在 php strtotime 中遇到格式不正确的数值

Question

如何解决这个错误？

  $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>';
}

注意：第 19 行 D:\xamp\htdocs\timetest.php

中遇到格式不正确的数值

注意：第 20 行 D:\xamp\htdocs\timetest.php 中遇到格式不正确的数值 02-01-1970 - 02-01-1970

第 19 行：

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

Answer 1

据我所知，为了在任何日期中添加一些天数或时间，strtotime 将第一个参数作为日期字符串和一些间隔。

所以代码应该是：

$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);

在 php strtotime 中遇到格式不正确的数值

A non well formed numeric value encountered in php strtotime

php

strtotime