Get difference between date returns zero - php

I have a date time in 'Y-m-d H:i:s', and i tried to substract the now date with the defined date +1 day to get remaining time in hours, minutes and seconds:
$time = '2017-10-05 14:54:03';
$now = date('Y-m-d H:i:s');
$endTransaction = date('Y-m-d H:i:s', strtotime($time. ' + 1 day'));
$dteDiff = $endTransaction - $now;
echo $dteDiff;
but i always get 0 as the result

You are doing it wrong. The date function returns string so PHP is not able to compare anything. Try with the DateTime class instead. Its diff method returns the DateInterval object with some public properties, like the days property among others, which is the positive integer number (rounded down) of days between two dates:
$now = new \DateTime();
$endTransaction = (new \DateTime('2017-12-05 14:54:03'))->modify('+1 day');
$diff = $endTransaction->diff($now);
printf(
'Difference in days: %d, hours: %d, minutes: %d, seconds: %d',
$diff->days,
$diff->h,
$diff->m,
$diff->s
);

You probably need to use this date_diff
$time = '2017-10-05 14:54:03';
$now = date_create(date('Y-m-d H:i:s'));
$endTransaction = date_create(date('Y-m-d H:i:s', strtotime($time. ' + 1 day')));
$dteDiff = date_diff($now, $endTransaction);
$date = new DateTime($dteDiff);
$result = $date->format('Y-m-d H:i:s');

According to above mentioned description please try executing following code snippet as a solution to it.
$time = '2017-10-05 14:54:03';
$now = strtotime(date('Y-m-d H:i:s'));
$endTransaction = strtotime(date('Y-m-d H:i:s', strtotime($time. ' + 1 day')));
$dteDiff = ($endTransaction - $now)/(24*60*60);
echo round($dteDiff);

$endTransaction and $now are strings.
$time = '2017-10-05 14:54:03';
$now = date('Y-m-d H:i:s');
$endTransaction = date('Y-m-d H:i:s', strtotime($time. ' + 1 day'));
echo($endTransaction."\n");
echo($now."\n");
It prints:
2017-10-06 14:54:03
2017-10-05 11:45:39
The subtraction is not a valid operation for strings. It can handle only numbers. The strings above are converted to numbers. The conversion uses only the leftmost digits present in the string, until it reaches the first character that is not a digit.
Both strings above produce 2017 when they are converted to numbers and their difference is, of course, 0.
The easiest way to work with dates in PHP is to use the DateTime and its related classes.
// Convert the input string to a DateTime object
$then = new DateTime('2017-10-05 14:54:03');
// Add 1 day
$then->add(new DateInterval('P1D'));
// Get the current date and time
$now = new DateTime('now');
// Compute the difference; it is a DateInterval object
$diff = $now->diff($then);
// Display the dates and the difference
echo('Then: '.$then->format("Y-m-d H:i:s\n"));
echo('Now : '.$now->format("Y-m-d H:i:s\n"));
echo('Remaining: '.$diff->format("%R%a days, %h hours, %i minutes, %s seconds.\n"));
The output:
Then: 2017-10-06 14:54:03
Now : 2017-10-05 12:36:25
Remaining: +1 days, 2 hours, 17 minutes, 38 seconds.

Related

PHP subtract unknown time from datetime

So, I'm creating a booking system. When I retrieve this booking from the database, I need to check if the current date and time is closer to the actual booked date and time.
On the admin dashboard, the admins specify how much time earlier the client can make a checkin, let's say for example, 30minutes. But this time can be different. Can be 1hour, 2hours, 10minutes.
When I get the result from the database I get them like this:
$date_schedule = '2021-03-25 15:40:00'; // Can be any date in the future as well;
$time_to_check = '00:30:00'; // Can be '01:05:00', whatever the admins set as time_to_check;
// Expected result
'2021-03-25 15:10:00';
I tried subtracting this but I didn't made it work.. This is what I did.
$current_date = date("Y-m-d H:i:s");
$booking_date = '2021-03-25 15:00:00'; // From database
$time_to_check = '00:30:00'; // From database
$hours = explode(':', $time_to_check);
$data_check = date($booking_date, strtotime('-' . $hours[0] . ' hour -' . $hours[1] . ' minutes'));
But with this, $data_check returns the same value as $booking_date, it's not subtracting the time.
Convert your date to DateTime to make some operations on it :
function changeDate($date, $interval) {
$datetime = new DateTime($date);
$values = explode(':', $interval);
$datetime->modify("-$values[0] hours -$values[1] minutes -$values[2] seconds");
return $datetime->format('Y-m-d H:i:s');
}
$date = changeDate('2021-03-25 15:00:00', '00:30:00');
echo $date; // 2021-03-25 14:30:00
$date = changeDate('2021-03-25 15:00:00', '01:30:00');
echo $date; // 2021-03-25 13:30:00
$date = changeDate('2021-03-25 15:00:00', '02:30:15');
echo $date; // 2021-03-25 12:29:45
You can find documentation here https://www.php.net/manual/en/book.datetime.php
Just convert the timestamp to Unix time, and then subtract 30 minutes in seconds (30 * 60) from the Unix time.
Like this:
$date = '2021-03-25 15:10:00';
$timestamp = strtotime($date);
$timestamp = ($timestamp - (30 * 60));
echo gmdate("Y-m-d H:i:s", $timestamp);
EDIT: If you are in an odd timezone, like me (GMT+0100) add or subtract difference;
$timestamp = (($timestamp - (30 * 60)) + (1 * 60 * 60))

Convert current date + 1 month in accountExpire AD attribut

I try to convert current date + 1 month to a final timestamp for Active Directory attribut accountExpire.
But the timestamp returned is wrong if someone can help me. Thanks
$now = strtotime(date('Y-m-d H:i:s'));
$final = date('Y-m-d H:i:s', strtotime('+1 month', $now));
$dateTime = new DateTime($final);
$timestamp = $dateTime->format('U');
echo $timestamp;
You only need
$timestamp = strtotime('+1 month');
I had to look it up, but it turns out that LDAP timestamp is not the same as UNIX timestamp. LDAP timestamp is counted in 100s of nanoseconds and is counted from year 1601. You can easily convert one into another by applying the difference and multiplying by 10000000
$UNIXtimestamp = strtotime('+1 month');
$LDAPTimestamp = ($UNIXtimestamp + 11644473600) * 10000000;

Calculating time difference for fixed hour

I want to calculate the time difference from now (lets say 18:30:00) till this evening at 20pm.
$today = date('Y-m-d', time());
$remain = strtotime($today. " 00:00:00 + 20 hours") - time();
$remain = date('H:i:s', $remain);
I get a result which is one hour larger (02:30:00) than the actual result (01:30:00). I tried setting time zones but it's always the same result.
Using the DateTime object, you can do this easily:
$d1 = new DateTime('2015-04-23 18:30');
$d2 = new DateTime('2015-04-23 20:00');
$interval = $d2->diff($d1);
echo $interval->format('%H:%i hours');

Substract two datetimes and get seconds

I tried to substract two datetimes to get result by seconds like that:
$created = "2015-01-16 07:26:55";
$newdate_created = date('Y-m-d H:i:s', strtotime('+2 month', strtotime($created)));
$now = date('Y-m-d h:i:s');
$interval = date_diff($now, $newdate_created);
$seconds = $interval * 60 * 60 * 12 ;
But i'm getting this error:
date_diff() expects parameter 1 to be DateTime
i dont know where is the problem to getting this working
I'm looking the output to be like that :
5259000 // something like that by seconds
Highly wasteful code. There is NO point in formatting your timestamps into strings, just to have to yank them back into timestamp format:
$newdate_created = strtotime('+2 month', strtotime($created));
$now = time();
$diff_in_seconds = $now - $newdate_created;

Adding days to $Date in PHP

I have a date returned as part of a MySQL query in the form 2010-09-17.
I would like to set the variables $Date2 to $Date5 as follows:
$Date2 = $Date + 1
$Date3 = $Date + 2
etc., so that it returns 2010-09-18, 2010-09-19, etc.
I have tried
date('Y-m-d', strtotime($Date. ' + 1 day'))
but this gives me the date before $Date.
What is the correct way to get my Dates in the format form 'Y-m-d' so that they may be used in another query?
All you have to do is use days instead of day like this:
<?php
$Date = "2010-09-17";
echo date('Y-m-d', strtotime($Date. ' + 1 days'));
echo date('Y-m-d', strtotime($Date. ' + 2 days'));
?>
And it outputs correctly:
2010-09-18
2010-09-19
If you're using PHP 5.3, you can use a DateTime object and its add method:
$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->add(new DateInterval('P1D')); // P1D means a period of 1 day
$Date2 = $date->format('Y-m-d');
Take a look at the DateInterval constructor manual page to see how to construct other periods to add to your date (2 days would be 'P2D', 3 would be 'P3D', and so on).
Without PHP 5.3, you should be able to use strtotime the way you did it (I've tested it and it works in both 5.1.6 and 5.2.10):
$Date1 = '2010-09-17';
$Date2 = date('Y-m-d', strtotime($Date1 . " + 1 day"));
// var_dump($Date2) returns "2010-09-18"
From PHP 5.2 on you can use modify with a DateTime object:
http://php.net/manual/en/datetime.modify.php
$Date1 = '2010-09-17';
$date = new DateTime($Date1);
$date->modify('+1 day');
$Date2 = $date->format('Y-m-d');
Be careful when adding months... (and to a lesser extent, years)
Here is a small snippet to demonstrate the date modifications:
$date = date("Y-m-d");
//increment 2 days
$mod_date = strtotime($date."+ 2 days");
echo date("Y-m-d",$mod_date) . "\n";
//decrement 2 days
$mod_date = strtotime($date."- 2 days");
echo date("Y-m-d",$mod_date) . "\n";
//increment 1 month
$mod_date = strtotime($date."+ 1 months");
echo date("Y-m-d",$mod_date) . "\n";
//increment 1 year
$mod_date = strtotime($date."+ 1 years");
echo date("Y-m-d",$mod_date) . "\n";
You can also use the following format
strtotime("-3 days", time());
strtotime("+1 day", strtotime($date));
You can stack changes this way:
strtotime("+1 day", strtotime("+1 year", strtotime($date)));
Note the difference between this approach and the one in other answers: instead of concatenating the values +1 day and <timestamp>, you can just pass in the timestamp as the second parameter of strtotime.
Here has an easy way to solve this.
<?php
$date = "2015-11-17";
echo date('Y-m-d', strtotime($date. ' + 5 days'));
?>
Output will be:
2015-11-22
Solution has found from here - How to Add Days to Date in PHP
Using a variable for Number of days
$myDate = "2014-01-16";
$nDays = 16;
$newDate = strtotime($myDate . '+ '.$nDays.' days');
echo new Date('d/m/Y', $newDate); //format new date
Here is the simplest solution to your query
$date=date_create("2013-03-15"); // or your date string
date_add($date,date_interval_create_from_date_string("40 days"));// add number of days
echo date_format($date,"Y-m-d"); //set date format of the result
This works. You can use it for days, months, seconds and reformat the date as you require
public function reformatDate($date, $difference_str, $return_format)
{
return date($return_format, strtotime($date. ' ' . $difference_str));
}
Examples
echo $this->reformatDate('2021-10-8', '+ 15 minutes', 'Y-m-d H:i:s');
echo $this->reformatDate('2021-10-8', '+ 1 hour', 'Y-m-d H:i:s');
echo $this->reformatDate('2021-10-8', '+ 1 day', 'Y-m-d H:i:s');
To add a certain number of days to a date, use the following function.
function add_days_to_date($date1,$number_of_days){
/*
//$date1 is a string representing a date such as '2021-04-17 14:34:05'
//$date1 =date('Y-m-d H:i:s');
// function date without a secrod argument returns the current datetime as a string in the specified format
*/
$str =' + '. $number_of_days. ' days';
$date2= date('Y-m-d H:i:s', strtotime($date1. $str));
return $date2; //$date2 is a string
}//[end function]
All have to use bellow code:
$nday = time() + ( 24 * 60 * 60);
echo 'Now: '. date('Y-m-d') ."\n";
echo 'Next Day: '. date('Y-m-d', $nday) ."\n";
Another option is to convert your date string into a timestamp and then add the appropriate number of seconds to it.
$datetime_string = '2022-05-12 12:56:45';
$days_to_add = 1;
$new_timestamp = strtotime($datetime_string) + ($days_to_add * 60 * 60 * 24);
After which, you can use one of PHP's various date functions to turn the timestamp into a date object or format it into a human-readable string.
$new_datetime_string = date('Y-m-d H:i:s', $new_timestamp);

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