How can we know the number of days between two given dates
using PHP?
- 8 Answers
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Answer / ash
$date1 = date('Y-m-d');
$date2 = "2007-07-05";
$days = (strtotime($date1) - strtotime($date2)) / (60 * 60
* 24);
echo " No of $days difference";
| Is This Answer Correct ? | 25 Yes | 4 No |
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
| Is This Answer Correct ? | 6 Yes | 0 No |
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";
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / dheeraj
<?php
$date1=date('Y-m-d');
$date2="2007-09-15";
echo (round($diff=strtotime($date1)-strtotime($date2)))/86400;
?>
| Is This Answer Correct ? | 5 Yes | 1 No |
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
| Is This Answer Correct ? | 5 Yes | 2 No |
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);
| Is This Answer Correct ? | 3 Yes | 2 No |
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