A non well formed numeric value encountered in php strtotime

Question

how to solve this error?

  $timeFirst  = strtotime('2011-05-12 18:20:20');
  $timeSecond = strtotime('2011-05-13 18:20:20');
  $differenceInSeconds = $timeSecond - $timeFirst;
  $hour=$differenceInSeconds/60/60; 
  $days=$hour/24;
  $days=ceil($days);


  $data=array();

  for($i=1; $i<=7; $i++){                               
    if($i==1){
      $next_travel_date=date('d-m-Y',$timeFirst);
      $next_endDate=date('d-m-Y',$timeSecond);
    }else if($i>1){     
      $string=" + $days days";
      $date1=strtotime($string, $next_travel_date);
      $date2= strtotime($string, $next_endDate);                                    
      $next_travel_date=date('d-m-Y',$date1 );
      $next_endDate=date('d-m-Y',$date2);                                   
    }
      $data['travel_date']=$next_travel_date;   
      $data['end_date']=$next_endDate;
      echo $data['travel_date'].' - '.$data['end_date'].'<br>';
}

Notice: A non well formed numeric value encountered in D:\xamp\htdocs\timetest.php on line 19

Notice: A non well formed numeric value encountered in D:\xamp\htdocs\timetest.php on line 20 02-01-1970 - 02-01-1970

line 19:

$date1=strtotime($string, $next_travel_date); $date2= strtotime($string, $next_endDate);

line 19 : $date1=strtotime($string, $next_travel_date); $date2= strtotime($string, $next_endDate); — kumar, Aug 24 '18 at 07:07
Please edit your question to include all info instead of posting it as comments. — M. Eriksson, Aug 24 '18 at 07:09
So where do the variables `$next_travel_date` and `$next_endDate` come from on those two lines? (Also, pick a form of casing and stick with it. Don't use different forms at the same time) — Loek, Aug 24 '18 at 07:09

score 1 · Accepted Answer · answered Aug 24 '18 at 07:12

As I understand, for adding some days or time in any date, strtotime takes first parameter as date string and some interval with it.

So code should be :

$string = " + $days days";
$date1 = strtotime($next_travel_date . $string);
$date2 = strtotime($next_endDate . $string);
$next_travel_date = date('d-m-Y', $date1);
$next_endDate = date('d-m-Y', $date2);

A non well formed numeric value encountered in php strtotime

1 Answers1