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.
Related
The timestamp in my database is 2015-03-03 00:25:39 (Take note that the type = timestamp and the correct current timestamp in my end is 2015-03-02 01:31:00. The difference should be around 23 hours. But now the problem is that the answers provided in the net will give me 30 hours instead of 23 hours. Some of the codes that I have tried are the following:
$target is the target date
CODE 1:
$then = strtotime($target);
$diff = $then - time();
echo sprintf("%s days and %s hours left", date('z', $diff), date('G', $diff));
But it gives me 1 days and 6 hours left. So 30 hours
CODE2:
$seconds = strtotime("$target") - time();
echo $seconds; exit();
$days = floor($seconds / 86400);
$seconds %= 86400;
$hours = floor($seconds / 3600);
echo $hours;
It gives me something like 107388 = 30 hours.
CODE 3:
//Convert to date
$datestr= $target;//Your date
$date=strtotime($datestr);//Converted to a PHP date (a second count)
//Calculate difference
$diff=$date-time();//time returns current time in seconds
$days=floor($diff/(60*60*24));//seconds/minute*minutes/hour*hours/day)
$hours=round(($diff-$days*60*60*24)/(60*60));
It gives me 6 hours
I don't know what I'm doing wrong, more like I have no idea how to do it.
This is now my last resort since I can't find the solution that will help me.
Hoping for your fast responses.
PHP's DateTime() (and DateInterval()) are much better for date math and returns the correct results:
$date = new DateTime('2015-03-03 00:25:39');
$now = new DateTime('2015-03-02 01:31:00');
$diff = $date->diff($now);
echo $diff->h, ' hours ', $diff->i, ' minutes';
Demo
This is a very late answer, but your question is a good one that will likely be searched for in the future.
Here is an online demo.
// this doesn't appreciate any timezone declarations, you'll need to add this if necessary
$target="2015-03-03 00:25:39"; // declare your input
$then=new DateTime($target); // feed input to DateTime
$now=new DateTime(); // get DateTime for Now
$diff=(array)$then->diff($now); // calculate difference & cast as array
$labels=array("y"=>"year","m"=>"month","d"=>"day","h"=>"hour","i"=>"minute","s"=>"second");
$readable=""; // declare as empty string
// filter the $diff array to only include the desired elements and loop
foreach(array_intersect_key($diff,$labels) as $k=>$v){
if($v>0){ // only add non-zero values to $readable
$readable.=($readable!=""?", ":"")."$v {$labels[$k]}".($v>1?"s":"");
// use comma-space as glue | show value | show unit | pluralize when necessary
}
}
echo "$readable";
// e.g. 2 years, 20 days, 1 hour, 10 minutes, 40 seconds
I am creating an application that charges a client based on time usage of a service. The problem is that the services can have double charge for a pre-specified time period of the day.
So let's say we have a service for printing documents and the charge for using the printer is 5€ per hour and 10€ between 23:00 and 02:00 in the morning. Also a client can rent the printer for as much time as he likes. This can be from 1 minute to months or even years.
Now the specific problem:
Let's say a client comes in my office to rent the printer for 55 hours. Also the rent starts at 20:00 at night.
So the charge must be for 43 hours in single charge and for 12 hours in double charge. Here are two example images:
Now, let me give you some extra info about the hours. In programming, each hour has a timestamp that it is time passed from January 1, 1970 00:00:00 to the time in seconds.
So the date July 05 2012 11:15:40 has the timestamp 1373022940 and the date July 05 2012 11:15:50 has the timestamp 1373022950
In the above example lets say that the first example placed in the date May 1, 2013, so the timestamp for 23:00 will be 1367449200 and the time stamp for three days later at the 02:00 the morning is 1367546400
Now the question:
Is there a way to extract the time duration of the double charged hours from a time frame? If so, what is the process?
Of course there is. You just need to count the interval between dates.
Let's say someone started using service since 8:00 and ended in 16:00.
Price from 8:00 - 16:00 = 2$
Price from 16:00 - 8:00 = 1$
So you need to convert the start of usage time and end of usage time to timestamp
date_default_timezone_set('UTC');
$day_start = '2011-06-22';
$day_end = '2011-06-22';
$start_usage = strtotime($day_start.' 8:00');
$end_usage = strtotime($day_end.' 17:00');
$price_low_rate = 1; //price for using service 16-8
$price_high_rate = 2; // price for using service since 8-16
$fee_for_eight_sixteen = 0; // total price for using service since 8-16
$fee_for_sixteen_eight = 0; // total price for using service 16-8
if($end_usage >strtotime($day_start.' 16:01'))
{
$fee_for_sixteen_eight = ($end_usage - strtotime($day_end.' 16:00'))/3600 * $price_low_rate;
}
if($start_usage >= strtotime($day_start.' 8:00'))
{
$fee_for_eight_sixteen = (strtotime($day_end.' 16:00') - $start_usage)/3600 * $price_high_rate;
}
echo $fee_for_eight_sixteen.' - '.$fee_for_sixteen_eight;
I've tested it and it works. Hope it helps.
Haven't tested this, but I hope it gets you on the right track:
<?php
$price = 0;
$start_timestamp = 1367449200;
$end_timestamp = 1367546400;
$start_time_of_day = 1367449200 % (24*60*60); // Number of seconds from start of the day
$end_time_of_day = 1367546400 % (24*60*60); // Number of seconds from start of the day
// How much time the time period spends in the first day (in seconds)
$first_day_time = (24*60*60) - $start_time_of_day;
// How much time the time period spends in the last day (in seconds)
$last_day_time = (24*60*60) - $end_time_of_day;
$full_days_time = $end_timestamp + $last_day_time - ($start_timestamp + $first_day_time);
$full_days = round($full_days_time/(24*60*60));
// You can calculate by hand how much one full 24-hour day from 00:00 to 00:00 costs
$price += $full_days * (2*10 + 21*5 + 1*10);
// so now the difficulty is the pieces of time on the first day and the last day.
