Answer / ash

$date1 = date('Y-m-d');
$date2 = "2007-07-05";
$days = (strtotime($date1) - strtotime($date2)) / (60 * 60
* 24);
echo " No of $days difference";

Answer / richa

Something we must know about strtotime

(try putting up a date grater than 30 years)

yes it will give an unexpected output, because it uses, unix
timestamp, and thus doesnt treat a date greater than 30 yrs,
as a correct value.

Amazing but true

Answer / savvy

"The function expects to be given a string containing a US
English date format and will try to parse that format into a
Unix timestamp (the number of seconds since January 1 1970
00:00:00 GMT), relative to the timestamp given in now, or
the current time if none is supplied." --- in php.net So Be
careful when calculating with strtotime(). Therefore in the
above ans it might be like this way
$date1 = date('M-d-Y');
$date2 = "August-07-2007";

Answer / dheeraj

<?php
$date1=date('Y-m-d');
$date2="2007-09-15";
echo (round($diff=strtotime($date1)-strtotime($date2)))/86400;
?>

Answer / rameshfromind

There is one function for gregoriantojd that means convert
to gregorian calendar date to julian date count.

Below the example for counting days between two date.

<?
$startDate = '01-01-2008' // mm-dd-yyyy/yy
$endDate = '01-30-2008' // mm-dd-yyyy/yy
echo gregoriantojd($endDate) - $gregoriantojd($startDate);
?

// OUTPUT

29

Answer / pushpendra thakur

the first 1 is correct answer

Answer / 44

44

Answer / sachin

If we know the date format and separator used in dates,
we can separate them out into day, month and year.
e.g.

$d1 = '29-12-1985'; //dd-mm-yyyy
$d2 = '31-12-1985'; //dd-mm-yyyy

$d1_pieces = explode('-',$d1);
$d2_pieces = explode('-',$d2);

$timestamp_d1 =
mktime(0,0,0,$d1_pieces[1],$d1_pieces[0],$d1_pieces[2],);

$timestamp_d2 =
mktime(0,0,0,$d2_pieces[1],$d2_pieces[0],$d2_pieces[2],);

$diff = $d1-$d2;

$num_days = $diff/(24*60*60);

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