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date_diff

(PHP 5 >= 5.3.0, PHP 7)

说明

此函数是该函数的别名： DateTime::diff()

User Contributed Notes

paez903 at hotmail dot com 24-Jul-2017 04:44


I do not have much programming in php and I hope I can help those that I want to do is that when entering in the form the date 1 and the date2 I calculate if between those two dates if they have passed 5 or more years and I add 3 more days taking As reference date 2, I do not know if I understand.



To see a theoretical example date = 10/01/2012 date2 = 23/07/2017



Between these two dates have passed 5 years, 6 months, and 13 days elapsed



Knowing this my serious conditional



If they are equal or more than 5 years but less than 10 years will be added 3 days

If they are equal or more than 10 years but less than 15 years will be added 6 days

If they are equal or more than 15 years but less than 20 years will be added 9 days

If they are equal to or more than 20 years will be added 12 days



Then having the conditionals



For this example would be case as it is greater than 5 years the result that I should show taking in days, having as date of departure the date2 = 23/07/2017 and to this date it is added 30 days that would be a constant and depending on the Years as the example happens 5 years would be: date2 = 23/07/2017 + 30 days = 08/30/2017 + 3 days

= End date to show = result = 02/09/2017.



But I still think how to do it if you can guide me I would appreciate a world, and everything should show without pressing buttons, if I press a button would be like to store in the database only the results otenidos as date1 date2 and result



<? Php

    

// $ date1 = $ _ POST ["date1"]; // this will be the first date or date of entry

// $ date2 = $ _ POST ["date2"]; // this will be the date with which you will calculate

    

// $ difference = $ date2 - $ date1;



// if ($ difference <= 5)

{

    // echo "number of days that correspond to it [". $ Date2 + 30 + 3. "].";



//} else {



// ($ difference <= 5)



{

// echo "number of days corresponding to it [" $ date2 + 30 "].";



}



 ?>

aleksandar at gvozden dot info 02-Dec-2016 04:20


diff from datetime 



    /**

     * diff from datetime

     * @param datetime $dt1

     * @param datetime $dt2

     * @return object $dtd (day, hour, min, sec / total)

     */

    static function datetimeDiff($dt1, $dt2){

        $t1 = strtotime($dt1);

        $t2 = strtotime($dt2);



        $dtd = new stdClass();

        $dtd->interval = $t2 - $t1;

        $dtd->total_sec = abs($t2-$t1);

        $dtd->total_min = floor($dtd->total_sec/60);

        $dtd->total_hour = floor($dtd->total_min/60);

        $dtd->total_day = floor($dtd->total_hour/24);



        $dtd->day = $dtd->total_day;

        $dtd->hour = $dtd->total_hour -($dtd->total_day*24);

        $dtd->min = $dtd->total_min -($dtd->total_hour*60);

        $dtd->sec = $dtd->total_sec -($dtd->total_min*60);

        return $dtd;

    }

vladimir-nt at mail dot ru 13-Oct-2015 12:11


My simple function to count up the number of weekdays between the dates (inclusive for both ends), one can add a holidays between the dates to take them into account as well as the weekends, the weekend days could also be changed.



function count_week_days($__date_from, $__date_to, $__holidays_between=array(), $__weekend_days=array(5,6)) {

   $total_days_count = $__date_to > $__date_from ? round(($__date_to - $__date_from)/(24*3600)) : 0;

   $full_weeks_count = floor($total_days_count/7);

   $weekend_days_count = $full_weeks_count*count($__weekend_days);

   $days_left_uncovered = $total_days_count - $full_weeks_count*7;

   for($i = 0; $i < $days_left_uncovered; $i++) {

      $date_to_check = $i ? strtotime("+{$i} day", $__date_from) : $__date_from;

      if(in_array(date('N', $date_to_check), $__weekend_days)) {

         $weekend_days_count++;

      }

   }

   $week_days_count = $total_days_count - $weekend_days_count - count($__holidays_between);

   return $week_days_count;

}



Tests:

print "\n 12.10.2015 to 10.10.2015 diff=".count_week_days(strtotime('12.10.2015'), strtotime('10.10.2015'));

print "\n 12.10.2015 to 13.10.2015 diff=".count_week_days(strtotime('12.10.2015'), strtotime('13.10.2015'));

print "\n 12.10.2015 to 15.10.2015 diff=".count_week_days(strtotime('12.10.2015'), strtotime('26.10.2015'));

print "\n 12.10.2015 to 13.10.2015 diff=".count_week_days(strtotime('12.10.2015'), strtotime('13.10.2016'));

print "\n 12.10.2015 to 13.10.2015 diff=".count_week_days(strtotime('12.10.2015'), strtotime('12.11.2015'));

print "\n 12.10.2015 to 13.10.2015 diff=".count_week_days(strtotime('12.10.2015'), strtotime('12.11.2015'), array(strtotime('07.11.2015')));



Results:

 12.10.2015 to 10.10.2015 diff=0

 12.10.2015 to 13.10.2015 diff=1

 12.10.2015 to 15.10.2015 diff=10

 12.10.2015 to 13.10.2015 diff=263

 12.10.2015 to 13.10.2015 diff=23

 12.10.2015 to 13.10.2015 diff=22

tuxedobob 23-Jul-2015 04:32


Even the top rated comment here, Sergio Abreu's, doesn't treat leap years entirely correctly. It should work between 1901 and 2099, but outside that it'll be a little off.



If you want to find out the number of days between two dates, use below. You can change to a different unit from that. It looks a little insane, but keep in mind the full set of rules for leap years:



If the year is divisible by 4, it's a leap year...

 - unless the year is divisible by 100, then it isn't...

 - - unless the year is divisible by 400, then it really is.



So in the functions below, we find the total numbers of days in full years since the mythical 1/1/0001, then add the number of days before the current one in the year passed. Do this for each date, then return the absolute value of the difference.



function days_diff($d1, $d2) {

    $x1 = days($d1);

    $x2 = days($d2);

    

    if ($x1 && $x2) {

        return abs($x1 - $x2);

    }

}



function days($x) {

    if (get_class($x) != 'DateTime') {

        return false;

    }

    

    $y = $x->format('Y') - 1;

    $days = $y * 365;

    $z = (int)($y / 4);

    $days += $z;

    $z = (int)($y / 100);

    $days -= $z;

    $z = (int)($y / 400);

    $days += $z;

    $days += $x->format('z');



    return $days;

}

Azuro 01-Apr-2015 10:12


I had to find the difference between two days (here is my solution without Date_diff()) :



$current_date = date("U") /* to have it in microseconds */

$selected_date_stamp = mktime(0,0,0,$your_month,$your_day,$your_year);

$selected_date = date("U",$selected_date_stamp);

$difference = round (($current_date - $selected_date)/(3600*24));

echo "The difference is :" . $difference . "<br/>";

maniarpratik at gmail dot com 30-Jan-2015 12:09


This function will return count of sunday between inputed dates.



<?php

function CountSunday($from,$to)

{

   

$from=date('d-m-Y',strtotime($from));

$to=date('d-m-Y',strtotime($to));

$cnt=0;

$nodays=(strtotime($to) - strtotime($from))/ (60 * 60 * 24); //it will count no. of days

$nodays=$nodays+1;

           for($i=0;$i<$nodays;$i++) 

            {      

                $p=0;

            list($d, $m, $y) = explode("-",$from);

            $datetime = strtotime("$d-$m-$y");           

            $nextday = date('d-m-Y',strtotime("+1 day", $datetime));  //this will add one day in from date (from date + 1)

            if($i==0)                           

                $p=date('w',strtotime($from));                           

            else

                $p=date('w',strtotime($nextday));

           

            if($p==0)            // check whethere value is 0 then its sunday

                $cnt++;                                //count variable of sunday                       

             $from=$nextday;         

             $p++;           

            }            

  return $cnt;

}

?>

qrworld.net 02-Dec-2014 02:32


Here you have in this post http://softontherocks.blogspot.com/2014/12/calcular-la-edad-con-php.html the code to get the age of a person specifying the date of birth:



function getAge($birthdate){

    $adjust = (date("md") >= date("md", strtotime($birthdate))) ? 0 : -1; // Si aún no hemos llegado al día y mes en este a?o restamos 1

    $years = date("Y") - date("Y", strtotime($birthdate)); // Calculamos el número de a?os 

    return $years + $adjust; // Sumamos la diferencia de a?os más el ajuste

}

jesushuertaarrabal at gmail dot com 03-Oct-2014 12:48


A way to verify a correct date getting your age



if (isset($_POST['birthday'])){

     if (preg_match("/^[0-9]{4}-[0-1][0-9]-[0-3][0-9]$/",$_POST['birthday'])){

          $items = explode("-", $_POST['birthday']);

           if (checkdate($items[1], $items[2], $items[0])){ //checkdate(m-d-y)

               //If you were born in a lip-year or lip - 1, then we have to add 5 days, else, we add 4 days 

                if ((0 == $items[0] % 4) && (0 != $items[0] % 100) || (0 == $items[0] % 400))

                 $bis = 4;

                else

                 $bis = 5;

            if (date_diff(date_create($_POST['birthday']) , date_create(date('Y-m-d')))->format("%R%a days") > (6569 + $bis)) //365*18

                 $print .= 'Birthday date: ' . $_POST['birthday'] . '<br>';

            else

                 $error[] = -1;

           }else

                $error[] = -2;

      }else

           $error[] = -3;

     }else

    $error[] = -4;

}

SunilKmCharde 21-May-2014 08:07


Powerful Function to get two date difference.



//////////////////////////////////////////////////////////////////////

//PARA: Date Should In YYYY-MM-DD Format

//RESULT FORMAT:

// '%y Year %m Month %d Day %h Hours %i Minute %s Seconds'        =>  1 Year 3 Month 14 Day 11 Hours 49 Minute 36 Seconds

// '%y Year %m Month %d Day'                                    =>  1 Year 3 Month 14 Days

// '%m Month %d Day'                                            =>  3 Month 14 Day

// '%d Day %h Hours'                                            =>  14 Day 11 Hours

// '%d Day'                                                        =>  14 Days

// '%h Hours %i Minute %s Seconds'                                =>  11 Hours 49 Minute 36 Seconds

// '%i Minute %s Seconds'                                        =>  49 Minute 36 Seconds

// '%h Hours                                                    =>  11 Hours

// '%a Days                                                        =>  468 Days

//////////////////////////////////////////////////////////////////////

function dateDifference($date_1 , $date_2 , $differenceFormat = '%a' )

{

    $datetime1 = date_create($date_1);

    $datetime2 = date_create($date_2);

    

    $interval = date_diff($datetime1, $datetime2);

    

    return $interval->format($differenceFormat);

    

}

Chiheb Nabil 09-Feb-2013 02:01


here a little solution of problem of missing date_diff function with php versions below 5.3.0 





<?php


function IntervalDays($CheckIn,$CheckOut){


$CheckInX = explode("-", $CheckIn);


$CheckOutX =  explode("-", $CheckOut);


$date1 =  mktime(0, 0, 0, $CheckInX[1],$CheckInX[2],$CheckInX[0]);


$date2 =  mktime(0, 0, 0, $CheckOutX[1],$CheckOutX[2],$CheckOutX[0]);


 $interval =($date2 - $date1)/(3600*24);





// returns numberofdays


return  $interval ;





}


?>

programacion at mundosica dot com 17-Apr-2012 10:36


Other data_diff aviable for php5.3>=



<?php

// Author: el pinche <fitorec>



function otherDiffDate($end='2020-06-09 10:30:00', $out_in_array=false){

        $intervalo = date_diff(date_create(), date_create($end));

        $out = $intervalo->format("Years:%Y,Months:%M,Days:%d,Hours:%H,Minutes:%i,Seconds:%s");

        if(!$out_in_array)

            return $out;

        $a_out = array();

        array_walk(explode(',',$out),

        function($val,$key) use(&$a_out){

            $v=explode(':',$val);

            $a_out[$v[0]] = $v[1];

        });

        return $a_out;

}

?>



#example 1

<?php

echo otherDiffDate();

?>

out1

       Years:08,Months:01,Days:22,Hours:17,Minutes:5,Seconds:26



example2 

<?php

print_r(otherDiffDate('2020-01-01 20:30:00',true));

?>

out2

Array

(

    [Years] => 07

    [Months] => 08

    [Days] => 15

    [Hours] => 03

    [Minutes] => 3

    [Seconds] => 48

)

Toine (contact at toine dot pro) 26-Aug-2011 07:22


This is a very simple function to calculate the difference between two timestamp values.


