How do I add 45 minutes to an existing time? - php

I am trying to add two time using strtotime.i use following code to add two time.
my expected output is 15:59:00 but it gives me 09:30:44. so what is the problem?
Thanks in advance.
$json['time'] = "15:14";
$total_duration = "45"; //in minutes
/* convert minutes into hours-minutes */
$hours = intval($total_duration / 60);
$mins = $total_duration % 60;
$service_duration = strtotime($hours."00:00")+strtotime("00:".$mins.":00");
/* end time = total service duration + start time */
$start_time = strtotime($json['time'].":00");
echo $service_duration."--".$start_time."==";
$end_time2 = $start_time+$service_duration;
$end_time = date('H:i:s', $end_time2);echo $end_time;exit;

You can do like
$json['time'] = "15:14";
$total_duration = "45"; //in minutes
$endTime = strtotime("+".$total_duration." minutes", strtotime($json['time']));
echo date('H:i:s', $endTime);
Output:
15:59:00

Related

Calculate Total time from array in php if total time greater than 24 hours

I want to get the sum of the time in array. There are a lot of questions asked before related this question. Only problem this solution work the only sum is less than 24 hours. After 24 hours it will start at 00:00:00. How do I get more than 24 hours as total?
<?php
$total = [
'00:02:55',
'00:07:56',
'01:03:32',
'15:13:34',
'02:13:44',
'03:08:53',
'13:13:54'
];
$sum = strtotime('00:00:00');
$sum2=0;
foreach ($total as $v){
$sum1=strtotime($v)-$sum;
$sum2 = $sum2+$sum1;
}
$sum3=$sum+$sum2;
echo date("H:i:s",$sum3);
?>
RESULT
11:04:28
Expected result
35:04:28
DEMO LINK
Try the following code
<?php
function explode_time($time) { //explode time and convert into seconds
$time = explode(':', $time);
$time = $time[0] * 3600 + $time[1] * 60;
return $time;
}
function second_to_hhmm($time) { //convert seconds to hh:mm
$hour = floor($time / 3600);
$minute = strval(floor(($time % 3600) / 60));
if ($minute == 0) {
$minute = "00";
} else {
$minute = $minute;
}
$time = $hour . ":" . $minute;
return $time;
}
$time = 0;
$time_arr = [
'00:02:55',
'00:07:56',
'01:03:32',
'15:13:34',
'02:13:44',
'03:08:53',
'13:13:54'
];
foreach ($time_arr as $time_val) {
$time +=explode_time($time_val); // this fucntion will convert all hh:mm to seconds
}
echo second_to_hhmm($time);
?>
With the external DateTime Extension dt you can add all times to a date.
With DateTime::diff you get the result:
$dt = dt::create("2000-1-1"); //fix Date
$dtsum = clone $dt;
foreach($total as $time){
$dtsum->addTime($time);
}
$diff = $dt->diff($dtsum);
printf('%d:%02d:%02d',$diff->days * 24 + $diff->h,$diff->i,$diff->s);
Output:
35:04:28
Update
Without a DateTime-Extension:
$dt = date_create("2000-1-1"); //fix Date
$dtsum = clone $dt;
foreach($total as $time){
$timeArr = explode(":",$time);
$secondsAdd = $timeArr[0] * 3600 + $timeArr[1] * 60 +$timeArr[2];
$dtsum->modify($secondsAdd." Seconds");
}
$diff = $dt->diff($dtsum);
printf('%d:%02d:%02d',$diff->days * 24 + $diff->h,$diff->i,$diff->s);
Look at what you are doing: using time to make computations ignoring date part.
Maybe considering things in another way : 1 hour = 60 seconds * 60 minutes. So convert all you iterations as seconds, do the sum at the end and write time you need yourself.
Or, or you will use some greater things from php documentation
<?php
$january = new DateTime('2010-01-01');
$february = new DateTime('2010-02-01');
$interval = $february->diff($january);
// %a will output the total number of days.
echo $interval->format('%a total days')."\n";
// While %d will only output the number of days not already covered by the
// month.
echo $interval->format('%m month, %d days');
Adapt to your needs, and I am sure it will work well.
Personally I would completely avoid touching any date functions because you're not working with dates. You could do something like:
// Input data
$data = [
'00:02:55',
'00:07:56',
'01:03:32',
'15:13:34',
'02:13:44',
'03:08:53',
'13:13:54'
];
// Total to hold the amount of seconds
$total = 0;
// Loop the data items
foreach($data as $item):
$temp = explode(":", $item); // Explode by the seperator :
$total+= (int) $temp[0] * 3600; // Convert the hours to seconds and add to our total
$total+= (int) $temp[1] * 60; // Convert the minutes to seconds and add to our total
$total+= (int) $temp[2]; // Add the seconds to our total
endforeach;
// Format the seconds back into HH:MM:SS
$formatted = sprintf('%02d:%02d:%02d', ($total / 3600),($total / 60 % 60), $total % 60);
echo $formatted; // Outputs 35:04:28
So we loop the items in the input array and explode the string by the : to get an array containing hours, minutes and seconds in indexes 0, 1, and 2.
We then convert each of those values to seconds and add to our total. Once we're done, we format back into HH:MM:SS format

