get it right?
Need this in urgent...really hope to get some assistance here...Thanks in advance!

function dateDiff($startDate, $endDate)

{

// Parse dates for conversion

$startArry = date_parse($startDate);

$endArry = date_parse($endDate);



// Convert dates to Julian Days

$start_date = gregoriantojd($startArry["month"], $startArry["day"], $startArry["year"]);

$end_date = gregoriantojd($endArry["month"], $endArry["day"], $endArry["year"]);



// Return difference

return round(($end_date - $start_date), 0);

} 

first date diff tech.. is good .It works proper

Using your example of 20th May 2009 + 2 months...

$d = date('Y-m-d', mktime(0,0,0,5+2,20,2009));

The mktime function is clever enough to understand what happens if, for example, you were to add 2 months to November 2000.... it will be clever enough to understand that 11+2 = 13 which it knows isn't a month and of course knows it should be january. In this example it will also be clever enough to add 1 to the years, giving a return date of January 2001. Same goes for if you were to add 7 days to May the 31st, it will correctly return June 7th.

The mktime function also accepts variables and other functions, I will use 2 examples below to demonstrate...

// Using Variables

$day = 20;

$month = 5;

$year = 2009;

$d = date('Y-m-d', mktime(0,0,0,$month+2,$day,$year));



// Using Date Function To Add 2 To Current Date

$d = date('Y-m-d', mktime(0,0,0,date('m')+2,date('d'),date('Y')));

Finally, the 'Y-m-d' part can be changed to output a large number of configurations. The above returns [Full 4 Digit Year]-[Month With Leading Zero]-[Day With Leading Zero]. For a full list please see http://uk3.php.net/manual/en/function.date.php

Hope this helps!

I tried
$days = (strtotime("2005-11-20") - strtotime("2005-11-26")) / (60 * 60 * 24);

and I get a number that is not a whole number, it has several decimal places, how can i fix that?

use round
round(strtotime("2005-11-26")) / (60 * 60 * 24));

i want to find . number of saterday in the given range of dates ?

Alexandar / Veselin has the example backwards.

$days_proc_time = (strtotime(date("Y-m-d")) - strtotime("2005-11-20") ) / (60 * 60 * 24);

Thanks a lot!!! I spent so time looking for this 2 lines!

ok...
how to list all date between this 2 date ?

Thank you very much

arse error: syntax error, unexpected ';' in /home/craig/public_html/page/billing.php on line 1

this doesnt work .

this is the easiest wa to calculate number of days:

$days = (strtotime("2005-11-20") - strtotime(date("Y-m-d"))) / (60 * 60 * 24);

print $days;