$expensive_time = 0; // Expensive time spent on the first and last day
$cheap_time = 0;
if ($start_time_of_day<2*60*60)
{
// Renting starts before 02:00
$expensive_time += 2*60*60 - $start_time_of_day;
$cheap_time += 21*60*60; // Full 21 hours of cheap time
$expensive_time += 1*60*60; // 1 hour of expensive time from 23:00 to midnight
}
elseif ($start_time_of_day<23*60*60)
{
// Renting starts after 02:00 and before 23:00
$cheap_time += 23*60*60 - $start_time_of_day;
$expensive_time += 1*60*60; // 1 hour of expensive time from 23:00 to midnight
}
else
{
// Renting starts after 23:00
$expensive_time += 24*60*60 - $start_time_of_day;
}
// !! Use a similar strategy for the $end_time_of_day here
$price += ceil($expensive_time/60/60) * 10;
$price += ceil($cheap_time/60/60) * 5;
echo $price." euro";
?>
I'm getting a list of items from my database, each has a CURRENT_TIMESTAMP which i have changed into 'x minutes ago' with the help of timeago. So that's working fine. But the problem is i also want a "NEW" banner on items which are less than 30 minutes old. How can i take the generated timestamp (for example: 2012-07-18 21:11:12) and say if it's less than 30 minutes from the current time, then echo the "NEW" banner on that item.
Use strtotime("-30 minutes") and then see if your row's timestamp is greater than that.
Example:
<?php
if(strtotime($mysql_timestamp) > strtotime("-30 minutes")) {
$this_is_new = true;
}
?>
I'm using strtotime() twice here to get unix timestamps for your mysql date, and then again to get what the timestamp was 30 minutes ago. If the timestamp from 30 mins ago is greater than the timestamp of the mysql record, then it must have been created more than 30 minutes go.
Try something like this, using PHP's DateTime object:
$now = new DateTime();
$then = DateTime($timestamp); // "2012-07-18 21:11:12" for example
$diff = $now->diff($then);
$minutes = ($diff->format('%a') * 1440) + // total days converted to minutes
($diff->format('%h') * 60) + // hours converted to minutes
$diff->format('%i'); // minutes
if ($minutes <= 30) {
echo "NEW";
}
Edit: Mike is right, I forgot that for whatever reason, only %a actually returns the total of its type (in this case days). All the others are for displaying time formatting. I've extended the above to actually work.
You can do like this also -
$currentTime=time();
to check the last updated time is 30 minute old or not
last_updated_at < $currentTime - (60*30)
I have a scenario in which the user selects a time and day (or multiple days) and that value must be converted to whatever that day and time would be in UTC time. I have the gmt offset amount for each user (the users set it when they signup). For instance:
A user in the eastern timezone selects:
3:15 pm, Monday, Tuesday, Friday
I need to know what time and days that information would be in UTC time. The solution has to take into situations such Monday in one timezone can be a different day in UTC time. Also, if the time can be converted to 24 hour format, that would be a plus.
For the sake of clarity, something along the lines of an array should be returned such as:
Array('<3:15 pm eastern adjusted for utc>', '<Monday adjusted for UTC>', '<Tuesday adjusted for UTC>', '<Friday adjusted for UTC>');
I don't need the result to be directly formatted into an array like that - that's just the end goal.
I am guessing it involves using strtotime, but I just can't quite my finger out how to go about it.
$timestamp = strtotime($input_time) + 3600*$time_adjustment;
The result will be a timestamp, here's an example:
$input_time = "3:15PM 14th March";
$time_adjustment = +3;
$timestamp = strtotime($input_time) + 3600*$time_adjustment;
echo date("H:i:s l jS F", $timestamp);
// 16:15:00 Monday 14th March
EDIT: kept forgetting little things, that should be working perfectly now.
Made a function to do the job:
<?
/*
* The function week_times() converts a a time and a set of days into an array of week times. Week times are how many seconds into the week
* the given time is. The $offset arguement is the users offset from GMT time, which will serve as the approximation to their
* offset from UTC time
*/
// If server time is not already set for UTC, uncomment the following line
//date_default_timezone_set('UTC');
function week_times($hours, $minutes, $days, $offset)
{
$timeUTC = time(); // Retrieve server time
$hours += $offset; // Add offset to user time to make it UTC time
if($hours > 24) // Time is more than than 24 hours. Increment all days by 1
{
$dayOffset = 1;
$hours -= 24; // Find out what the equivelant time would be for the next day
}
else if($hours < 0) // Time is less than 0 hours. Decrement all days by 1
{
$dayOffset = -1;
$hours += 24; // Find out what the equivelant time would be for the prior day
}
$return = Array(); // Times to return
foreach($days as $k => $v) // Iterate through each day and find out the week time
{
$days[$k] += $dayOffset;
// Ensure that day has a value from 0 - 6 (0 = Sunday, 1 = Monday, .... 6 = Saturday)
if($days[$k] > 6) { $days[$k] = 0; } else if($days[$k] < 0) { $days[$k] = 6; }
$days[$k] *= 1440; // Find out how many minutes into the week this day is
$days[$k] += ($hours*60) + $minutes; // Find out how many minutes into the day this time is
}
return $days;
}
?>
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.