I have a date which is say like this
$given_date = '2014-12-25'; //Y-m-d format
Now i want to get the midnight timestamp of the given date, so I am doing this method
$midnight = strtotime(date('Y-m-d',$given_date).' 00:00:00');
Am I doing it right??
or I can use something like this?
$midnight = strtotime("midnight $given_date");
Which is better?
I would prefer a more OO approach instead of fiddling around with strings:
$date = new DateTime($given_date);
$date->setTime(0,0,0);
// echo $date->format('Y-m-d H:i:s');
echo $date->getTimestamp();
Using the static method createFromFormat from DateTime you can force the time-parts to be reset to 0 using '|':
$date = DateTime::createFromFormat('Y-m-d|', $given_date);
echo $date->format('Y-m-d H:i:s');
It is also possible to do it with:
list($y, $m, $d) = explode('-', $given_date);
$midnight = mktime(0, 0, 0, $m, $d, $y);
Related
I a using PHP and would like the date of third day before the script is called in the format of YYYYMMDD. How can I do this?
Try this
$result = date('Y.m.d',strtotime("-3 days"));
Use DateTime:
$date = new \DateTime('-3 days');
echo $date->format('Ymd');
// Alternatively, store the string in a variable
$result = $date->format('Ymd');
try this easy way
echo date('Ymd', strtotime("-3 days"));
Atleast visit php.net once before such questions
<?php
$date = mktime(0, 0, 0, date("m") , date("d")-3, date("Y"));
echo date('Ymd', strtotime($date));
?>
I am letting a user input a date. How should I return a date from their input and add 3 years to it? I tried to use mktime and date but it did not work out very well.
$input_date = 2010-03-28
My solution currently is just basic math for a given date 3 years ahead.
$input_date = $input_date + 3000;
Let's say I would want to give a date 4 years and 4 months 10 days
$future_date1 _datum = mktime(0, 0, 0, date("m")-2, date("d"), date("Y")+3);
$future_date2 = mktime(0, 0, 0, date("m"), date("d"), date("Y")+3);
You can use strtotime('+ 3 years') or DateInterval for an object oriented approach.
Use the strtotime function:
$input_date = '2010-03-28';
$future_date = strtotime("+3 years", strtotime($input_date);
http://www.php.net/manual/en/function.strtotime.php
Returns a timestamp, if you want to return YYYY-mm-dd: $future_date = date("Y-m-d", $future_date);
DateTime() offers multiple ways to do this. It's the way PHP recommends doing date math.
You can use DateTime::modify():
$date = new DateTime($input_date);
$date->modify('+3 months');
echo $date->format('Y-m-d');
// one-liner
echo (new DateTime($input_date))->modify('+3 months')->format('Y-m-d');
or with DateInterval()
$date = new DateTime($input_date);
$date->add(new DateInterval('P3M'));
echo $date->format('Y-m-d');
// one-liner
echo (new DateTime($input_date))->add(new DateInterval('P3M'))->format('Y-m-d');
In PHP if I have a variable ($getTimeStamp) that follows the format 0000-00-00 00:00:00 (i.e. 2013-09-26 13:06:00).
What is the easiest way to get the date ($getDate), hour ($getHour) and minute ($getMinute) as separate variables?
The easiest way is to use PHP DateTime class
$getTimeStamp = '2013-09-26 13:06:00';
$date = new \DateTime($getTimeStamp);
$dateString = $date->format('Y-m-d');
$hourString = $date->format('H');
$minuteString = $date->format('i');
It's not timestamp ;) Check what time() function returns, that's how timestamp looks like.
You can use something like that:
$time = strtotime($getTimeStamp);
$getDate = date('Y-m-d', $time);
$getHour = date('H', $time);
$getMinute = date('i', $time);
You can simply achieve this using -
$time = strtotime($getTimeStamp);
$getDate = date('Y-m-d', $time);
$getHour = date('H', $time);
$getMinute = date('i', $time);
Have a look at time() and date().
Well, since all you want is really to parse some text this would be the shortest way:
list($date, $hour, $minute) = preg_split('/[ :]/', $getTimeStamp);
There is no real reason to involve the date/time classes if you are not manipulating dates or calculating timestamps.
Live example.
Just one answer using the DateTime class
$dt = DateTime::createFromFormat('Y-m-d H:i:s', $getTimeStamp);
$date = $dt->format('Y-m-d ');
$hour = $dt->format('H');
$minutes = $dt->format('i');
I have a date in this format 030512 (ddmmyy).
But I'm having trouble with converting this to a date usable format I can add days to.
Basically.. I extracted the date above from a text file, Now I need to be able to add a number of days to it. But I am having trouble parsing the date in this format.
Is there another way of doing this rather then something like this:
// I have a date in this format
$date = '030512'; // 03 May 2012
$day = substr($date,0,2);
$month = substr($date, 2,2);
$year = substr($date, 4,2);
$date_after = $day . "-" . $month . "-".$year;
// Now i need to add x days to this
$total_no_nights = 010; // must be this format
$days_to_add = ltrim($total_no_nights,"0"); // 10, remove leading zero
// how do i add 10 days to this date.
You can do this (php >= 5.3):
$date = DateTime::createFromFormat('dmy', '030512');
$date->modify('+1 day');
echo $date->format('Y-m-d');
http://www.php.net/manual/en/datetime.createfromformat.php
For php < 5.3 :
$dateArray = str_split('030512', 2);
$dateArray[2] += 2000;
echo date("d/m/Y", strtotime('+1 day', strtotime(implode('-', array_reverse($dateArray)))));
try this using the month/day/year you already have:
$date = "$month/$day/$year";
$change = '+10 day';
echo date("d/m/Y", strtotime($change, strtotime($date)));
Assuming the date will always be in the future (or at least after 1st Jan 2000), you're not far wrong:
// I have a date in this format
$date = '030512'; // 03 May 2012
$day = substr($date,0,2);
$month = substr($date, 2,2);
$year = substr($date, 4,2);
// dd-mm-yy is not a universal format but we can use mktime which also gives us a timestamp to use for manipulation
$date_after = mktime( 0, 0, 0, $month, $day, $year );
// Now i need to add x days to this
$total_no_nights = "010"; // must be this format
$days_to_add = intval( $total_no_nights ); // No need to use ltrim
// Here's the "magic". Again it returns a timestamp
$new_date = strtotime( "+$days_to_add days", $date_after );
Using the DateTime object would be easier but you say you're not on PHP5.3.
You can't do date manipulation with strings becase, well, they are not dates. In PHP, you can use Unix timestamps (which are actually integers...) or DateTime objects. In you case:
$timestamp = strtotime('-10 days', mktime(0, 0, 0, $month, $day, $year));
echo date('r', $timestamp);
... or:
$object = new DateTime("$year-$month-$day");
$object->modify('-10 days');
echo $object->format('r');
Also, please note that 010 is an octal number that corresponds to 8 in decimal.
using the convert function in sql the date can be obtained in appropriate format.
anter that operations can be performed in php to manipulate the date. i hope that answers your query.
I've got a date in this format:
2009-01-01
How do I return the same date but 1 year earlier?
You can use strtotime:
$date = strtotime('2010-01-01 -1 year');
The strtotime function returns a unix timestamp, to get a formatted string you can use date:
echo date('Y-m-d', $date); // echoes '2009-01-01'
Use strtotime() function:
$time = strtotime("-1 year", time());
$date = date("Y-m-d", $time);
Using the DateTime object...
$time = new DateTime('2099-01-01');
$newtime = $time->modify('-1 year')->format('Y-m-d');
Or using now for today
$time = new DateTime('now');
$newtime = $time->modify('-1 year')->format('Y-m-d');
an easiest way which i used and worked well
date('Y-m-d', strtotime('-1 year'));
this worked perfect.. hope this will help someone else too.. :)
On my website, to check if registering people is 18 years old, I simply used the following :
$legalAge = date('Y-m-d', strtotime('-18 year'));
After, only compare the the two dates.
Hope it could help someone.
// set your date here
$mydate = "2009-01-01";
/* strtotime accepts two parameters.
The first parameter tells what it should compute.
The second parameter defines what source date it should use. */
$lastyear = strtotime("-1 year", strtotime($mydate));
// format and display the computed date
echo date("Y-m-d", $lastyear);
Although there are many acceptable answers in response to this question, I don't see any examples of the sub method using the \Datetime object: https://www.php.net/manual/en/datetime.sub.php
So, for reference, you can also use a \DateInterval to modify a \Datetime object:
$date = new \DateTime('2009-01-01');
$date->sub(new \DateInterval('P1Y'));
echo $date->format('Y-m-d');
Which returns:
2008-01-01
For more information about \DateInterval, refer to the documentation: https://www.php.net/manual/en/class.dateinterval.php
You can use the following function to subtract 1 or any years from a date.
function yearstodate($years) {
$now = date("Y-m-d");
$now = explode('-', $now);
$year = $now[0];
$month = $now[1];
$day = $now[2];
$converted_year = $year - $years;
echo $now = $converted_year."-".$month."-".$day;
}
$number_to_subtract = "1";
echo yearstodate($number_to_subtract);
And looking at above examples you can also use the following
$user_age_min = "-"."1";
echo date('Y-m-d', strtotime($user_age_min.'year'));