Question

我有两个时间戳，我会根据

来获取差异

如果超过60秒可以得到分钟
如果超过60分钟的时间差异

在我的代码中我有

 $diff = ($truckHistory->created_at - $model->created_at);

 return $this->convertToSecMinHour($diff)

  public function convertToSecMinHour($diff){
    switch($diff){
      case ($diff <= 60):{
            return $diff. "sec"
             break;
          } 

       //stuck for minutes and hours cases
    }


  }

以上以秒为单位返回所以我试过了

import sys, os, time

Answer 1

我有这个功能，但它包括其他测量时间的方法。你可以简单地删除那些你不需要的东西。

public function timeAgo($timestamp) {
    $estimateTime = time() - $timestamp;

    if ($estimateTime < 1) {
        return 'less than 1 second ago';
    }

    $condition = array( 
        12 * 30 * 24 * 60 * 60  =>  'year',
        30 * 24 * 60 * 60       =>  'month',
        24 * 60 * 60            =>  'day',
        60 * 60                 =>  'hour',
        60                      =>  'minute',
        1                       =>  'second'
    );

    foreach ($condition as $secs => $str) {
        $d = $estimateTime / $secs;

        if($d >= 1) {
            $r = round( $d );
            return $r . ' ' . $str . ( $r > 1 ? 's' : '' ) . ' ago';
        }
    }
}

Answer 2

您可以快速使用此功能，只需将秒数传递给它。 Example here

$time1 = '1504802148'; // Unix timeStamp
$timeDiff = ceil((time() - $time1));

echo secondsToTime($timeDiff);

function secondsToTime($seconds) {
  $dtF = new \DateTime('@0');
  $dtT = new \DateTime("@$seconds");
  $makeTime = $dtF->diff($dtT)->format('%ad,%hh,%im,%ss');
  $data = explode(",", $makeTime);
  $array = array('0d', '0h', '0m');
  $time = array_diff($data, $array);
  return implode("", $time);
}

Answer 3

使用此功能：

<?php
$diff = (time() - time()+60);
$diff2 = (time() - time()+2000);
$diff3 = (time() - time()+5000);

echo convertToSecMinHour($diff)."<br>";
echo convertToSecMinHour($diff2)."<br>";
echo convertToSecMinHour($diff3)."<br>";

function convertToSecMinHour($diff, $decimals = 2) {
    switch($diff){
        case ($diff <= 60):
            $result = $diff;
            $unit = "sec";
            break;
        case ($diff > 60 && $diff < 3600):
            $result =  ($diff / 60);
            $unit = "min";
            break;
        case ($diff >= 3600):
            $result = ($diff / (60 * 60));
            $unit = "hrs";
            break;
        default:
            $result = 0;
            $unit = "";
    }
    return round($result, $decimals).$unit;
}

如果差异小于60，则以秒为单位返回差异（即20秒），如果差异在60到3600秒之间，则以分钟（即20分钟）的差异，以及小时差异（即20小时）如果差异大于3600.第二个参数是结果中所需的小数位数（默认为2）。

Demo

Answer 4

我认为您想要$ result，还有其他变量可供使用

$result;
$hour = $diff / 3600;
if ($hour == 0) {
    $min = ($diff % 3600) / 60;
    if ($min == 0) {
        $sec = ($diff % 3600) % 60;
        $result = $sec;
    }else{
        $result = $min;
    }
}else{
    $result = $hour;
}

在php中以分钟，小时或秒为单位获取两个时间戳之间的差异

4 个答案: