$bought_months = 6;
$currentDate = date('Y-m-d');
$expiring = strtotime('+'.$bought_months.' month', $currentDate);
This is what i tried.
I am trying to get a unix timestamp value of 6 months ahead from today.
How can i add months to my unix timestamp right? I tried above, since my other thought - calculating by seconds like one month in seconds: 2 629 743.83 * how many months + current timestamp, will not be precise.
(ofcourse because months have different number of days)
I get "A non well formed numeric value encountered" for the code above.
How can i do this?
Since you are using the current timestamp, you can omit the $currentDate variable altogether and it should make the notice go away too.
$bought_months = 6;
$expiring = strtotime('+'.$bought_months.' month');
To make the date correctly you'd need to use something like:
mktime( 0, 0, 0, ( $month + 6 ), $day, $year );
For maximum compatibility with post-2038 dates, use php's builtin DateTime class:
$d = new DateTime();
$d->modify("+6 months");
echo $d->format("U");
Related
I want to add an x number of week days (e.g. 48 weekday hours) to the current timestamp. I am trying to do this using the following
echo (strtotime('2 weekdays');
However, this doesn't seem to take me an exact 48 hours ahead in time. For example, inputting the current server time of Tuesday 18/03/2014 10:47 returns Thursday 20/03/2014 00:00. using the following function:
echo (strtotime('2 weekdays')-mktime())/86400;
It can tell that it's returning only 1.3 weekdays from now.
Why is it doing this? Are there any existing functions which allow an exact amount of weekday hours?
Given you want to preserve the weekdays functionality and not loose the hours, minutes and seconds, you could do this:
$now = new DateTime();
$hms = new DateInterval(
'PT'.$now->format('H').'H'.
$now->format('i').'M'.
$now->format('s').'S'.
);
$date = new DateTime('2 weekdays');
$date->add($hms);//add hours here again
The reason why weekday doesn't add the hours is because, if you add 1 weekday at any point in time on a monday, the next weekday has to be tuesday.
The hour simply does not matter. Say your date is 2014-01-02 12:12:12, and you want the next weekday, that day starts at 2014-01-03 00:00:00, so that's what you get.
My last solution works though, and here's how: I use the $now instance of DateTime, and its format method to construct a DateInterval format string, to be passed to the constructor. An interval format is quite easy: it starts with P, for period, then a digit and a char to indicate what that digit represents: 1Y for 1 Year, and 2D for 2 Days.
However, we're only interested in hours, minutes and seconds. Actual time, which is indicated using a T in the interval format string, hence we start the string with PT (Period Time).
Using the format specifiers H, i and s, we construct an interval format that in the case of 12:12:12 looks like this:
$hms = new DateInterval(
'PT12H12M12S'
);
Then, it's a simple matter of calling the DateTime::add method to add the hours, minutes and seconds to our date + weekdays:
$weekdays = new DateTime('6 weekdays');
$weekdays->add($hms);
echo $weekdays->format('Y-m-d H:i:s'), PHP_EOL;
And you're there.
Alternatively, you could just use the basic same trick to compute the actual day-difference between your initial date, and that date + x weekdays, and then add that diff to your initial date. It's the same basic principle, but instead of having to create a format like PTXHXMXS, a simple PXD will do.
Working example here
I'd urge you to use the DateInterface classes, as it is more flexible, allows for type-hinting to be used and makes dealing with dates just a whole lot easier for all of us. Besides, it's not too different from your current code:
$today = new DateTime;
$tomorrow = new DateTime('tomorrow');
$dayAfter = new DateTime('2 days');
In fact, it's a lot easier if you want to do frequent date manipulations on a single date:
$date = new DateTime();//or DateTime::createFromFormat('Y-m-d H:i:s', $dateString);
$diff = new DateInterval('P2D');//2 days
$date->add($diff);
echo $date->format('Y-m-d H:i:s'), PHP_EOL, 'is the date + 2 days', PHP_EOL;
$date->sub($diff);
echo $date->format('Y-m-d H:i:s'), PHP_EOL, 'was the original date, now restored';
Easy, once you've spent some time browsing through the docs
I think I have found a solution. It's primitive but after some quick testing it seems to work.
The function calculates the time passed since midnight of the current day, and adds it onto the date returned by strtotime. Since this could fall into a weekend day, I've checked and added an extra day or two accordingly.
function weekDays($days) {
$tstamp = (strtotime($days.' weekdays') + (time() - strtotime("today")));
if(date('D',$tstamp) == 'Sat') {
$tstamp = $tstamp + 86400*2;
}
elseif(date('D',$tstamp) == 'Sun') {
$tstamp = $tstamp + 86400;
}
return $tstamp;
}
Function strtotime('2 weekdays') seems to add 2 weekdays to the current date without the time.
