I don't know how to explain this correctly but just some sample for you guys so that you can really get what Im trying to say.
Today is April 09, 2010
7 days from now is April 16,2010
Im looking for a php code, which can give me the exact date giving the number of days interval prior to the current date.
I've been looking for a thread which can solve or even give a hint on how to solve this one but I found none.
If you are using PHP >= 5.2, I strongly suggest you use the new DateTime object, which makes working with dates a lot easier:
<?php
$date = new DateTime("2006-12-12");
$date->modify("+7 day");
echo $date->format("Y-m-d");
?>
Take a look here - http://php.net/manual/en/function.strtotime.php
<?php
// This is what you need for future date from now.
echo date('Y-m-d H:i:s', strtotime("+7 day"));
// This is what you need for future date from specific date.
echo date('Y-m-d H:i:s', strtotime('01/01/2010 +7 day'));
?>
The accepted answer is not wrong but not the best solution:
The DateTime class takes an optional string in the constructor, which can define the same logic as the modify method.
<?php
$date = new DateTime("+7 day");
For example:
<?php
namespace DateTimeExample;
$now = new \DateTime("now");
$inOneWeek = new \DateTime("+7 day");
printf("Now it's the %s", $now->format('Y-m-d'));
printf("In one week it's the %s", $inOneWeek->format('Y-m-d'));
For a list of available relative formats (for the DateTime constructor) take a look at http://php.net/manual/de/datetime.formats.relative.php
If you are using PHP >= 5.3, this could be an option.
<?php
$date = new DateTime( "2006-12-12" );
$date->add( new DateInterval( "P7D" ) );
?>
You will have to look into strtotime(). I'd imagine your final code would look something like this:
$future_date = "April 16,2010";
$seconds = strtotime($future_date) - time();
$days = $seconds /(60 * 60* 24);
echo $days; //Returns "6.0212962962963"
You can use mktime with date. (http://php.net/manual/en/function.date.php)
Date gives you the current date. This is better than simply adding/subtracting to a timestamp since it can take into account daylight savings time.
<?php
# this gets you 7 days earlier than the current date
$lastWeek = mktime(0, 0, 0, date("m") , date("d")-7, date("Y"));
# now pretty-print it out (eg, prints April 2, 2010.)
echo date("F j, Y.", $lastWeek), "\n";
?>
Related
I am working on a task where I need the user to provide exact number of months it will take them to complete a task and then i need to convert that number to an exact date, so lets suppose a user enters 6, this should give me a date 6 months from now.
I tried the following code looking at different examples on line but I have a feeling the following examples treats the $monthNum as the actual month of a year rather than what I need it to do.
$monthNum = 5;
$monthName = date("F", mktime(0, 0, 0, $monthNum, 10));
echo $monthName;
I will really appreciate any assistance here.
Demo here
Pop in your month in the modify() method.
$monthNum = 6;
$date = new DateTime();
$date->modify(" +{$monthNum} month");
echo $date->format("Y-m-d");
Outputs
2015-05-14
You can try:
$time = new \DateTime('+5 months');
You can use strtotime:
$date = date("Y-m-d", strtotime("+5 months"));
in php 5.4+
echo (new DateTime())->modify('+6 months')->format('d M Y');
Assume the date is:
$date = "2011-08-28";
It need to calculate 3 months previous to $date - how can that be done?
$new_timestamp = strtotime('-3 months', strtotime($date));
You can then use the new timestamp with the php date() function to display the new date how you wish, for example:
echo date("Y-m-d",$new_timestamp);
For me this way is much better because the code is more readable.
$datetime = Datetime::createFromFormat('Y-m-d', "2011-08-28");
$datetime->modify('-3 months');
echo $datetime->format('Y-m-d');
edit: I'm an idiot. You could do the above as follows
$datetime = new Datetime("2011-08-28");
$datetime->modify('-3 months');
echo $datetime->format('Y-m-d');
edit: As pointed out by calumbrodie, you can use the sub method instead of inverting the interval and adding it to the date object
I was trying to do something similar to the original question. I needed to subtract 1 day from a DateTime object. I know the other solutions work, but here's another way I liked better. I used this function:
function getPreviousDay($dateObject){
$interval = new DateInterval('P1D');
$dateObject->sub($interval);
return $dateObject;
}
$dateObject is a DateTime object that can have any date you wan't, but as I wanted the current date, I wrote:
$dateObject = new DateTime('now');
What my function does, is subtract 1 day from the date it receives, but you can modify it so it subtracts 3 months modifying the DateInterval constructor like this:
$interval = new DateInterval('P3M');
$dateObject->sub($interval);
return $dateObject;
You can find the explanation on the string format used in the DateInterval constructor here
DateInterval constructor documentation
There you'll se that letter 'P' (period) is mandatory, the you use an int to specify period length and then specify the period type, in the case of months 'M'. It looks as if there was an error in the documentation, it says that "M" is used for months and minutes, but it actually works for months. If you need to use minutes, you must specify "PTM", for example "PT3M" for 3 minutes, outside the table it says you must use the "T" identifier for time intervals.
