Stuck on this question for college!
subtract one time from another. The times are in the format “HH:MM:SS:FF” where:
HH = hours – there are 24 hours in a day
MM = minutes – there are 60 minutes in a hour
SS = seconds – there are 60 seconds in a minute
FF = frames – there are 25 frames in a second
The time is always 11 characters long and each element will always be 2 characters – padded with zeros if required.
subtract $y from $x
'$x = “03:14:59:20”;
$y = “01:16:01:02”;
I have done this code
<?php
$now = new DateTime(); // current date/time
$now = new DateTime("03:14:59:20");
$ref = new DateTime("01:16:01:02");
$diff = $now->diff($ref);
printf('%d hours, %d minutes %d seconds %d frames', $diff->h, $diff->i, $diff->s, $diff->f);
but the frames part does not work
Thanks in advance
Related
I'm having problem when getting the exact number of days. Given I have date/time which consider hours in counting number of days below the code give me zero days
$fisrstDate = new DateTime("2018-03-07 04:46:00");
$secondDate = new DateTime("2018-03-07 11:10:00");
$days=$fisrstDate->diff($secondDate)->days;
another example is this it should give me 2 days but shows only 1 days my idea is when 24 hours exceed I want to add another 1 days so that it would give me an output of 2 days
$fisrstDate = new DateTime("2018-03-07 04:46:00");
$secondDate = new DateTime("2018-03-08 05:00:00");
$days=$fisrstDate->diff($secondDate)->days;
You can use strtotime to get the exact seconds between two time stamps and then convert it to days followed by ceil to make it work. Eg:
$fisrstDate = strtotime("2018-03-07 04:46:00");
$secondDate = strtotime("2018-03-07 11:10:00");
$days = abs(ceil((abs($fisrstDate - $secondDate)/ (60 * 60 * 24)) - (1 / 24)));
echo $days;
Isn't it just date2 - date1 + 1?
I want to display a number of hours in days and hours as a human readable string but I need that 1 day be equal to 7 hours, so a working day.
I found this solution, but based on 1 day equal to 24 hours :
function secondsToTime($seconds) {
$dtF = new \DateTime('#0');
$dtT = new \DateTime("#$seconds");
return $dtF->diff($dtT)->format('%a days, %h hours, %i minutes and %s seconds');
}
source: Convert seconds into days, hours, minutes and seconds
DateTime won't help here. You need to count it on your own.
But this is simple math. Integer division for days and modulo for hours.
$hours = 12345;
$days = floor($hours/7);
$restHours = $hours%7;
echo "{$days} days, {$restHours} hours"; //1763 days, 4 hours
Custom made code to deal with this (probably on of 1000 ways on can implement this, I don't claim this to be the best, or good or anything like that).
function secondsToTime($seconds) {
$totalMinutes = intval($seconds / 60);
$totalHours = intval($totalMinutes / 60);
$totalDays = intval($totalHours / 7);
echo "$totalDays days and ".($totalHours%7)." hours ".($totalMinutes%60)." minutes and ".($seconds%60)." seconds";
}
e.g.
secondsToTime(8*60*60);
prints
1 days and 1 hours 0 minutes and 0 seconds
im trying to get difference of date/time from a field type datetime to "right now" using php and mysql as database
this code is working fine, returns the output beautifully ok as required
$datetime1 = new DateTime('mydate1');
$datetime2 = new DateTime();
$interval = $datetime1->diff($datetime2);
$elapsed = $interval->format('%d days %h hours %i minutes');
that is ok so far, no issues as this function is for php 5.3 and i have it on server
my need is 4 small things actually
1) how to eliminate the need for days?
i want to have (25 hours 10 minutes) instead of (1 day 1 hours 10 minutes)
2) how i can make $elapsed be bold or colored if the value is more than 5 hours for example!? simple IF logic will not work as the output is not actually a predefined value...
3) if the days or hours are 0, then want to remove them!
- For example if showing (0 days 10 hours 40 mins) then no need to display the (0 days), should show (10 hours 40 mins) that is enough
- Another example: 0 days 0 hours 45 minutes then to show only "45 minutes" no need for days and hours!
