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I want to convert the date like this
$original_date = date("M/d/Y"); // the output of that is Aug/27/2013
to "08/27/2013"
Like this:
date('m/d/Y', strtotime($original_date));
strtotime can convert pretty much anything reasonable you give it into a Unix timestamp, even stuff like "Next Friday".
EDIT
Funny, but seems that strtime doesn't work with a date formatted like that... The first thing that came to my mind was to replace those slashes with spaces, using str_replace or implode/explode or whatever works for you...
$newDate = date('m/d/Y', strtotime(str_replace('/', ' ', $origDate)));
$newDate = date('m/d/Y', strtotime(implode(' ', explode('/', $origDate))));
Simply by passing it back to the date function
$original_date = date("M/d/Y");
$new_date = date('m/d/Y', strtotime($original_date));
Simply change the format string:
$originalDate = date("m/d/Y"); // = 08/27/2013
Or use DateTime and specify the input format to avoid ambiguity:-
$dateString = \DateTime::createFromFormat('M/d/Y', $origanalDate)->format('m/d/Y');
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can you please help me?
i need to change
2017-12-31 11:45:00
to
2017-12-31T11:45
using php function.
I need this datetime-local format for putting in a input value.
This might do it for you:
<?php
$date = '2017-12-31 11:45:00';
$newdate = date('Y-m-d\TH:i', strtotime($date));
echo $newdate; // 2017-12-31T11:45
;?>
Documentation about strtotime: http://php.net/manual/en/function.strtotime.php
NOTE:
strtotime max date is '19 Jan 2038 03:14:07 UTC'.
New solution: PHP 5 >= 5.2.0, PHP 7
Please consider to use the new DateTime function: http://php.net/manual/en/datetime.construct.php
Example new DateTime:
<?php
$d = new DateTime("9999-12-31");
$d->format("Y-m-d"); // "9999-12-31"
$d = new DateTime("0000-12-31");
$d->format("Y-m-d"); // "0000-12-31"
$d = new DateTime("-9999-12-31");
$d->format("Y-m-d"); // "-9999-12-31"
?>
This can be done quite simply using the DateTime object provided with PHP
<?php
$dt = DateTime::createFromFormat('Y-m-d H:i:s', '2017-12-31 11:45:00');
$newFormat = $dt->format('Y-m-d\TH:i');
echo $newFormat;
Result:
2017-12-19T11:45
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My date string value is 1478025000. want to get year and month in the string format itself from this string,is it possible? the day should be 0 and next month and next year like that.
my desired output should be 1477852200
Try something like this:
$epoch = 1478025000;
$dt = new DateTime("#$epoch"); // convert UNIX timestamp to PHP DateTime
echo $dt->format('Y-m'); // Display as year and month: YYYY-MM
echo $dt->format('m-Y'); // Display as month and year: MM-YYYY
If you want to get just the year and month as an epoch, try something like:
$epoch = 1478025000;
$dt = new DateTime("#$epoch"); // convert UNIX timestamp to PHP DateTime
$year = $dt->format('Y');
$day = $dt->format('n');
$answer = new DateTime();
$answer->setDate($year, $month, 0);
$answer->setTime(0, 0, 0);
echo $answer->getTimestamp();
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I need to create DateTime interval to compare it with date difference. Is something like this possible?
$week = new \DateTime("1 week");
Well, you can use DateInterval class instead. While you cannot pass the relative timeformats such as '1 week' into its constructor directly, there's a useful static method createFromDateString() that supports such syntax:
$weekInterval = \DateInterval::createFromDateString('1 week');
The problem is that DateInterval instances are not comparable in PHP (check this feature request for details):
$someInterval = (new \DateTime('2015-09-14'))->diff(new \DateTime('2015-09-13'));
var_dump($weekInterval > $someInterval); // false somehow
DateTime instances are comparable, however. So one possible workaround is making a comparison between the results of adding two DateInterval to the same fixed date. For example:
$now = new \DateTimeImmutable();
var_dump($now->add($weekInterval) > $now->add($someInterval)); // true
Eval.in demo.
I think what you are looking for is the diff(); function in the DateTime class.
$dateOne = new DateTime('2015-10-11');
$dateTwo = new DateTime('2014-10-13');
$interval = $dateOne->diff($dateTwo);
echo $interval->format('%R%a days');
Comparing UNIX TimeStamp
$dateTime = new DateTime("#" . $someOtherTimeStamp);
$endTime = new DateTime("#" . time());
$interval = $dateTime->diff($endTime);
Reading Material
diff();
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I have a form with an input box named "date[]" and a button that allows you to create more rows to add more dates, all with the name "date[]". The date picker format in my text input box is m-d-Y (03-15-2015). When I submit the form I need to convert the array of dates to mysql Date format Y-m-d (2015-03-15). Iv successfully captured the array of dates in an array in my php script and I'm trying to do a foreach loop and convert each date[]value into a timestamp using 'strtotime', that i will then convert to the correct Y-m-d format. The problem is strtotime expects a string not an array.
My question is how can I go about iterating through each date[] value and converting it into a timestamp? There is not a fixed amount of dates, it can change with each user.
Thank you in advance.
You can do that with array_map() or array_walk()
Here's an array_walk example that let's you specify the formats at call-time.
$dates = ['03-15-2015', '05-29-2015'];
array_walk($dates, function(&$date, $i, $formats)
{
$datetime = DateTime::createFromFormat($formats['from'], $date);
$date = $datetime->format($formats['to']);
}, ['from' => 'm-d-Y', 'to' => 'Y-m-d']);
Use foreach :
$dates = $_POST['date']; // in case you are using POST request
foreach($dates as $date){
$datetime = DateTime::createFromFormat('m-d-Y', $date);
$cDates[] = $datetime->format('Y-m-d');
}
Now all the converted dates exist inside $cDates array.
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I have a variable $dob which returns a date, in the following format: 1970-02-01 00:00:00
How can I use php to calculate the age of the person?
Your question is explained in detail in the PHP documentation, it bascially goes like this:
// create a datetime object for a given birthday
$birthday = new DateTime("2012-12-12");
// substract your timestamp from it
$diff = $birthday->diff($dob);
// output the difference in years
echo $diff->format('%Y');
Try this:
<?php
$age = '1970-02-01 00:00:00';
echo (int)((time()-strtotime($age))/31536000);
Output: 43, in years.
$dob = '1970-02-01 00:00:00';
$date = new DateTime($dob);
$diff = $date->diff(new DateTime);
echo $diff->format('%R%a days');
Basically stolen from here: http://uk1.php.net/manual/en/datetime.diff.php
Formatting options: http://uk1.php.net/manual/en/dateinterval.format.php
One-liner:
echo date_create('1970-02-01')->diff(date_create('today'))->y;
Demo.
i would try this
$user_dob = explode('-',$dob);
$current_year= date('Y');
$presons_age = $current_year - $user_dob[0];
This is an untested code but i feel u should get the logic.
strtotime() will convert your date into a timestamp, then subject that from time() and the result is the age in seconds.
$age_in_seconds = time() - strtotime('1970-02-01');
To display the age in years (+- 1 day), then divide by the number of seconds in a year:
echo "Age in whole years is " . floor($age_in_seconds / 60 / 60 / 24 / 365.25);