edit:
To give a complete example, you have to use this format for a full date time interval:
$format = 'P1Y2M4DT2H1M40S';
This will represent an interval of 1 year, 2 months, 4 days, 2 hours,1 minute and 40 seconds
Hope this helps someone
<?php
$threemonthsago = mktime(0, 0, 0, date("m")-3, date("d"), date("Y"));
?>
<?php
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('10 days'));
echo date_format($date, 'Y-m-d');
?>
I want to get the date with specific day and time in PHP, like i want the date of next day and time of 9.30 am i.e "2011-06-02 09:30:00".
the code i was using get to do that,
<?php
$next_day_date = date("Y")."-".date("m")."-".(date("d")+1)." 09:30:00";
$new_trig_time_stamp = strtotime($next_day_date);
$trigger_date_time = date("Y-m-d H:i:s",$new_trig_time_stamp);
echo $trigger_date_time;
?>
the code above works fine but fails on 31 day, on 31st it returns "1970-01-01 05:30:00".
Is there any other way to do so.
When shifting dates by a fixed number, it's better to use mktime(), because it handles invalid dates well (e.g. it knows that January 32 is in fact February 1)
$trigger_date_time = date("Y-m-d H:i:s", mktime(9,30,0, date('n'), date('j')+1, date('Y'));
strtotime() is very useful here.
$trigger_date_time = date( "Y-m-d H:i:s", strtotime( "tomorrow 9:30" ) );
Calculate it via unix timestamp - much less annoyance
<?php
$trigger_date_time = date("Y-m-d 09:30:00",time() + 60*60*24);
echo $trigger_date_time;
?>
echo date('Y-m-d', strtotime('+1 day')) . ' 09:30:30';
Have a look at date_add.
In more detail, something like...
$myDate = new DateTime("Some time");
$myDate->add(new DateInterval("P1D"));
You can then use $myDate->format(…) to extract formatted representations.
add a day to date, so I can store tomorrow's date in a variable.
$tomorrow = date("Y-m-d")+86400;
I forgot.
date returns a string, whereas you want to be adding 86400 seconds to the timestamp. I think you're looking for this:
$tomorrow = date("Y-m-d", time() + 86400);
I'd encourage you to explore the PHP 5.3 DateTime class. It makes dates and times far easier to work with:
$tomorrow = new DateTime('tomorrow');
// e.g. echo 2010-10-13
echo $tomorrow->format('d-m-Y');
Furthermore, you can use the + 1 day syntax with any date:
$xmasDay = new DateTime('2010-12-24 + 1 day');
echo $xmasDay->format('Y-m-d'); // 2010-12-25
date() returns a string, so adding an integer to it is no good.
First build your tomorrow timestamp, using strtotime to be not only clean but more accurate (see Pekka's comment):
$tomorrow_timestamp = strtotime("+ 1 day");
Then, use it as the second argument for your date call:
$tomorrow_date = date("Y-m-d", $tomorrow_timestamp);
Or, if you're in a super-compact mood, that can all be pushed down into
$tomorrow = date("Y-m-d", strtotime("+ 1 day"));
Nice and obvious:
$tomorrow = strtotime('tomorrow');
You can use the add method datetime class.
Eg, you want to add one day to current date and time.
$today = new DateTime();
$today->add(new DateInterval('P1D'));
Further reference php datetime add
Hope this helps.
I find mktime() most useful for this sort of thing. E.g.:
$tomorrow=date("Y-m-d", mktime(0, 0, 0, date("m"), date("d")+1, date("Y")));
Given a time, how can I find the time one month ago.
strtotime( '-1 month', $timestamp );
http://php.net/manual/en/function.strtotime.php
In php you can use strtotime("-1 month"). Check out the documentation here: http://ca3.php.net/strtotime
We can achieve same by using PHP's modern date handling. This will require PHP 5.2 or better.
// say its "2015-11-17 03:27:22"
$dtTm = new DateTime('-1 MONTH', new DateTimeZone('America/Los_Angeles')); // first argument uses strtotime parsing
echo $dtTm->format('Y-m-d H:i:s'); // "2015-10-17 03:27:22"
Hope this adds some more info for this question.
<?php
$date = new DateTime("18-July-2008 16:30:30");
echo $date->format("d-m-Y H:i:s").'<br />';
date_sub($date, new DateInterval("P1M"));
echo '<br />'.$date->format("d-m-Y").' : 1 Month';
?>
PHP 5.2=<
$date = new DateTime(); // Return Datetime object for current time
$date->modify('-1 month'); // Modify to deduct a month (Also can use '+1 day', '-2 day', ..etc)
echo $date->format('Y-m-d'); // To set the format
Ref: http://php.net/manual/en/datetime.modify.php
This code is for getting 1 month before not 30 days
$date = "2016-03-31";
$days = date("t", strtotime($date));
echo date("Y-m-d", strtotime( "-$days days", strtotime($date) ));
These answers were driving me nuts. You can't subtract 31 days and have a sane result without skipping short months.
I'm presuming you only care about the month, not the day of the month, for a case like filtering/grouping things by year and month.
I do something like this:
$current_ym = date('ym',strtotime("-15 days",$ts));