4) if output less than 5 minutes in total (0days 0 hours 1-5mins), then wanna make it show like "a while ago" only no need for any days, hours or minutes... then after 6 minutes.. go like "6 mins"
shortly something like facebook!?
okay, what i searched tried is different combination of workarounds but never worked as you know this interval is for php 5.3 and still seem not widely used?
any hint for one or more parts of this long question is appreciated,
M, Derik
tried to answer all your questions. so read the comments because i didnt write numbers for the problems. sorry
$datetime1 = new DateTime('mydate1');
$datetime2 = new DateTime();
$minutes = round(abs($datetime1 - $datetime2) / 60,2); //to calculate total time in MINUTES
if($minutes < 5) // for awhile ago problem.
{
return "awhile ago";
}
elseif($minutes > 6 && < 60)
{
return $minutes." minutes ago"; //for 6 minutes and after.
}
elseif($minutes>60) // for the hours..
{
$hours = floor($final_time_saving / 60);
$minutes = $final_time_saving % 60;
$string = $hours. " hours and" . $minutes . " minutes ago.";
if($hours>5) //for bolding characters after
{
return "<b>".$hours." hours and".$minutes." minutes ago.</b>"; //for bolding character
}
else
return $hours." hours and".$minutes." minutes ago.";
}
hope this helps.
How can I calculate the nearest hours to midnight time 00:00 regardless of date in PHP. For example:
If time is 22:00 then 2 hours are required to reach 00:00
If time is 04:00 then -4 hours are the nearest to reach 00:00
Currently I have the following PHP function:
<?php
$ts1 = strtotime('00:00');
$ts2 = strtotime('04:00');
$diff = ($ts1 - $ts2) / 3600;
?>
But this won't be helpful much in the above.
If you have the php Datetime class available you can calculate the difference between two DateTimes.
$time1 = new \DateTime('00:00');
$time2 = new \DateTime('04:00');
$diff = $time1->diff($time2, true);
$hourDifference = 0;
if ($diff->h < 12) {
$hourDifference = -$diff->h;
} elseif ($diff->h > 12) {
$hourDifference = 24 - $diff->h;
} else {
$hourDifference = 12; // kann be positive or negative
}
And you'll get a DateInverall object where you can access, hours, minuts, seconds and compare them with normal php operators.
If you'r not too interested in minutes;
1. Extract minutes.
check if minutes is > or <=30
if greater, 'store' 1
2. Extract hour
check if hour is greater than 12
if not, add 12 (store flag also to say it will be minus)
3. if greater (ref. Step 1), add 1 to extracted hour.
4. 24 - extracted hour is your interval.
Please note, this may be reduced/ simplified greatly.
Your interval (should) be correct to the nearest half hour
The answer depends on the date (not only the time). This is because of daylight saving time changes. For example might 02:59 being closer to 00:00 then 21:01 on the time where daylight saving time will set back hour.
Suppose the target time is 4.30 pm and the current time is 3.25 pm , how will i calculate the minutes remaining to reach the target time ? I need the result in minutes.
session_start();
$m=30;
//unset($_SESSION['starttime']);
if(!$_SESSION['starttime']){
$_SESSION['starttime']=date('Y-m-d h:i:s');
}
$stime=strtotime($_SESSION['starttime']);
$ttime=strtotime((date('Y-m-d h:i:s',strtotime("+$m minutes"))));-->Here I want to calcuate the target time; the time is session + 30 minutes. How will i do that
echo round(abs($ttime-$stime)/60);
Krishnik
A quick calculation of the difference between two times can be done like this:
$start = strtotime("4:30");
$stop = strtotime("6:30");
$diff = ($stop - $start); //Diff in seconds
echo $diff/3600; //Return 2 hours. Divide by something else to get in mins etc.
Edit*
Might as well add the answer to your problem too:
$start = strtotime("3:25");
$stop = strtotime("4:30");
$diff = ($stop - $start);
echo $diff/60; //Echoes 65 min
Oh and one more edit:) If the times are diffent dates, like start is 23:45 one day and end is 0:30 the next you need to add a date too to the strtotime.