Question

如果我有一些字符串$startDate和$endDate设置为（例如）"2011/07/01"和"2011/07/17"（意味着2011年7月1日和2011年7月17日）。我如何计算从开始日期到结束日期的天数？在给出的例子中，它将是17天。

Answer 1

这是实现这一目标的原始方式

$startTimeStamp = strtotime("2011/07/01");
$endTimeStamp = strtotime("2011/07/17");

$timeDiff = abs($endTimeStamp - $startTimeStamp);

$numberDays = $timeDiff/86400;  // 86400 seconds in one day

// and you might want to convert to integer
$numberDays = intval($numberDays);

Answer 2

Use DateTime::diff（又名date_diff）：

$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);

或者：

$datetime1 = date_create('2009-10-11');
$datetime2 = date_create('2009-10-13');
$interval = date_diff($datetime1, $datetime2);

然后，您可以通过调用$interval->days来获取整数的间隔。

Answer 3

PHP有一个date_diff（）函数来执行此操作。

Answer 4

如果您的DateTime还有小时：分钟：秒，您仍然希望有天数..

   /**
     * Returns the total number of days between to DateTimes, 
     * if it is within the same year
     * @param $start
     * @param $end
     */
    public function dateTimesToDays($start,$end){
       return intval($end->format('z')) - intval($start->format('z')) + 1;
    }

https://github.com/dukeatcoding/timespan-converter

Answer 5

<?php
$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days');
?>

来源：http://www.php.net/manual/en/datetime.diff.php

Answer 6

这些解决方案都不适合我。对于仍然使用PHP 5.2的人（5.3中引入了DateTime :: diff），此解决方案可以工作：

function howDays($from, $to) {
    $first_date = strtotime($from);
    $second_date = strtotime($to);
    $offset = $second_date-$first_date; 
    return floor($offset/60/60/24);
}

Answer 7

如果您想知道天数（如果有），小时数（如果有的话），minutues（如果有的话）和秒数，您可以执行以下操作：

$previousTimeStamp = strtotime("2011/07/01 21:12:34");
$lastTimeStamp = strtotime("2013/09/17 12:34:11");

$menos=$lastTimeStamp-$previousTimeStamp;

$mins=$menos/60;
if($mins<1){
$showing= $menos . " seconds ago";
}
else{
$minsfinal=floor($mins);
$secondsfinal=$menos-($minsfinal*60);
$hours=$minsfinal/60;
if($hours<1){
$showing= $minsfinal . " minutes and " . $secondsfinal. " seconds ago";

}
else{
$hoursfinal=floor($hours);
$minssuperfinal=$minsfinal-($hoursfinal*60);
$days=$hoursfinal/24;
if($days<1){
$showing= $hoursfinal . "hours, " . $minssuperfinal . " minutes and " . $secondsfinal. " seconds ago";

}
else{
$daysfinal=floor($days);
$hourssuperfinal=$hoursfinal-($daysfinal*24);
$showing= $daysfinal. "days, " .$hourssuperfinal . " hours, " . $minssuperfinal . " minutes and " . $secondsfinal. " seconds ago";
}}}

echo $showing;

如果要添加月份和年份，可以使用相同的逻辑。

Answer 8

简单的计算方法是，

$currentdate = date('Y-m-d H:i:s');
$after1yrdate =  date("Y-m-d H:i:s", strtotime("+1 year", strtotime($data)));

$diff = (strtotime($after1yrdate) - strtotime($currentdate)) / (60 * 60 * 24);

echo "<p style='color:red'>The difference is ".round($diff)." Days</p>";

Answer 9

$date1 = date_create("2017-04-15");
$date2 = date_create("2017-05-18");

//difference between two dates
$diff = date_diff($date1,$date2);

//count days
echo 'Days Count - '.$diff->format("%a");

Answer 10

我创建了一个函数，如果您传递两个日期，它将返回按天计算的值。为了更好地理解，请查看开始日期的输出： 2018-11-12 11:41:19 和结束日期 2018-11-16 12:07:26

private function getTimeData($str1,$str2){
$datetime1 = strtotime($str1);
$datetime2 = strtotime($str2);
$myArray = array();
if(date('d', $datetime2) != date('d', $datetime1) ||  date('m', $datetime2) != date('m', $datetime1) || date('y', $datetime2) != date('y', $datetime1)){
    $exStr1 = explode(' ',$str1);
    $exStr2 = explode(' ',$str2);
    $datediff = strtotime($exStr2[0]) - strtotime($exStr1[0]);
    $totalDays = round($datediff / (60 * 60 * 24));
    $actualDate1 = $datetime1;
    $actualDate2 = date('Y-m-d', $datetime1)." 23:59:59";
    $interval  = abs(strtotime($actualDate2)-$actualDate1);
    $minutes   = round($interval / 60);
    $myArray[0]['startDate'] = date('Y-m-d H:i:s', $actualDate1); 
    $myArray[0]['endDate'] = $actualDate2;
    $myArray[0]['minutes'] = $minutes;
    $i = 1;
    if($totalDays > 1){    
        for($i=1; $i<$totalDays; $i++){
            $dayString = "+".$i." day";
            $edate = strtotime($dayString, $actualDate1);
            $myArray[$i]['startDate'] = date('Y-m-d', $edate)." 00:00:00"; 
            $myArray[$i]['endDate'] = date('Y-m-d', $edate)." 23:59:59"; 
            $myArray[$i]['minutes'] = 1440;
        }
    }
    $actualSecDate1 = date('Y-m-d', $datetime2)." 00:00:00";
    $actualSecDate2 = $datetime2;
    $interval  = abs(strtotime($actualSecDate1)-$actualSecDate2);
    $minutes   = round($interval / 60);
    $myArray[$i]['startDate'] = $actualSecDate1; 
    $myArray[$i]['endDate'] = date('Y-m-d H:i:s', $actualSecDate2); 
    $myArray[$i]['minutes'] = $minutes;
}
else{
    $interval  = abs($datetime2-$datetime1);
    $minutes   = round($interval / 60);
    $myArray[0]['startDate'] = date('Y-m-d H:i:s', $datetime1); 
    $myArray[0]['endDate'] = date('Y-m-d H:i:s', $datetime2);
    $myArray[0]['minutes'] = $minutes;
}

return $myArray;
}

输出

Array
(
[0] => Array
    (
        [startDate] => 2018-11-12 11:41:19
        [endDate] => 2018-11-12 23:59:59
        [minutes] => 739
    )

[1] => Array
    (
        [startDate] => 2018-11-13 00:00:00
        [endDate] => 2018-11-13 23:59:59
        [minutes] => 1440
    )

[2] => Array
    (
        [startDate] => 2018-11-14 00:00:00
        [endDate] => 2018-11-14 23:59:59
        [minutes] => 1440
    )

[3] => Array
    (
        [startDate] => 2018-11-15 00:00:00
        [endDate] => 2018-11-15 23:59:59
        [minutes] => 1440
    )

[4] => Array
    (
        [startDate] => 2018-11-16 00:00:00
        [endDate] => 2018-11-16 12:07:26
        [minutes] => 727
    )
)

Answer 11

您可以使用date_diff来计算两个日期之间的差异：

$date1 = date_create("2013-03-15");
$date2 = date_create("2013-12-12");
$diff =  date_diff($date1 , $date2);
echo  $diff->format("%R%a days");

如何计算PHP中两个日期之间的天数？

11 个答案: