I have today date and i want to reduce hours\days from it. i get the "hours to reduce interval" in int that indicate number of days.
I tried something like this:
$today_date = date('Y-m-d H:i:s');
$temp_interval_date = $settings->days_back;
$interval_date = date('H',$temp_interval_date*24);
$final = $temp_interval_date - $interval_date;
My final goal is to get todaydate - interval period in this format
'Y-m-d H:i:s'
I am c# dude :)
Thanks
I'm not entirely clear on what you're asking but I think this is what you're looking for.
$date = new DateTime();
$date->sub(new DatePeriod('P'.$settings->days_back.'D'));
echo $date->format('Y-m-d H:i:s');
You can also do (if you're using PHP 5.2)
$date = new DateTime();
$date->modify('-' . $settings->days_back . ' days'));
echo $date->format('Y-m-d H:i:s');
reference
DateTime()
DatePeriod()
maybe this could be helpful:
echo date('Y-m-d', strtotime('-1 day', date('Y-m-d') ));
Another way to do it
<?php
$temp_interval_date = 2; //in hours (for example =2)
echo date('Y-m-d H:i:s', strtotime('-'.($temp_interval_date*24).' hours',strtotime(date('Y-m-d H:i:s'))));
?>
Related
In my PHP program, I'm using $_SERVER to log the page's date visited:
$dateStamp = $_SERVER['REQUEST_TIME'];
The result is that the $dateStamp variable contains a Unix timestamp like:
1385615749
What's the simplest way to convert it into a human-readable date/time (with year, month, day, hour, minutes, seconds)?
This number is called Unix time. Functions like date() can accept it as the optional second parameter to format it in readable time.
Example:
echo date('Y-m-d H:i:s', $_SERVER['REQUEST_TIME']);
If you omit the second parameter the current value of time() will be used.
echo date('Y-m-d H:i:s');
Your functional approch to convert timestamp into Human Readable format are as following
function convertDateTime($unixTime) {
$dt = new DateTime("#$unixTime");
return $dt->format('Y-m-d H:i:s');
}
$dateVarName = convertDateTime(1385615749);
echo $dateVarName;
Output :-
2013-11-28 05:15:49
Working Demo
<?php
$date = new DateTime();
$dateStamp = $_SERVER['REQUEST_TIME'];
$date->setTimestamp($dateStamp);
echo $date->format('U = Y-m-d H:i:s') . "\n";
?>
you can try this
<?php
$date = date_create();
$dateStamp = $_SERVER['REQUEST_TIME'];
date_timestamp_set($date, $dateStamp);
echo date_format($date, 'U = D-M-Y H:i:s') . "\n";
?>
this code will work for you
$dateStamp = $_SERVER['REQUEST_TIME'];
echo date('d-M-Y H:i:s',strtotime($dateStamp));
REQUEST_TIME - It is unix timestamp - The timestamp of the start of the request.
$dateStamp = $_SERVER['REQUEST_TIME'];
echo date('d m Y', $dateStamp);
OR
$date = new DateTime($dateStamp);
echo $date->format('Y-m-d');
I'm really stuck with adding X minutes to a datetime, after doing lots of google'ing and PHP manual reading, I don't seem to be getting anywhere.
The date time format I have is:
2011-11-17 05:05: year-month-day hour:minute
Minutes to add will just be a number between 0 and 59
I would like the output to be the same as the input format with the minutes added.
Could someone give me a working code example, as my attempts don't seem to be getting me anywhere?
$minutes_to_add = 5;
$time = new DateTime('2011-11-17 05:05');
$time->add(new DateInterval('PT' . $minutes_to_add . 'M'));
$stamp = $time->format('Y-m-d H:i');
The ISO 8601 standard for duration is a string in the form of P{y}Y{m1}M{d}DT{h}H{m2}M{s}S where the {*} parts are replaced by a number value indicating how long the duration is.
For example, P1Y2DT5S means 1 year, 2 days, and 5 seconds.
In the example above, we are providing PT5M (or 5 minutes) to the DateInterval constructor.
PHP's DateTime class has a useful modify method which takes in easy-to-understand text.
$dateTime = new DateTime('2011-11-17 05:05');
$dateTime->modify('+5 minutes');
You could also use string interpolation or concatenation to parameterize it:
$dateTime = new DateTime('2011-11-17 05:05');
$minutesToAdd = 5;
$dateTime->modify("+{$minutesToAdd} minutes");
$newtimestamp = strtotime('2011-11-17 05:05 + 16 minute');
echo date('Y-m-d H:i:s', $newtimestamp);
result is
2011-11-17 05:21:00
Live demo is here
If you are no familiar with strtotime yet, you better head to php.net to discover it's great power :-)
You can do this with native functions easily:
strtotime('+59 minutes', strtotime('2011-11-17 05:05'));
I'd recommend the DateTime class method though, just posted by Tim.
I don't know why the approach set as solution didn't work for me.
