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php subtract days from string date

Posted on 2009-04-01
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Last Modified: 2013-12-12
i have a string, in european date format like this: 24/11/2009
i would like to subtract 28 days from the date and as output get a string variable with value 27/10/2009 (output in this format exactly)

i tried smth like the code below, but it does not work, could you please help me get the code right?
date_default_timezone_set('Europe/London');

$mudate = new DateTime("24/11/2009");

date_sub($mudate, new DateInterval("P28D"));

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Question by:KristjanLaane
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Expert Comment

by:Rurne
ID: 24041068
Several things.

1.  Line 2 can fail, depending on your host.  DateTime expects a string as strtotime() can parse; according to the documentation, that's US English-style, so your switching of days and months might cause some problems.
2.  DateTime::sub doesn't actually modify the original DateTime object, but instead returns a new copy:
<?php

$mudate = new DateTime('2009-11-24', 'Europe/London');

$muinterval = new DateInterval('P28D');

$mudate->sub($muinterval);

echo $mudate->format('d/m/Y')."\n";    // prints '24/11/2009';

$newmudate = $mudate->sub($muinterval);

echo $newmudate->format('d/m/Y')."\n"; // prints '27/11/2009';

?>

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Assisted Solution

by:pghn
pghn earned 100 total points
ID: 24041179
Try something like this:

(Since strtotime() I've found to be very good and reliable on parsing any date formate and coming up with the proper timestamp.)
$mudate = new DateTime("24/11/2009");

$newtimestamp = strtotime("-28 days",strtotime($mudate));

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

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Expert Comment

by:pghn
ID: 24041192
Just a correction since you want in the format d/m/Y;

Line 3 it should be:
$newtimestamp = date("d/m/Y");

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Author Comment

by:KristjanLaane
ID: 24061483
the second line of Rurne code and the first line of pghn code causes an error, so the subsequent lines are of no use at the moment. i do not seem to be able to declare a date, do you have ideas how i could get around that?

and not even the following works:


$mudate = date_create('2008-08-03 14:52:10');

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Author Comment

by:KristjanLaane
ID: 24061513
might it be because my server is PHP Version 5.0.3 ? I cannot update the version at the moment as im not admin ...
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Accepted Solution

by:
Rurne earned 400 total points
ID: 24061947
Yes.  DateInterval is not available for versions before 5.3.0.  Your alternative is to work something around strtotime(), get the timestamp, and manipulate it directly.  The below code is a fairly verbose example:
// setting up constants for explanation purposes only

define("SECONDS_PER_MINUTE", 60);

define("MINUTES_PER_HOUR", 60);

define("HOURS_PER_DAY", 24);
 

$day_interval = 28;  // aka DateInterval('P28D');
 

$beginning_timestamp = strtotime('2009-11-24');

$interval = $day_interval * SECONDS_PER_MINUTE * MINUTES_PER_HOUR * HOURS_PER_DAY;

$new_timestamp = $beginning_timestamp - $interval;

echo date('m/d/Y', $new_timestamp)."\n";  // prints "10/27/2009"

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Author Comment

by:KristjanLaane
ID: 24064669
thanks, this works great ! strtotime seems to have read the date in in US format, so i changed my initial requirement of european data input and converted it to US date format first.
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