I am trying to add a certain amount of days to a set date in PHP. However, all of the code I use is not working. Here is the code I am currently experiencing problems with:
echo date("2013-12-01", strtotime("+7 days"));
I wanting to add 7 days to the date above. When I echo out this code, it just prints '2013-12-01'. Is there a way to do this?
Thanks
For the sake of completeness, here's how you do it with DateTime():
$datetime = new DateTime("2013-12-01");
$datetime->add(new DateInterval('P7D'));
echo $datetime->format('Y-m-d');
or
$datetime = new DateTime("2013-12-01");
$datetime->modify('+7 days');
echo $datetime->format('Y-m-d');
You can use date_add() function:
$date = date_create('2013-12-01');
date_add($date, date_interval_create_from_date_string('7 days'));
echo date_format($date, 'Y-m-d');
This will output 2013-12-08
It must be like this:
$NewDate = date('Y-m-d', strtotime("2013-12-01" . " +7 days"));
echo $NewDate;
Related
Is there a way of calculating the EXACT date after/before 6 months in Php taking into consideration the dynamic change in the days of the month i.e. consider some months have 30 days and some have 31 and 28 of course. I know it can be done with MySQL but I want to know if there is an option with Php as well.
Thanks in advance.
you need to date_create, date_interval_create_from_date_string and date_format to complete the job.
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('6 months'));
echo date_format($date, 'Y-m-d'); // 2000-07-01
With Object Oriented:
Adding
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P6M'));
echo $date->format('Y-m-d') . "\n"; // 2000-07-01
Subtracting
$date = new DateTime('2000-01-01');
$date->sub(new DateInterval('P6M'));
echo $date->format('Y-m-d') . "\n"; // 1999-07-01
get date before 6 month from current month:
echo date("F 1, Y", strtotime("-6 months"));
I want to calculate EXACT past 30 days time period in php from now (for example 30 aug 14 23:06) to 30 days back (for example 1 aug 14 23:06). I wrote this where current datetime goes in $d1 and past 30 days datetime goes in $d2 but somehow i am not getting correct results. Any idea?
$url=$row["url"];
$pageid=getPageID($url);
$date=date('y-m-d g:i');
$d1=strtotime($date);
$d2=date(strtotime('today - 30 days'));
Thanks
The problem is likely caused by the malformed date() call. The first argument passed to date() should be the format (as shown in the Docs) and the second should be an optional timestamp.
Try this:
$d2 = date('c', strtotime('-30 days'));
PHPFiddle
As a short aside, the whole snippet can be simplified as follows:
$url = $row["url"];
$pageid = getPageID($url);
$date = date('y-m-d g:i');
$d1 = time();
$d2 = date('y-m-d g:i', strtotime('-30 days'));
You can also use the DateTime class's sub() method together with an DateInterval:
$now = new DateTime();
$back = $now->sub(DateInterval::createFromDateString('30 days'));
echo $back->format('y-m-d g:i');
if you would like to get out put as 2014-08-01 then try the below code. thanks
$date = '2014-08-30 23:06';
$new_date = date('Y-m-d G:i', strtotime($date.' - 29 days'));
echo "30 days back is " . $new_date;
From your brief description and example given, I believe that you want the date to be 30 days back and time to be the same as of now. The below code will serve this purpose. Thanks.
<?php
$date=date('y-m-d g:i');
$time=date('g:i');
echo "Todays date:" . $date. "<br>";
$d2 = date('y-m-d', strtotime('-30 days'));
echo "30 days back:" . $d2 . ' ' .$time;
?>
Try:
echo date("Y-m-d h:i:s",strtotime('-30 days'));
For more detail click here
Very simple two lines of code
$date = new DateTime();
echo $date->modify('-30 day')->format('y-m-d g:i');
I know you said with PHP, however, I can't imagine not getting the records from a DB. If you want to do so from the DB,use:
$sql='SELECT * FROM myTable WHERE date > CURRENT_DATE - INTERVAL 30 DAY';
$pdo->query($sql);
Very simple one lines of code:
echo (new DateTime())->modify('-30 day')->format('y-m-d g:i');
In the example below, it makes no sense if the variable $date is not
used anywhere else!
$date = new DateTime();
echo $date->modify('-30 day')->format('y-m-d g:i');
Sample answer is
$dateBack30Days=date('Y-m-d g:i', strtotime('-30 days'));
I had gone through various stackoverflow solutions and other blogs but still it doesn't fix my problem.
Let's say that the date today is: 2013-12-28 and I want to get the date after 1 month and it is supposed to display 2014-01-28.
$date = date('o-m-d');
$final = date('o-m-d', strtotime("+1 month", $date));
echo $final;
Above is my code. It returns 02/01/1970.
I have also tried the mktime method but still it displays the 1970 output.
What am I doing wrong?
BTW. I am working this on a hosted server.
Thanks ahead. :)
Use DateTime function modify
$date = new DateTime( 'o-m-d' );
echo $date->modify( '+1 month' )->format('o-m-d');
If you want the current date +1 month use:
$final = date('o-m-d', strtotime("+1 month"));
Or with a given date:
$date = date('o-m-d');
$final = date('o-m-d', strtotime($date . " +1 month"));
echo $final;
If you want to use the second parameter of strtotime it has to be a timestamp.
Go the OOP way..
<?php
$date = new DateTime('2013-12-28');
$date->add(new DateInterval('P1M'));
echo $date->format('Y-m-d'); //prints 2014-01-28
<?php
$date = "04-15-2013";
$date = strtotime($date);
$date = strtotime("+1 day", $date);
echo date('m-d-Y', $date);
?>
This is driving me crazy and seems so simple. I'm pretty new to PHP, but I can't figure this out. The echo returns 01-01-1970.