So I'm posting here what worked for me in hope it can help anybody:
$startTime = date("Y-m-d H:i:s");
//display the starting time
echo '> '.$startTime . "<br>";
//adding 2 minutes
$convertedTime = date('Y-m-d H:i:s', strtotime('+2 minutes', strtotime($startTime)));
//display the converted time
echo '> '.$convertedTime;
I thought this would help some when dealing with time zones too. My modified solution is based off of #Tim Cooper's solution, the correct answer above.
$minutes_to_add = 10;
$time = new DateTime();
**$time->setTimezone(new DateTimeZone('America/Toronto'));**
$time->add(new DateInterval('PT' . $minutes_to_add . 'M'));
$timestamp = $time->format("Y/m/d G:i:s");
The bold line, line 3, is the addition. I hope this helps some folks as well.
A bit of a late answer, but the method I would use is:
// Create a new \DateTime instance
$date = DateTime::createFromFormat('Y-m-d H:i:s', '2015-10-26 10:00:00');
// Modify the date
$date->modify('+5 minutes');
// Output
echo $date->format('Y-m-d H:i:s');
Or in PHP >= 5.4
echo (DateTime::createFromFormat('Y-m-d H:i:s', '2015-10-26 10:00:00'))->modify('+5 minutes')->format('Y-m-d H:i:s')
If you want to give a variable that contains the minutes.
Then I think this is a great way to achieve this.
$minutes = 10;
$maxAge = new DateTime('2011-11-17 05:05');
$maxAge->modify("+{$minutes} minutes");
Use strtotime("+5 minute", $date);
Example:
$date = "2017-06-16 08:40:00";
$date = strtotime($date);
$date = strtotime("+5 minute", $date);
echo date('Y-m-d H:i:s', $date);
As noted by Brad and Nemoden in their answers above, strtotime() is a great function. Personally, I found the standard DateTime Object to be overly complicated for many use cases. I just wanted to add 5 minutes to the current time, for example.
I wrote a function that returns a date as a string with some optional parameters:
1.) time:String | ex: "+5 minutes" (default = current time)
2.) format:String | ex: "Y-m-d H:i:s" (default = "Y-m-d H:i:s O")
Obviously, this is not a fully featured method. Just a quick and simple function for modifying/formatting the current date.
function get_date($time=null, $format='Y-m-d H:i:s O')
{
if(empty($time))return date($format);
return date($format, strtotime($time));
}
// Example #1: Return current date in default format
$date = get_date();
// Example #2: Add 5 minutes to the current date
$date = get_date("+5 minutes");
// Example #3: Subtract 30 days from the current date & format as 'Y-m-d H:i:s'
$date = get_date("-30 days", "Y-m-d H:i:s");
one line mysql datetime format
$mysql_date_time = (new DateTime())->modify('+15 minutes')->format("Y-m-d H:i:s");
One more example of a function to do this: (changing the time and interval formats however you like them according to this for function.date, and this for DateInterval):
(I've also written an alternate form of the below function.)
// Return adjusted time.
function addMinutesToTime( $dateTime, $plusMinutes ) {
$dateTime = DateTime::createFromFormat( 'Y-m-d H:i', $dateTime );
$dateTime->add( new DateInterval( 'PT' . ( (integer) $plusMinutes ) . 'M' ) );
$newTime = $dateTime->format( 'Y-m-d H:i' );
return $newTime;
}
$adjustedTime = addMinutesToTime( '2011-11-17 05:05', 59 );
echo '<h1>Adjusted Time: ' . $adjustedTime . '</h1>' . PHP_EOL . PHP_EOL;
Without using a variable:
$yourDate->modify("15 minutes");
echo $yourDate->format( "Y-m-d H:i");
With using a variable:
$interval= 15;
$yourDate->modify("+{$interval } minutes");
echo $yourDate->format( "Y-m-d H:i");
I want to add 5 minutes to this date: 2011-04-8 08:29:49
$date = '2011-04-8 08:29:49';
When I use strtotime I am always getting 1970-01-01 08:33:31
How do I add correctly 5 minutes to 2011-04-8 08:29:49?
$date = '2011-04-8 08:29:49';
$currentDate = strtotime($date);
$futureDate = $currentDate+(60*5);
$formatDate = date("Y-m-d H:i:s", $futureDate);
Now, the result is 2011-04-08 08:34:49 and is stored inside $formatDate
Enjoy! :)
Try this:
echo date('Y-m-d H:i:s', strtotime('+5 minutes', strtotime('2011-04-8 08:29:49')));
$expire_stamp = date('Y-m-d H:i:s', strtotime("+5 min"));
$now_stamp = date("Y-m-d H:i:s");
echo "Right now: " . $now_stamp;
echo "5 minutes from right now: " . $expire_stamp;
Results in:
2012-09-30 09:00:03
2012-09-30 09:05:03
$date = '2011-04-8 08:29:49';
$newDate = date("Y-m-d H:i:s",strtotime($date." +5 minutes"))
For adding
$date = new DateTime('2014-02-20 14:20:00');
$date->add(new DateInterval('P0DT0H5M0S'));
echo $date->format('Y-m-d H:i:s');
It add 5minutes
For subtracting
$date = new DateTime('2014-02-20 14:20:00');
$date->sub(new DateInterval('P0DT0H5M0S'));
echo $date->format('Y-m-d H:i:s');
It subtract 5 minutes
If i'm right in thinking.
