What I'm trying to do is to return the rest of this simple equation as date.
$time = (time()+(60*60*12)) - time();
$date = date('d H:i:s', $time);
echo $date;
As you see i have added 12 hours, but this added 1 day + 2 hours. And i stuck here :/.
Output is 01 14:00:00
Expected is 00 12:00:00
So what i'm doing wrong?
EDIT
I have tried date_default_timezone_set() function but it seems to not work for me :/
NEW EDIT
I realize when i do this.
$time = time() - time();
$date = date('d H:i:s', $time);
echo $date;
Print out 01 02:00:00 what is that?
You can do this using DateTime-
$start = new DateTime();
$end = new DateTime();
$end->modify('+12 hour');
$interval = $end->diff($start);
$elapsed = $interval->format('%a days %h hours %i minutes %s seconds');
echo $elapsed;
Not the answer to your question, but take a look at Carbon, "A simple PHP API extension for DateTime.". Very useful for these kind of DateTime-problems.
Related
I currently have php returning the current date/time like so:
$now = date("Y-m-d H:m:s");
What I'd like to do is have a new variable $new_time equal $now + $hours, where $hours is a number of hours ranging from 24 to 800.
Any suggestions?
You may use something like the strtotime() function to add something to the current timestamp. $new_time = date("Y-m-d H:i:s", strtotime('+5 hours')).
If you need variables in the function, you must use double quotes then like strtotime("+{$hours} hours"), however better you use strtotime(sprintf("+%d hours", $hours)) then.
An other solution (object-oriented) is to use DateTime::add
Example:
<?php
$now = new DateTime(); //now
echo $now->format('Y-m-d H:i:s'); // 2021-09-11 01:01:55
$hours = 36; // hours amount (integer) you want to add
$modified = (clone $now)->add(new DateInterval("PT{$hours}H")); // use clone to avoid modification of $now object
echo "\n". $modified->format('Y-m-d H:i:s'); // 2021-09-12 13:01:55
Run script
DateTime::add PHP doc
DateInterval::construct PHP doc
You can use strtotime() to achieve this:
$new_time = date("Y-m-d H:i:s", strtotime('+3 hours', $now)); // $now + 3 hours
Correct
You can use strtotime() to achieve this:
$new_time = date("Y-m-d H:i:s", strtotime('+3 hours', strtotime($now))); // $now + 3 hours
You can also use the unix style time to calculate:
$newtime = time() + ($hours * 60 * 60); // hours; 60 mins; 60secs
echo 'Now: '. date('Y-m-d') ."\n";
echo 'Next Week: '. date('Y-m-d', $newtime) ."\n";
Um... your minutes should be corrected... 'i' is for minutes. Not months. :) (I had the same problem for something too.
$now = date("Y-m-d H:i:s");
$new_time = date("Y-m-d H:i:s", strtotime('+3 hours', $now)); // $now + 3 hours
I use this , its working cool.
//set timezone
date_default_timezone_set('GMT');
//set an date and time to work with
$start = '2014-06-01 14:00:00';
//display the converted time
echo date('Y-m-d H:i',strtotime('+1 hour +20 minutes',strtotime($start)));
for add 2 hours to "now"
$date = new DateTime('now +2 hours');
or
$date = date("Y-m-d H:i:s", strtotime('+2 hours', $now)); // as above in example
or
$now = new DateTime();
$now->add(new DateInterval('PT2H')); // as above in example
You can try lib Ouzo goodies, and do this in fluent way:
echo Clock::now()->plusHours($hours)->format("Y-m-d H:m:s");
API's allow multiple operations.
For a given DateTime, you can add days, hours, minutes, etc. Here's some examples:
$now = new \DateTime();
$now->add(new DateInterval('PT24H')); // adds 24 hours
$now->add(new DateInterval('P2D')); // adds 2 days
PHP: DateTime::add - Manual https://www.php.net/manual/fr/datetime.add.php
$to = date('Y-m-d H:i:s'); //"2022-01-09 12:55:46"
$from = date("Y-m-d H:i:s", strtotime("$to -3 hours")); // 2022-01-09 09:55:46
$date_to_be-added="2018-04-11 10:04:46";
$added_date=date("Y-m-d H:i:s",strtotime('+24 hours', strtotime($date_to_be)));
A combination of date() and strtotime() functions will do the trick.
$now = date("Y-m-d H:i:s");
date("Y-m-d H:i:s", strtotime("+1 hours $now"));
I need to calculate the datetime difference in minutes using PHP. I am explaining my code below .
$date='10-03-2018 03:44 PM';
$endTime = strtotime($date);
$currentDate=date("d-m-Y h:i A");//10-03-2018 03:53 PM
$currentTime = strtotime($currentDate);
echo (round(abs($currentTime - $endTime) / 60,2));//25344617
Here I need to calculate the difference in minutes but the differnce value is more where the expected time difference should be 9 but as per my code I am getting the wrong value.
Let the PHP DateTime class with diff() method do the work with time calculations.
$now = '10-03-2018 03:53 PM'; // or use simply 'now' for current time
$endTime = '10-03-2018 03:44 PM';
$datetime1 = new DateTime($now);
$datetime2 = new DateTime($endTime);
$interval = $datetime1->diff($datetime2);
echo $interval->format('%i minutes'); // 9 minutes
See it live: https://eval.in/969615
I have this datetime code below. I want to subtract 23 hours and 59 minutes in $begin variable:
$date = 2016-03-14 23:59:59;
$given = new DateTime($date, new DateTimeZone("Asia/Tokyo"));
$given->setTimezone(new DateTimeZone("UTC"));
$end = $given->format("Y-m-d H:i:s");
$begin = date('Y-m-d H:i:s',strtotime('-23 hours',strtotime($end)));
Now The output is:
$end is: 2016-03-14 14:59:59
$begin is: 2016-03-13 15:59:59
The output I want is:
$end is: 2016-03-14 14:59:59
$begin is: 2016-03-13 15:00:00
How can I subtract the minutes in seconds in begin? Or there is a best way to do it?
As you just want to change the hour, you can make this to get the desire result, but you also can do something else like Amit Roy did or ceejayoz suggest. This is a pretty simple solution, just do it: reset the min and sec to 00 so that your output shows like you want.
echo $begin = date('Y-m-d H:00:00',strtotime('-23 hours', strtotime($end)));
Use the DateInterval class with the DateTime::sub method.
Example:
$begin_datetime = $given->sub(new DateInterval('PT23H59M00S'));
$begin = $begin_datetime->format('Y-m-d H:i:s');
http://php.net/manual/en/datetime.sub.php
http://php.net/manual/en/dateinterval.construct.php
[As Frayne make the comment.] As you just want to change the hour, not the min or sec, you have to set the min and sec to 00:00 as your desire output.
echo $begin = date('Y-m-d H:00:00',strtotime('-23 hours', strtotime($end)));
You can did this thing in many different ways, and this is one way.
You have to do this to get desired result. You were subtracting hours alone
$begin = date('Y-m-d H:i:s',strtotime('-23 hours',strtotime($end)));
You have to subtract min and sec too
$begin = date('Y-m-d H:i:s',strtotime('-23 hours -59 minutes -59 seconds',strtotime($end)));
Finally you get this code:
$date = "2016-03-14 23:59:59";
$given = new DateTime($date, new DateTimeZone("Asia/Tokyo"));
$given->setTimezone(new DateTimeZone("UTC"));
$end = $given->format("Y-m-d H:i:s");
echo $begin = date('Y-m-d H:i:s',strtotime('-23 hours -59 minutes -59 seconds',strtotime($end))); //2016-03-13 15:00:00
i want to find out the date after days from the given time.
for example. we have date 29 may 2015
and i want to cqlculate the date after 2 days of 25 may 2015
$Timestamp = 1432857600 (unix time of 29-05-2015)
i have tried to do it with following code but it is not working
$TotalTimeStamp = strtotime('2 days', $TimeStamp);
Missed the + - strtotime('2 days', $TimeStamp); .
Add the + to + 2 days.
Use date & strtotime for this - You can try this -
echo date('d-m-Y',strtotime(' + 2 day', strtotime('2015-05-16')));
$Timestamp & $TimeStamp are not same(may be typo). For your code -
$Timestamp = strtotime(date('Y-m-d'));
$TotalTimeStamp = strtotime('+ 2 days', $Timestamp);
echo date('d-m-Y', $TotalTimeStamp);
Php does have a pretty OOP Api to deal with date and time.
This will create a \DateTime instance using as reference the 25 May 2015 and then you can call the modify method on that instance to add 2 days.
$date = new \DateTime('2015-05-25');
$date->modify('+2 day');
echo $date->format('Y-m-d');
You may find this resource useful:
http://code.tutsplus.com/tutorials/dates-and-time-the-oop-way--net-35395
You can also just add seconds to your timestamp if you have a timestamp ready:
$NewDateStamp = $Timestamp + (60*60*24 * 2);
In the above, sec * min * hours = day -- or 86400 seconds. * 2 = 2 days.
In PHP 5 you can also use D
<?php
$date = date_create('2015-05-16');
date_add($date, date_interval_create_from_date_string('2 days'));
echo date_format($date, 'Y-m-d');
?>
OR
<?php
$date = new DateTime('2015-05-16');
$date->add(new DateInterval('2 days'));
echo $date->format('Y-m-d') . "\n";
?>
I want to calculate EXACT past 30 days time period in php from now (for example 30 aug 14 23:06) to 30 days back (for example 1 aug 14 23:06). I wrote this where current datetime goes in $d1 and past 30 days datetime goes in $d2 but somehow i am not getting correct results. Any idea?
$url=$row["url"];
$pageid=getPageID($url);
$date=date('y-m-d g:i');
$d1=strtotime($date);
$d2=date(strtotime('today - 30 days'));
Thanks
The problem is likely caused by the malformed date() call. The first argument passed to date() should be the format (as shown in the Docs) and the second should be an optional timestamp.
Try this:
$d2 = date('c', strtotime('-30 days'));
PHPFiddle
As a short aside, the whole snippet can be simplified as follows:
$url = $row["url"];
$pageid = getPageID($url);
$date = date('y-m-d g:i');
$d1 = time();
$d2 = date('y-m-d g:i', strtotime('-30 days'));
You can also use the DateTime class's sub() method together with an DateInterval:
$now = new DateTime();
$back = $now->sub(DateInterval::createFromDateString('30 days'));
echo $back->format('y-m-d g:i');
if you would like to get out put as 2014-08-01 then try the below code. thanks
$date = '2014-08-30 23:06';
$new_date = date('Y-m-d G:i', strtotime($date.' - 29 days'));
echo "30 days back is " . $new_date;
From your brief description and example given, I believe that you want the date to be 30 days back and time to be the same as of now. The below code will serve this purpose. Thanks.
<?php
$date=date('y-m-d g:i');
$time=date('g:i');
echo "Todays date:" . $date. "<br>";
$d2 = date('y-m-d', strtotime('-30 days'));
echo "30 days back:" . $d2 . ' ' .$time;
?>
Try:
echo date("Y-m-d h:i:s",strtotime('-30 days'));
For more detail click here
Very simple two lines of code
$date = new DateTime();
echo $date->modify('-30 day')->format('y-m-d g:i');
I know you said with PHP, however, I can't imagine not getting the records from a DB. If you want to do so from the DB,use:
$sql='SELECT * FROM myTable WHERE date > CURRENT_DATE - INTERVAL 30 DAY';
$pdo->query($sql);
Very simple one lines of code:
echo (new DateTime())->modify('-30 day')->format('y-m-d g:i');
In the example below, it makes no sense if the variable $date is not
used anywhere else!
$date = new DateTime();
echo $date->modify('-30 day')->format('y-m-d g:i');
Sample answer is
$dateBack30Days=date('Y-m-d g:i', strtotime('-30 days'));