If you want to add 48 hours why not adding 2*24*60*60 to mktime()?
echo(date('Y-m-d', mktime()+2*24*60*60));
The currently accepted solution works, but it will fail when you want to add weekdays to a timestamp that is not now. Here's a simpler snippet that will work for any given point in time:
$start = new DateTime('2021-09-29 15:12:10');
$start->add(date_interval_create_from_date_string('+ 3 weekdays'));
echo $start->format('Y-m-d H:i:s'); // 2021-10-04 15:12:10
Note that this will also work for a negative amount of weekdays:
$start = new DateTime('2021-09-29 15:12:10');
$start->add(date_interval_create_from_date_string('- 3 weekdays'));
echo $start->format('Y-m-d H:i:s'); // 2021-09-24 15:12:10
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 am getting a date back from a mysql query in the format YYYY-MM-DD.
I need to determine if that is more than three months in the past from the current month.
I currently have this code:
$passwordResetDate = $row['passwordReset'];
$today = date('Y-m-d');
$splitCurrentDate = explode('-',$today);
$currentMonth = $splitCurrentDate[1];
$splitResetDate = explode('-', $passwordResetDate);
$resetMonth = $splitResetDate[1];
$diferenceInMonths = $splitCurrentDate[1] - $splitResetDate[1];
if ($diferenceInMonths > 3) {
$log->lwrite('Need to reset password');
}
The problem with this is that, if the current month is in January, for instance, giving a month value of 01, and $resetMonth is November, giving a month value of 11, then $differenceInMonths will be -10, which won't pass the if() statement.
How do I fix this to allow for months in the previous year(s)?
Or is there a better way to do this entire routine?
Use strtotime(), like so:
$today = time(); //todays date
$twoMonthsLater = strtotime("+3 months", $today); //3 months later
Now, you can easily compare them and determine.
I’d use PHP’s built-in DateTime and DateInterval classes for this.
<?php
// create a DateTime representation of your start date
// where $date is date in database
$resetDate = new DateTime($date);
// create a DateIntveral representation of 3 months
$passwordExpiry = new DateInterval('3M');
// add DateInterval to DateTime
$resetDate->add($passwordExpiry);
// compare $resetDate to today’s date
$difference = $resetDate->diff(new DateTime());
if ($difference->m > 3) {
// date is more than three months apart
}
I would do the date comparison in your SQL expression.
Otherwise, PHP has a host of functions that allow easy manipulation of date strings:
PHP: Date/Time Functions - Manual
I want to get the $registratiedag and count a couple of days extra, but I always get stuck on the fact that it needs to be a UNIX timestamp? I did some google-ing, but I really don't get it.
I hope someone can help me figure this out. This is what I got so far.
$registratiedag = $oUser['UserRegisterDate'];
$today = strtotime('$registratiedag + 6 days');
echo $today;
echo $registratiedag;
echo date('Y-m-d', $today);
There's obviously something wrong with the strtotime('$registratiedag + 6 days'); part, because I always get 1970-01-01
You probably want this:
// Store as a timestamp
$registratiedag = strtotime($oUser['UserRegisterDate']);
$new_date = strtotime('+6 days', $registratiedag);
// You'll need to format for printing $new_date
echo date('Y-m-d', $new_date);
// I think you want to compare $new_date against
// today's date. I'd recommend a string comparison here,
// As time() includes the time as well
// time() is implied as the second argument to date,
// But we'll put it anyways just to be clearer
if( date('Y-m-d', $new_date) == date('Y-m-d', time()) ) {
// The dates are equal, do something here
}
else if($new_date < time()) {
// if the new date is earlier than today
}
// etc.
First it converts $registratiedag to a timestamp, then it adds 6 days
EDIT: You probably should change $today to something less misleading like $modified_date or something
try:
$today = strtorime($registratiedag);
$today += 86400 * 6; // seconds in 1 day * 6 days
at least one of your problems is that PHP does not expand variables in single quotes.
$today = strtotime("$registratiedag + 6 days");
//use double quotes and not single quotes when embedding a php variable in a string
If you want to include the value of variable $registratiedag right into the text passed as parameter of strtotime, you have to enclose that parameter with ", not with '.
What is the best (date format independent way) in PHP to calculate difference in days between two dates in specified format.
I tried the following function:
function get_date_offset($start_date, $end_date)
{
$start_time = strtotime($start_date);
$end_time = strtotime($end_date);
return round(($end_time-$start_time)/(3600*24));
}
It works ok on linux box, but when running under windows strtotime returns ''.
EDIT:
Input date is in mm/dd/yyyy format, but I would like to make it accept $format as a parameter.
I need only difference in days.
If you do not want or you cannot use Zend Framework or Pear package try this function i hope this would help:
function dateDifference($date1, $date2)
{
$d1 = (is_string($date1) ? strtotime($date1) : $date1);
$d2 = (is_string($date2) ? strtotime($date2) : $date2);
$diff_secs = abs($d1 - $d2);
$base_year = min(date("Y", $d1), date("Y", $d2));
$diff = mktime(0, 0, $diff_secs, 1, 1, $base_year);
return array
(
"years" => abs(substr(date('Ymd', $d1) - date('Ymd', $d2), 0, -4)),
"months_total" => (date("Y", $diff) - $base_year) * 12 + date("n", $diff) - 1,
"months" => date("n", $diff) - 1,
"days_total" => floor($diff_secs / (3600 * 24)),
"days" => date("j", $diff) - 1,
"hours_total" => floor($diff_secs / 3600),
"hours" => date("G", $diff),
"minutes_total" => floor($diff_secs / 60),
"minutes" => (int) date("i", $diff),
"seconds_total" => $diff_secs,
"seconds" => (int) date("s", $diff)
);
}
The PEAR Date class offers all kinds of features for finding the differences between dates and about 1000 other things as well. The docs for it are here...