If you convert your date to a unix timestamp via strtotime(), then just add 300 (5min * 60 seconds) to that number.
$timestamp = strtotime($date) + (5*60)
Hope this helps
more illustrative for simple and clear solution
$date = '2011-04-8 08:29:49';
$newtimestamp = strtotime($date. ' + 5 minute');//gets timestamp
//convert into whichever format you need
$newdate = date('Y-m-d H:i:s', $newtimestamp);//it prints 2011-04-08 08:34:49
In certain situations I want to add 1 day to the value of my DATETIME formatted variable:
$start_date = date('Y-m-d H:i:s', strtotime("{$_GET['start_hours']}:{$_GET['start_minutes']} {$_GET['start_ampm']}"));
What is the best way to do this?
There's more then one way to do this with DateTime which was introduced in PHP 5.2. Unlike using strtotime() this will account for daylight savings time and leap year.
$datetime = new DateTime('2013-01-29');
$datetime->modify('+1 day');
echo $datetime->format('Y-m-d H:i:s');
// Available in PHP 5.3
$datetime = new DateTime('2013-01-29');
$datetime->add(new DateInterval('P1D'));
echo $datetime->format('Y-m-d H:i:s');
// Available in PHP 5.4
echo (new DateTime('2013-01-29'))->add(new DateInterval('P1D'))->format('Y-m-d H:i:s');
// Available in PHP 5.5
$start = new DateTimeImmutable('2013-01-29');
$datetime = $start->modify('+1 day');
echo $datetime->format('Y-m-d H:i:s');
If you want to do this in PHP:
// replace time() with the time stamp you want to add one day to
$startDate = time();
date('Y-m-d H:i:s', strtotime('+1 day', $startDate));
If you want to add the date in MySQL:
-- replace CURRENT_DATE with the date you want to add one day to
SELECT DATE_ADD(CURRENT_DATE, INTERVAL 1 DAY);
The DateTime constructor takes a parameter string time. $time can be different things, it has to respect the datetime format.
There are some valid values as examples :
'now' (the default value)
2017-10-19
2017-10-19 11:59:59
2017-10-19 +1day
So, in your case you can use the following.
$dt = new \DateTime('now +1 day'); //Tomorrow
$dt = new \DateTime('2016-01-01 +1 day'); //2016-01-02
Use strtotime to convert the string to a time stamp
Add a day to it (eg: by adding 86400 seconds (24 * 60 * 60))
eg:
$time = strtotime($myInput);
$newTime = $time + 86400;
If it's only adding 1 day, then using strtotime again is probably overkill.
You can use
$now = new DateTime();
$date = $now->modify('+1 day')->format('Y-m-d H:i:s');
You can use as following.
$start_date = date('Y-m-d H:i:s');
$end_date = date("Y-m-d 23:59:59", strtotime('+3 days', strtotime($start_date)));
You can also set days as constant and use like below.
if (!defined('ADD_DAYS')) define('ADD_DAYS','+3 days');
$end_date = date("Y-m-d 23:59:59", strtotime(ADD_DAYS, strtotime($start_date)));
I suggest start using Zend_Date classes from Zend Framework. I know, its a bit offtopic, but I'll like this way :-)
$date = new Zend_Date();
$date->add('24:00:00', Zend_Date::TIMES);
print $date->get();
Using server request time to Add days. Working as expected.
25/08/19 => 27/09/19
$timestamp = $_SERVER['REQUEST_TIME'];
$dateNow = date('d/m/y', $timestamp);
$newDate = date('d/m/y', strtotime('+2 day', $timestamp));
Here '+2 days' to add any number of days.
One liner !
echo (new \DateTime('2016-01-01 +1 day'))->format('Y-m-d H:i:s');
Let me know :)
$add_date = date ("Y-m-d H:m:s");
$expiry_date = 'how?';
How to insert into db the $expiry_date for 60 days. mysql format is datetime
Use strtotime():
$start_date = date('Y-m-d H:m:s');
$end_date = date('Y-m-d H:m:s', strtotime("+60 days"));
or more simply:
$end_date = date('Y-m-d H:m:s', time() + 86400 * 60);
A method avoiding time conversions:
$time = date('Y-m-d H:m:s', time()+3600*24*60)
EDIT
However, it may be less readable and the time saved is probably irrelevant. Plus cletus just edited a similar method into his answer
If you are using PHP >= 5.2 I strongly suggest you use the new DateTime object. For example like below:
$add_date = date("Y-m-d H:m:s");
$expiry_date = new DateTime($add_date);
$expiry_date ->modify("+60 days");
echo $expiry_date ->format("Y-m-d H:m:s");
Live Demo