The $date will be coming from a POST in the format m-d-Y, I need to add one day and have it as a new variable to be used later.
Do I have to convert $date to Y-m-d, add 1 day, then convert back to m-d-Y?
Would I be better off learning how to use DateTime?
there you go
$date = "04-15-2013";
$date1 = str_replace('-', '/', $date);
$tomorrow = date('m-d-Y',strtotime($date1 . "+1 days"));
echo $tomorrow;
this will output
04-16-2013
Documentation for both function
date
strtotime
$date = DateTime::createFromFormat('m-d-Y', '04-15-2013');
$date->modify('+1 day');
echo $date->format('m-d-Y');
See it in action
Or in PHP 5.4+
echo (DateTime::createFromFormat('m-d-Y', '04-15-2013'))->modify('+1 day')->format('m-d-Y');
reference
DateTime::createFromFormat()
$date = strtotime("+1 day");
echo date('m-d-y',$date);
use http://www.php.net/manual/en/datetime.add.php like
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('1 days'));
echo date_format($date, 'Y-m-d');
output
2000-01-2
The format you've used is not recognized by strtotime(). Replace
$date = "04-15-2013";
by
$date = "04/15/2013";
Or if you want to use - then use the following line with the year in front:
$date = "2013-04-15";
Actually I wanted same alike thing,
To get one year backward date, for a given date! :-)
With the hint of above answer from #mohammad mohsenipur
I got to the following link, via his given link!
Luckily, there is a method same as date_add method, named date_sub method! :-)
I do the following to get done what I wanted!
$date = date_create('2000-01-01');
date_sub($date, date_interval_create_from_date_string('1 years'));
echo date_format($date, 'Y-m-d');
Hopes this answer will help somebody too! :-)
Good luck guys!
I have a variable called $effectiveDate containing the date 2012-03-26.
I am trying to add three months to this date and have been unsuccessful at it.
Here is what I have tried:
$effectiveDate = strtotime("+3 months", strtotime($effectiveDate));
and
$effectiveDate = strtotime(date("Y-m-d", strtotime($effectiveDate)) . "+3 months");
What am I doing wrong? Neither piece of code worked.
Change it to this will give you the expected format:
$effectiveDate = date('Y-m-d', strtotime("+3 months", strtotime($effectiveDate)));
This answer is not exactly to this question. But I will add this since this question still searchable for how to add/deduct period from date.
$date = new DateTime('now');
$date->modify('+3 month'); // or you can use '-90 day' for deduct
$date = $date->format('Y-m-d h:i:s');
echo $date;
I assume by "didn't work" you mean that it's giving you a timestamp instead of the formatted date, because you were doing it correctly:
$effectiveDate = strtotime("+3 months", strtotime($effectiveDate)); // returns timestamp
echo date('Y-m-d',$effectiveDate); // formatted version
You need to convert the date into a readable value. You may use strftime() or date().
Try this:
$effectiveDate = strtotime("+3 months", strtotime($effectiveDate));
$effectiveDate = strftime ( '%Y-%m-%d' , $effectiveDate );
echo $effectiveDate;
This should work. I like using strftime better as it can be used for localization you might want to try it.
Tchoupi's answer can be made a tad less verbose by concatenating the argument for strtotime() as follows:
$effectiveDate = date('Y-m-d', strtotime($effectiveDate . "+3 months") );
(This relies on magic implementation details, but you can always go have a look at them if you're rightly mistrustful.)
The following should work,Please Try this:
$effectiveDate = strtotime("+1 months", strtotime(date("y-m-d")));
echo $time = date("y/m/d", $effectiveDate);
Following should work
$d = strtotime("+1 months",strtotime("2015-05-25"));
echo date("Y-m-d",$d); // This will print **2015-06-25**
Add nth Days, months and years
$n = 2;
for ($i = 0; $i <= $n; $i++){
$d = strtotime("$i days");
$x = strtotime("$i month");
$y = strtotime("$i year");
echo "Dates : ".$dates = date('d M Y', "+$d days");
echo "<br>";
echo "Months : ".$months = date('M Y', "+$x months");
echo '<br>';
echo "Years : ".$years = date('Y', "+$y years");
echo '<br>';
}
As of PHP 5.3, DateTime along with DateInterval could be a feasible option to achieve the desired result.
$months = 6;
$currentDate = new DateTime();
$newDate = $currentDate->add(new DateInterval('P'.$months.'M'));
echo $newDate->format('Y-m-d');
If you want to subtract time from a date, instead of add, use sub.
Here are more examples on how to use DateInterval:
$interval = new DateInterval('P1Y2M3DT4H5M6S');
// This creates an interval of 1 year, 2 months, 3 days, 4 hours, 5 minutes, and 6 seconds.
$interval = new DateInterval('P2W');
// This creates an interval of 2 weeks (which is equivalent to 14 days).
$interval = new DateInterval('PT1H30M');
// This creates an interval of 1 hour and 30 minutes (but no days or years, etc.).
The following should work, but you may need to change the format:
echo date('l F jS, Y (m-d-Y)', strtotime('+3 months', strtotime($DateToAdjust)));
public function getCurrentDate(){
return date("Y-m-d H:i:s");
}
public function getNextDateAfterMonth($date1,$monthNumber){
return date('Y-m-d H:i:s', strtotime("+".$monthNumber." months", strtotime($date1)));
}