The problem with PHP is that it doesn't have a definite DateTime type. You can use a Unix timestamp, or the built-in DateTime class, but they are pretty limited in their functionality. I expect that there should be some 3rd party classes with more extensive support for date-time calculations, but I haven't looked for it.
Using Unix timestamps for date (not time) calculations is also tricky. You'd have to discard the time part, but simply resetting to 00:00 is not safe because of daylight savings time (DST). DST has the effect that there are two days every year that don't have exactly 24 hours. Thus, when adding/subtracting dates you might end up with a value that does not divide evenly with 3600*24.
I'd suggest looking for some 3rd party class that has proper support for all this stuff. Date/Time calculations are awesome in their ugliness. :P
The Zend Framework has the class Zend_Date for dealing with "date math". It works around system specific timestamp limits by using the BCMath extension, or if that's not available limits the timestamps by max float value for your system.
// example printing difference in days
require('Zend/Date.php');
$date1 = new Zend_Date();
$date1->set(2, Zend_Date::MONTH);
$date1->set(27, Zend_Date::DAY);
$date1->set(2008, Zend_Date::YEAR);
$date2 = new Zend_Date();
$date2->set(3, Zend_Date::MONTH);
$date2->set(3, Zend_Date::DAY);
$date2->set(2008, Zend_Date::YEAR);
echo ($date2->getTimestamp() - $date1->getTimestamp()) / (3600*24);
I'm not sure what is considered best, since there is no built-in function in PHP for doing this, but some people have used gregoriantojd(), for example in this forum post.
gregoriantojd() gives the same results as using strtotime(), see this blogpost for how to do it:
http://www.phpro.org/examples/Calculate-Age-With-PHP.html
The following works for me. Believe I found it on the php.net docs somewhere.
*Edit - Woops, didn't see csl's post. This is the exact function from his link, must have been where I found it. ;)
//Find the difference between two dates
function dateDiff($startDate, $endDate)
{
// Parse dates for conversion
$startArry = date_parse($startDate);
$endArry = date_parse($endDate);
// Convert dates to Julian Days
$start_date = gregoriantojd($startArry["month"], $startArry["day"], $startArry["year"]);
$end_date = gregoriantojd($endArry["month"], $endArry["day"], $endArry["year"]);
// Return difference
return round(($end_date - $start_date), 0);
}
I was trying to calculate the difference of two dates for the purpose of showing the duration of an event. Most of the functions given on the problem fails if the event has a duration form friday at 17:00 to sunday at 15:00. My goal was to find the difference between the dates like:
date('Ymd',$end)-date('Tmd',$begining);
But that is likly to fail because there isn't 99 month in a year and 99 days in a month. I could convert the date string to UNIX timestamp and divide by 60*60*12, but some days have a greater or lesser number of hours, sometimes there's eaven a leap secound. So I made my own function using getdate() a function that returns an array of innformation about the timestamp.
/*
* datediff($first,$last)
* $first - unix timestamp or string aksepted by strtotime()
* $last - unix timestamp or string aksepted by strtotime()
*
* return - the difference in days betveen the two dates
*/
function datediff($first,$last){
$first = is_numeric($first) ? $first : strtotime($first);
$last = is_numeric($last ) ? $last : strtotime($last );
if ($last<$first){
// Can't calculate negative difference in dates
return -1;
}
$first = getdate($first);
$last = getdate($last );
// find the difference in days since the start of the year
$datediff = $last['yday'] - $first['yday'];
// if the years do not match add the appropriate number of days.
$yearCountedFrom = $first['year'];
while($last['year'] != $yearCountedFrom ){
// Take leap years into account
if( $yearCountedFrom % 4 == 0 && $yearCountedFrom != 1900 && $yearCountedFrom != 2100 ){
//leap year
$datediff += 366;
}else{
$datediff += 365;
}
$yearCountedFrom++;
}
return $datediff;
}
Concerning the GregorianToJD() function, it might work, but I feel a little bit uneasy since I do not understand how it work.
Let's not overengineer, guys.
$d1 = strtotime($first_date);
$d2 = strtotime($second_date);
$delta = $d2 - $d1;
$num_days = ($delta / 86400);
Calculate the difference between two Dates (and time) using Php. The following page provides a range of different methods (7 in total) for performing date / time calculations using Php, to determine the difference in time (hours, munites), days, months or years between two dates.
See Php Date Time - 7 Methods to Calculate the Difference between 2 dates
PHP 5.2 introduces the DateTime and DateInterval classes which makes this easy:
function get_date_offset($start_date, $end_date)
{
$start_time = new DateTime($start_date);
$end_time = new DateTime($end_date);
$interval = $end_time->diff($start_time);
return $interval->format('%a days');
}