<?php


function diff($start,$end = false) {


    /*


    * For this function, i have used the native functions of PHP. It calculates the difference between two timestamp.


    *


    * Author: Toine


    *


    * I provide more details and more function on my website


    */ 





    // Checks $start and $end format (timestamp only for more simplicity and portability)


    if(!$end) { $end = time(); }


    if(!is_numeric($start) || !is_numeric($end)) { return false; }


    // Convert $start and $end into EN format (ISO 8601)


    $start  = date('Y-m-d H:i:s',$start);


    $end    = date('Y-m-d H:i:s',$end);


    $d_start    = new DateTime($start);


    $d_end      = new DateTime($end);


    $diff = $d_start->diff($d_end);


    // return all data


    $this->year    = $diff->format('%y');


    $this->month    = $diff->format('%m');


    $this->day      = $diff->format('%d');


    $this->hour     = $diff->format('%h');


    $this->min      = $diff->format('%i');


    $this->sec      = $diff->format('%s');


    return true;


}





/*


 * How use it?


 *


 * Call your php class (myClass for this example) and use the function :


*/


$start  = strtotime('1985/02/09 13:54:17');


$end    = strtotime('2012/12/12 17:30:21');


$myClass = new myClass();


$myClass->Diff($start,$end);


// Display result


echo 'Year: '.$myClass->Year;


echo '<br />Month: '.$myClass->Month;


echo '<br />Day: '.$myClass->Day;


echo '<br />Hour: '.$myClass->Hour;


echo '<br />Min: '.$myClass->Min;


echo '<br />Sec: '.$myClass->Sec;


// Display only month for all duration


$month = ($myClass->Year * 12) + $myClass->Month;


echo '<br />Total month: '.$month;


// if you want you can use this function without $end value :


$myClass->Diff($start);


// Automatically the end is the current timestamp


?>

vglebov at gmail dot com 25-May-2011 11:59


Get the difference between the dates without days off





<?php


function get_date_diff($date1, $date2) {


  $holidays = 0;


  for ($day = $date2; $day < $date1; $day += 24 * 3600) {


    $day_of_week = date('N', $day);


    if($day_of_week > 5) {


      $holidays++;


    }


  }


  return $date1 - $date2 - $holidays * 24 * 3600;


}





function test_get_date_diff()


{


  $datas = array(


    array('Fri 20 May 2011 14:00:00', 'Fri 20 May 2011 13:00:00', 1 * 3600),


    array('Sat 21 May 2011 15:00:00', 'Fri 20 May 2011 13:00:00', 2 * 3600),


    array('Sun 22 May 2011 16:00:00', 'Fri 20 May 2011 13:00:00', 3 * 3600),


    array('Mon 23 May 2011 14:00:00', 'Fri 20 May 2011 13:00:00', 25 * 3600),


    array('Fri 27 May 2011 13:00:00', 'Fri 13 May 2011 13:00:00', 24 * 10 * 3600),


  );


  foreach ($datas as &$data) {


    $actual = get_date_diff(strtotime($data[0]), strtotime($data[1]));


    if ($actual != $data[2]) {


      echo "Test for get_date_diff faled expected {$data[2]} but was {$actual}, date1: {$data[0]}, date2: {$data[1]}.<br>";


    }


  }


}


test_get_date_diff($data);


?>

kshegunov at gmail dot com 10-Jan-2011 09:26


Here is how I solved the problem of missing date_diff function with php versions below 5.3.0


The function accepts two dates in string format (recognized by strtotime() hopefully), and returns the date difference in an array with the years as first element, respectively months as second, and days as last element.


It should be working in all cases, and seems to behave properly when moving through February.