take minutes only from the time format in php before the record update

$aak is show 00:17:00 and i just want get 17
$aak=$data2["jamwaktukerja"];
i use this code for get minute,but still wrong
$time = date("i ", $aak);
var_dump($time);
$masuk2 = 'UPDATE presensi SET menitwaktukerja ="'.$time.'" where
tanggal="'.$cellValueB.'" and nip="'.$data1["nip"].'" ';
$query4 = mysqli_query($con,$masuk2);
There are several ways to get the minutes of that input string.
Using explode - use only if the input is always the same format:
<?php
$aak = "00:17:00";
$time = explode(':', $aak);
$minutes = $time[1];
echo $minutes; // 17
Using DateTime Class:
$aak = "00:17:00";
// either with specified format:
$d = DateTime::createFromFormat("H:i:s",$aak);
// or without:
$d = new DateTime($aak);
$minutes = $d->format("i");
echo $minutes; // 17
// as a oneliner:
$minutes = (new DateTime($aak))->format("i");
Go the way you did, via date(), but provide it with a valid timestamp created with strtotime():
$aak = "00:17:00";
$timestamp = strtotime($aak);
$minutes = date("i", $timestamp);
echo $minutes; // 17

PHP - How to deduct a time to a total time without converting to decimal?

I'm currently working with a timekeeping system which computes the sum of the basic hours of the week and deduct certain time if there's a late record.
Given that the employee has a total hours rendered for this week is 45 hours (45:00), and he she/has a total late record for that week of 50 minutes (00:50),
Using, PHP. How can I deduct the late record to the total hours rendered without converting time to decimal? The desired output for the above sample is 44:10 since 00:50 is deducted to 45:00.
I see so your goal is to subtract durations ex.
45:00 - 00:50 = 44:10
1: Create a function that convert them into hours
function convertToHours($duration) {
$duration = explode(':',$duration);
$hours+= (int)$duration[0];
$hours+= (int)$duration[1] / 60;
return $hours;
}
2: Create a funciton thats convert from seconds to duration hours:seconds
function secondsToDuration($seconds) {
$H = floor($seconds / 3600);
$i = ($seconds / 60) % 60;
$s = $seconds % 60;
return sprintf("%02d:%02d:%02d", $H, $i, $s);
}
Convert them into hours using function created
$duration1 = convertToHours("25:00");
$duration2 = convertToHours("00:50");
Then subtract them
$difference = $duration1 - $duration2;
Lastly use the created method which convert them back into duration
$duration = secondsToDuration($difference * 3600);
See Demo here
Hope it helps you
You can convert the string to a date and get the difference.
$d1 = "00:45:00";
$d2 = "00:00:50";
date_default_timezone_set("utc");
$fakedate = '01/01/2017';
$d1 = $fakedate . ' ' . $d1;
$d2 = $fakedate . ' ' . $d2;
$dt1 = new DateTime($d1);
$dt2 = new DateTime($d2);
$diff = $dt1->diff($dt2);
echo $diff->format("%H:%I:%S");
The output will be: 00:44:10

PHP sum two different minutes

i have two different break time
default break time
extra break time
here i want to sum of two times and display 12 hrs format
EX :
$default_time = "00:30";
$extra_time = "00:25";
my expected output : 00:55
but now display 01:00
this is my code
$default_time = $work_data->break_time;
$break_time = $work_data->extra_time;
$total_break = strtotime($default_time)+strtotime($break_time);
echo date("h:i",strtotime($total_break));
Here is the function you can calculate total time by passing the arguments to functions.
$hours, $min are supposed variable which is zero
$default_time = "00:30";
$break_time = "00:25";
function calculate_total_time() {
$i = 0;
foreach(func_get_args() as $time) {
sscanf($time, '%d:%d', $hour, $min);
$i += $hour * 60 + $min;
}
if( $h = floor($i / 60) ) {
$i %= 60;
}
return sprintf('%02d:%02d', $h, $i);
}
// use example
echo calculate_total_time($default_time, $break_time); # 00:55
There is one function call to strtotime function too much.
You should leave out the strtotime() call in the last line, as $total_break already is a UNIX timestamp:
$total_break = strtotime($default_time)+strtotime($break_time);
echo date("h:i",$total_break);
The problem is that you're trying to add too specific timestamps, but what you're trying to achieve is adding two durations. So you need to convert those timestamps into durations. For that you need a base, which in your case is 00:00.
$base = strtotime("00:00");
$default_time = $work_data->break_time;
$default_timestamp = strtotime($default_time);
$default_duration = $default_timestamp - $base; // Duration in seconds
$break_time = $work_data->extra_time;
$break_timestamp = strtotime($break_time);
$break_duration = $break_timestamp - $base; // Duration in seconds
$total_break = $default_duration + $break_duration; // 55 min in seconds
// If you want to calculate the timestamp 00:55, just add the base back to it
echo date("H:i", $base + $total_break);
Consider using standard DateTime and DateInterval classes. All you will need is to convert your second variable value to interval_spec format (see http://php.net/manual/en/dateinterval.construct.php for details):
$defaultTime = "00:30";
$breakTime = "PT00H25M"; // Or just 'PT25M'
$totalBreak = (new DateTime($defaultTime))->add($breakTime);
echo $totalBreak->format('H:i');
You could try the following code fragment:
$time1 = explode(":", $default_time);
$time2 = explode(":", $break_time);
$fulltime = ($time1[0] + $time2[0]) * 60 + $time1[1] + $time2[1];
echo (int)($fulltime / 60) . ":" . ($fulltime % 60);
<?php
$time = "00:30";
$time2 = "00:25";
$secs = strtotime($time2)-strtotime("00:00:00");
$result = date("H:i:s",strtotime($time)+$secs);
print_r($result);
?>
Use below code you will definitely get your answers.
$default_time = "00:30:00";
$extra_time = "00:25:00";
$secs = strtotime($extra_time)-strtotime("00:00:00");
$result = date("H:i:s A",strtotime($default_time)+$secs);
echo $result;die;
You can modify above code as per your need.
You could try the following:
$default_time = $work_data->break_time;
$date_start = new DateTime($default_time);
$break_time = $work_data->extra_time;
$interval = new DateInterval("PT" . str_replace(":", "H", $break_time) . "M");
$date_end = $date_start->add($interval);
echo $date_end->format("H:i");
Note that this doesn't account for times which span a 24 hour period

PHP: Time/HH:Minute:Seconds convert to Seconds

I have a code where it will subtract the Total Duration and the Total Time, and after that the result for the computation will be converted into seconds...
Assuming in my Total Duration is "02:00:00"
then for Total Time is "01:30:00"
For computation...
02:00:00 - 01:30:00 = 00:30:00
then for the result, "00:30:00" will be converted to seconds and the result is "1800"
How can I convert it?
Thanks for the help...
Use strtotime function. It returns the UNIX timestamp (number of seconds since January 1st 1970 00:00:00). If you'll pass the hour format HH:MM:SS to it, you can easily do the math
$to = strtotime('02:00:00');
$from = strtotime('01:30:00');
$seconds = $to - $from; // outputs 30
You assumed that the format is minutes:seconds:miliseconds and you wanted to receive 30 seconds in your case. Actually the output is 30 minutes. Miliseconds are separated with a dot.
Your hours should probably look like this:
$to = strtotime('00:02:00');
$from = strtotime('00:01:30');
How about splitting the Time-String into three substrings with the function (returns an array of substrings)
$substrings = new Array();
$substrings = explode(":", $timeString);
Now the array $substrings contains three substrings (hours, minutes, seconds).
you could compute the seconds just by multiplicating:
$hours = intval($substrings[0]);
$minutes = intval($substrings[1]);
$seconds = intval($substrings[2]);
$seconds = $hours * 3600 + $minutes * 60 + $seconds;
Can you try this,
$start = '01:30:00';
$end = '02:00:00';
$workingHours = (strtotime($end) - strtotime($start));
$res= date("i", $workingHours);
echo "DIFF: ". $res; //OP 30 Minutes
echo $resFull= date("H:i:s", $workingHours); //OP 00:30:00
If you use format HH:MM:SS then you can convert it to seconds by next code
$timestr = "00:30:00";
$temp = explode(":", $timestr);
if ($temp && is_array($temp) && count($temp) == 3) {
$time = intval($temp[0]) * 3600 + intval($temp[1]) * 60 + intval($temp[1]);
} else {
$time = null;
}
Alternative with PHP 5.3:
<?php
try {
$date1 = new DateTime('02:00:00');
$date2 = new DateTime('01:30:00');
$diff = $date1->diff($date2);
echo $diff->format('H:i:s');
} catch (Exception $e) {
echo $e->getMessage();
exit(1);
}

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