I have a PHP date in a database, for example 8th August 2011. I have this date in a strtotime() format so I can display it as I please.
I need to adjust this date to make it 8th August 2013 (current year). What is the best way of doing this? So far, I've been racking my brains but to no avail.
Some of the answers you have so far have missed the point that you want to update any given date to the current year and have concentrated on turning 2011 into 2013, excluding the accepted answer. However, I feel that examples using the DateTime classes are always of use in these cases.
The accepted answer will result in a Notice:-
Notice: A non well formed numeric value encountered......
if your supplied date is the 29th February on a Leapyear, although it should still give the correct result.
Here is a generic function that will take any valid date and return the same date in the current year:-
/**
* #param String $dateString
* #return DateTime
*/
function updateDate($dateString){
$suppliedDate = new \DateTime($dateString);
$currentYear = (int)(new \DateTime())->format('Y');
return (new \DateTime())->setDate($currentYear, (int)$suppliedDate->format('m'), (int)$suppliedDate->format('d'));
}
For example:-
var_dump(updateDate('8th August 2011'));
See it working here and see the PHP manual for more information on the DateTime classes.
You don't say how you want to use the updated date, but DateTime is flexible enough to allow you to do with it as you wish. I would draw your attention to the DateTime::format() method as being particularly useful.
strtotime( date( 'd M ', $originaleDate ) . date( 'Y' ) );
This takes the day and month of the original time, adds the current year, and converts it to the new date.
You can also add the amount of seconds you want to add to the original timestamp. For 2 years this would be 63 113 852 seconds.
You could retrieve the timestamp of the same date two years later with strtotime() first parameter and then convert it in the format you want to display.
<?php
$date = "11/08/2011";
$time = strtotime($date);
$time_future = strtotime("+2 years", $time);
$future = date("d/m/Y", $time_future);
echo "NEW DATE : " . $future;
?>
You can for instance output it like this:
date('2013-m-d', strtotime($myTime))
Just like that... or use
$year = date('Y');
$myMonthDay = date('m-d', strtotime($myTime));
echo $year . '-' . $myMonthDay;
Use the date modify function Like this
$date = new DateTime('2011-08-08');
$date->modify('+2 years');
echo $date->format('Y-m-d') . "\n";
//will give "2013-08-08"
Related
I have SQLite DB one table contains datetime field
with datatype "timestamp" REAL value is 18696.0
attach image for table structure
So, I want this 18696.0 value to be converted into MySQL Y-m-d format and result should be 2021-03-10
I have didn't found any solution online. any help would be appreciated.
SQLite timestamp converted into MySQL timestamp.
EDIT: Thankyou for updating your question with the correct number and what date it should represent.
You can achieve what you need with a function that adds the days onto the Unix Epoch date:
function realDateToYmd($real, $outputFormat='Y-m-d')
{
$date = new DateTime('1970-01-01');
$date->modify('+' . intval($real) . ' days');
return $date->format($outputFormat);
}
echo realDateToYmd('18696.0');
// returns 2021-03-10
SQLite dates stored in REAL data type stores dates as a Julian Day.
From https://www.sqlite.org/datatype3.html
REAL as Julian day numbers, the number of days since noon in Greenwich on November 24, 4714 B.C. according to the proleptic Gregorian calendar.
PHP has a jdtogregorian function, in which one comment has a handy function to convert to ISO8601 dates:
function JDtoISO8601($JD) {
if ($JD <= 1721425) $JD += 365;
list($month, $day, $year) = explode('/', jdtogregorian($JD));
return sprintf('%+05d-%02d-%02d', $year, $month, $day);
}
echo JDtoISO8601('17889.0');
// Results in -4664-11-16
The results don't exactly look right, is it definitely 17889.0 in SQLite?
If this float number 18696.0 represents the number of days since 1970-01-01 then the date can also be calculated like this:
$days = 18696.0;
$dt = date_create('#'.((int)($days * 86400)));
$mysqlDate = $dt->format('Y-m-d'); //"2021-03-10"
background information
Or simply with gmdate:
$mySqlDate = gmdate('Y-m-d',$days*86400);
The days are simply converted into seconds to get a valid timestamp for gmdate.
Try this:
<?php
echo date('Y-m-d H:i:s', 17889);
?>
Output:
1970-01-01 04:58:09
$date_raw = '05/05/1995';
$newDate = (date('j F Y', strtotime('-192years -14months -2days', strtotime($date_raw))));
print "New Date: $newDate <br>";
I'm trying to subtract 100+ years from a given date. but the value i get is real till 92 years only. after that i don't get correct subtraction. whats the reason?
If you need to work with dates that fall outside the range of a 32-bit signed unix timestamp (1901-12-13 to 2038-01-19), then start using DateTime objects
$date_raw = '05/05/1995';
$newDate = (new DateTime($date_raw))
->sub(new DateInterval('P192Y14M2D'))
->format('j F Y');
echo $newDate, PHP_EOL;
gives
3 March 1802
or you can do this in the following way
// set your date here
$mydate = "2018-06-27";
/* strtotime accepts two parameters.
The first parameter tells what it should compute.
The second parameter defines what source date it should use. */
$lastHundredyear = strtotime("-100 year", strtotime($mydate));
// format and display the computed date
echo date("Y-m-d", $lastHundredyear);
this will give you following output
1918-06-27
I'm trying to figure out how to accept a date/time from a form, which is consequently in the user's timezone, and change it to UTC before inserting it into the database. For some reason, no amount of searching has netted me an answer.
My form will POST whatever date is selected by the user to my code, so I expect to be able to do something like this. Note: the $userDate may be relative to any number of timezones based on user's location
$userDate = $_POST['user_date'] // 2014-05-15 16:37:23
I anticipate using Date().getTimezoneOffset() on my form to also submit the users UTC offset (as detailed here).
$userOffset = $_POST['user_offset']
Then before inserting the date into my database, I would like to convert it to UTC -- but I am stumped on how to do that with PHP (I'm actually using Laravel so if you know of a way using Carbon, that would be even easier, but I couldn't find it in their docs).
I've been half tempted to manually parse the offset and convert it to number of seconds and add or subtract it to strtotime() output of the $userDate and then convert it back into a date format using date() -- but there has to be a better way!
What am I missing here? Does PHP have a function I just don't know about that lets me do something like:
$userDate = '2014-05-15 16:37:23';
$userOffset = '+04:00';
$utcDate = date_apply_offset($userDate, $userOffset);
echo $utcDate; // Outputs: 2014-05-15 20:37:23
Or am I making this harder than it has to be?
EDIT
Based on the solution provided by #vascowhite, I went with the following (added into question to improve answers for those seeking guidance)
I ended up using a function from moment.js since I was already using it to convert UTC to user's timezone on display.
HTML:
<input id="user_offset" type="hidden" name="user_offset" value="">
Javascript:
var offset = moment().format('ZZ');
$('#user_offset').val(offset);
PHP (in a custom date class):
class MyDate {
/**
* Convert Date to UTC
*
* #param string $date Any date parsable with strtotime()
* #param string $offset UTC offset of date
*/
public static function toUTC($date, $offset = '+0:00')
{
if ($timestamp = strtotime($date) && ! empty($offset) )
{
$newDate = date('Y-m-d H:i:s', $timestamp);
$newDate = new \DateTime($date . ' ' . $offset);
$newDate->setTimezone(new DateTimeZone('UTC'));
$date = $newDate->format('Y-m-d H:i:s');
}
return $date;
}
}
// To convert
$userDate = trim($_POST['user_offset']);
$userOffset = trim($_POST['user_date']);
$utc = MyDate::toUTC($userDate, $userOffset)
That class method isn't perfect, and in the event something goes wrong, it just returns the date back -- when really it should throw an exception.
This is a simple task with the DateTime classes:-
$userDate = '2014-05-15 16:37:23';
$userOffset = '+04:00';
$date = new \DateTime($userDate . ' ' . $userOffset);
var_dump($date);
$date->setTimezone(new \DateTimeZone('UTC'));
var_dump($date);
You can then format the date as you wish for output eg:-
echo $date->format('Y-m-d H:i:s');
or:-
$utcDate = $date->format('Y-m-d H:i:s');
echo $utcDate; // Outputs: 2014-05-15 20:37:23
See it working.
If you are doing any work with dates and times in PHP it is worth taking the time to become familiar with these extremely useful classes.
For all sorts of date/time manipulations you can make use of moment.php
For your example all what is needed are two lines of code:
$m = new \Moment\Moment('2014-05-15 16:37:23', '+0400');
echo $m->setTimezone('UTC')->format(); // 2014-05-15T12:37:23+0000
There is much more which helps to deal with date/time issues: https://github.com/fightbulc/moment.php
Cheers
I'm trying to use DateTime to check if a credit card expiry date has expired but I'm a bit lost.
I only want to compare the mm/yy date.
Here is my code so far
$expmonth = $_POST['expMonth']; //e.g 08
$expyear = $_POST['expYear']; //e.g 15
$rawExpiry = $expmonth . $expyear;
$expiryDateTime = \DateTime::createFromFormat('my', $rawExpiry);
$expiryDate = $expiryDateTime->format('m y');
$currentDateTime = new \DateTime();
$currentDate = $currentDateTime->format('m y');
if ($expiryDate < $currentDate) {
echo 'Expired';
} else {
echo 'Valid';
}
I feel i'm almost there but the if statement is producing incorrect results. Any help would be appreciated.
It's simpler than you think. The format of the datess you are working with is not important as PHP does the comparison internally.
$expires = \DateTime::createFromFormat('my', $_POST['expMonth'].$_POST['expYear']);
$now = new \DateTime();
if ($expires < $now) {
// expired
}
You can use the DateTime class to generate a DateTime object matching the format of your given date string using the DateTime::createFromFormat() constructor.
The format ('my') would match any date string with the string pattern 'mmyy', e.g. '0620'. Or for dates with 4 digit years use the format 'mY' which will match dates with the following string pattern 'mmyyyy', e.g. '062020'. It's also sensible to specify the timezone using the DateTimeZone class.
$expiryMonth = 06;
$expiryYear = 20;
$timezone = new DateTimeZone('Europe/London');
$expiryTime = \DateTime::createFromFormat('my', $expiryMonth.$expiryYear, $timezone);
See the DateTime::createFromFormat page for more formats.
However - for credit/debit card expiry dates you will also need to take into account the full expiry DATE and TIME - not just the month and year.
DateTime::createFromFormat will by default use todays day of the month (e.g. 17) if it is not specified. This means that a credit card could appear expired when it still has several days to go. If a card expires 06/20 (i.e. June 2020) then it actually stops working at 00:00:00 on 1st July 2020. The modify method fixes this. E.g.
$expiryTime = \DateTime::createFromFormat('my', $expiryMonth.$expiryYear, $timezone)->modify('+1 month first day of midnight');
The string '+1 month first day of midnight' does three things.
'+1 month' - add one month.
'first day of' - switch to the first day of the month
'midnight' - change the time to 00:00:00
The modify method is really useful for many date manipulations!
So to answer the op, this is what you need — with a slight adjustment to format to cater for single digit months:
$expiryMonth = 6;
$expiryYear = 20;
$timezone = new DateTimeZone('Europe/London');
$expiryTime = \DateTime::createFromFormat(
'm-y',
$expiryMonth.'-'.$expiryYear,
$timezone
)->modify('+1 month first day of midnight');
$currentTime = new \DateTime('now', $timezone);
if ($expiryTime < $currentTime) {
// Card has expired.
}
An addition to the above answers.
Be aware that by default the days will also be in the calculation.
For example today is 2019-10-31 and if you run this:
\DateTime::createFromFormat('Ym', '202111');
It will output 2021-12-01, because day 31 does not exist in November and it will add 1 extra day to your DateTime object with a side effect that you will be in the month December instead of the expected November.
My suggestion is always use the day in your code.
For op's question:
$y=15;
$m=05;
if(strtotime( substr(date('Y'), 0, 2)."{$y}-{$m}" ) < strtotime( date("Y-m") ))
{
echo 'card is expired';
}
For others with full year:
$y=2015;
$m=5;
if(strtotime("{$y}-{$m}") < strtotime( date("Y-m") ))
{
echo 'card is expired';
}
Would it not be simpler to just compare the string "201709" to the current year-month? Creating datetime objects will cost php some effort, I suppose.
if($_POST['expYear']. str_pad($_POST['expMonth'],2,'0', STR_PAD_LEFT ) < date('Ym')) {
echo 'expired';
}
edited as Adam states
The best answer is provided by John Conde above. It it does the minimum amount of processing: creates two correct DateTime objects, compares them and that's all it needs.
It could work also as you started but you must format the dates in a way that puts the year first.
Think a bit about it: as dates, 08/15 (August 2015) is after 12/14 (December 2014) but as strings, '08 15' is before '12 14'.
When the year is in front, even as strings the years are compared first and then, only when the years are equal the months are compared:
$expiryDate = $expiryDateTime->format('y m');
$currentDate = $currentDateTime->format('y m');
if ($expiryDate < $currentDate) {
echo 'Expired';
} else {
echo 'Valid';
}
Keep it simple, as the answer above me says except you need to string pad to the left:
isCardExpired($month, $year)
{
$expires = $year.str_pad($month, 2, '0', STR_PAD_LEFT);
$now = date('Ym');
return $expires < $now;
}
No need to add extra PHP load using DateTime
If you are using Carbon, which is a very popular Datetime extension library. Then this should be:
$expMonth = $_POST['month'];
$expYear = $_POST['year'];
$format_m_y = str_pad($expMonth,2,'0', STR_PAD_LEFT).'-'.substr($expYear, 2);
$date = \Carbon\Carbon::createFromFormat('m-y', $format_m_y)
->endOfMonth()
->startOfDay();
if ($date->isPast()) {
// this card is expired
}
Also take into consideration the exact date and time expiration:
Credit cards expire at the end of the month printed as its expiration date, not at the beginning. Many cards actually technically expire one day after the end of that month. In any case, unless they list a specific day of expiration along with month and year, they should work all the way through the end of their expiration month. Cardholders should not wait until the last moment to secure a replacement card. Source
I need to create functions in PHP that let me step up/down given datetime units. Specifically, I need to be able to move to the next/previous month from the current one.
I thought I could do this using DateTime::add/sub(P1M). However, when trying to get the previous month, it messes up if the date value = 31- looks like it's actually trying to count back 30 days instead of decrementing the month value!:
$prevMonth = new DateTime('2010-12-31');
Try to decrement the month:
$prevMonth->sub(new DateInterval('P1M')); // = '2010-12-01'
$prevMonth->add(DateInterval::createFromDateString('-1 month')); // = '2010-12-01'
$prevMonth->sub(DateInterval::createFromDateString('+1 month')); // = '2010-12-01'
$prevMonth->add(DateInterval::createFromDateString('previous month')); // = '2010-12-01'
This certainly seems like the wrong behavior. Anyone have any insight?
Thanks-
NOTE: PHP version 5.3.3
(Credit actually belongs to Alex for pointing this out in the comments)
The problem is not a PHP one but a GNU one, as outlined here:
Relative items in date strings
The key here is differentiating between the concept of 'this date last month', which, because months are 'fuzzy units' with different numbers of dates, is impossible to define for a date like Dec 31 (because Nov 31 doesn't exist), and the concept of 'last month, irrespective of date'.
If all we're interested in is the previous month, the only way to gaurantee a proper DateInterval calculation is to reset the date value to the 1st, or some other number that every month will have.
What really strikes me is how undocumented this issue is, in PHP and elsewhere- considering how much date-dependent software it's probably affecting.
Here's a safe way to handle it:
/*
Handles month/year increment calculations in a safe way,
avoiding the pitfall of 'fuzzy' month units.
Returns a DateTime object with incremented month/year values, and a date value == 1.
*/
function incrementDate($startDate, $monthIncrement = 0, $yearIncrement = 0) {
$startingTimeStamp = $startDate->getTimestamp();
// Get the month value of the given date:
$monthString = date('Y-m', $startingTimeStamp);
// Create a date string corresponding to the 1st of the give month,
// making it safe for monthly/yearly calculations:
$safeDateString = "first day of $monthString";
// Increment date by given month/year increments:
$incrementedDateString = "$safeDateString $monthIncrement month $yearIncrement year";
$newTimeStamp = strtotime($incrementedDateString);
$newDate = DateTime::createFromFormat('U', $newTimeStamp);
return $newDate;
}
Easiest way to achieve this in my opinion is using mktime.
Like this:
$date = mktime(0,0,0,date('m')-1,date('d'),date('Y'));
echo date('d-m-Y', $date);
Greetz Michael
p.s mktime documentation can be found here: http://nl2.php.net/mktime
You could go old school on it and just use the date and strtotime functions.
$date = '2010-12-31';
$monthOnly = date('Y-m', strtotime($date));
$previousMonth = date('Y-m-d', strtotime($monthOnly . ' -1 month'));
(This maybe should be a comment but it's to long for one)
Here is how it works on windows 7 Apache 2.2.15 with PHP 5.3.3:
<?php $dt = new DateTime('2010-12-31');
$dt->sub(new DateInterval('P1M'));
print $dt->format('Y-m-d').'<br>';
$dt->add(DateInterval::createFromDateString('-1 month'));
print $dt->format('Y-m-d').'<br>';
$dt->sub(DateInterval::createFromDateString('+1 month'));
print $dt->format('Y-m-d').'<br>';
$dt->add(DateInterval::createFromDateString('previous month'));
print $dt->format('Y-m-d').'<br>'; ?>
2010-12-01
2010-11-01
2010-10-01
2010-09-01
So this does seem to confirm it's related to the GNU above.
Note: IMO the code below works as expected.
$dt->sub(new DateInterval('P1M'));
Current month: 12
Last month: 11
Number of Days in 12th month: 31
Number of Days in 11th month: 30
Dec 31st - 31 days = Nov 31st
Nov 31st = Nov 1 + 31 Days = 1st of Dec (30+1)