<?php


        function dateDifference($startDate, $endDate)


        {


            $startDate = strtotime($startDate);


            $endDate = strtotime($endDate);


            if ($startDate === false || $startDate < 0 || $endDate === false || $endDate < 0 || $startDate > $endDate)


                return false;


                


            $years = date('Y', $endDate) - date('Y', $startDate);


            


            $endMonth = date('m', $endDate);


            $startMonth = date('m', $startDate);


            


            // Calculate months


            $months = $endMonth - $startMonth;


            if ($months <= 0)  {


                $months += 12;


                $years--;


            }


            if ($years < 0)


                return false;


            


            // Calculate the days 


                        $offsets = array();


                        if ($years > 0)


                            $offsets[] = $years . (($years == 1) ? ' year' : ' years');


                        if ($months > 0)


                            $offsets[] = $months . (($months == 1) ? ' month' : ' months');


                        $offsets = count($offsets) > 0 ? '+' . implode(' ', $offsets) : 'now';





                        $days = $endDate - strtotime($offsets, $startDate);


                        $days = date('z', $days);    


                        


            return array($years, $months, $days);


        }


?>

Sergio Abreu 26-Jun-2010 07:22


<?php 

/* 

 * A mathematical decimal difference between two informed dates 

 *

 * Author: Sergio Abreu

 * Website: http://sites.sitesbr.net

 *

 * Features: 

 * Automatic conversion on dates informed as string.

 * Possibility of absolute values (always +) or relative (-/+)

*/



function s_datediff( $str_interval, $dt_menor, $dt_maior, $relative=false){



       if( is_string( $dt_menor)) $dt_menor = date_create( $dt_menor);

       if( is_string( $dt_maior)) $dt_maior = date_create( $dt_maior);



       $diff = date_diff( $dt_menor, $dt_maior, ! $relative);

       

       switch( $str_interval){

           case "y": 

               $total = $diff->y + $diff->m / 12 + $diff->d / 365.25; break;

           case "m":

               $total= $diff->y * 12 + $diff->m + $diff->d/30 + $diff->h / 24;

               break;

           case "d":

               $total = $diff->y * 365.25 + $diff->m * 30 + $diff->d + $diff->h/24 + $diff->i / 60;

               break;

           case "h": 

               $total = ($diff->y * 365.25 + $diff->m * 30 + $diff->d) * 24 + $diff->h + $diff->i/60;

               break;

           case "i": 

               $total = (($diff->y * 365.25 + $diff->m * 30 + $diff->d) * 24 + $diff->h) * 60 + $diff->i + $diff->s/60;

               break;

           case "s": 

               $total = ((($diff->y * 365.25 + $diff->m * 30 + $diff->d) * 24 + $diff->h) * 60 + $diff->i)*60 + $diff->s;

               break;

          }

       if( $diff->invert)

               return -1 * $total;

       else    return $total;

   }



/* Enjoy and feedback me ;-) */

?>

Flavio Tubino 26-May-2010 05:37


This is a very simple function to calculate the difference between two datetime values, returning the result in seconds. To convert to minutes, just divide the result by 60. In hours, by 3600 and so on.



Enjoy.



<?php

function time_diff($dt1,$dt2){

    $y1 = substr($dt1,0,4);

    $m1 = substr($dt1,5,2);

    $d1 = substr($dt1,8,2);

    $h1 = substr($dt1,11,2);

    $i1 = substr($dt1,14,2);

    $s1 = substr($dt1,17,2);    



    $y2 = substr($dt2,0,4);

    $m2 = substr($dt2,5,2);

    $d2 = substr($dt2,8,2);

    $h2 = substr($dt2,11,2);

    $i2 = substr($dt2,14,2);

    $s2 = substr($dt2,17,2);    



    $r1=date('U',mktime($h1,$i1,$s1,$m1,$d1,$y1));

    $r2=date('U',mktime($h2,$i2,$s2,$m2,$d2,$y2));

    return ($r1-$r2);



}

?>

tom at knapp2meter dot tk 16-Apr-2009 08:06


A simple way to get the time lag (format: <hours>.<one-hundredth of one hour>).



Hier ein einfacher Weg zur Bestimmung der Zeitdifferenz (Format: <Stunden>.<hundertstel Stunde>).



<?php



function GetDeltaTime($dtTime1, $dtTime2)

{

  $nUXDate1 = strtotime($dtTime1->format("Y-m-d H:i:s"));

  $nUXDate2 = strtotime($dtTime2->format("Y-m-d H:i:s"));



  $nUXDelta = $nUXDate1 - $nUXDate2;

  $strDeltaTime = "" . $nUXDelta/60/60; // sec -> hour

            

  $nPos = strpos($strDeltaTime, ".");

  if (nPos !== false)

    $strDeltaTime = substr($strDeltaTime, 0, $nPos + 3);



  return $strDeltaTime;

}



?>

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