我有两个表格日期:

Start Date: 2007-03-24 
End Date: 2009-06-26

现在我需要通过以下形式找到这两者之间的区别:

2 years, 3 months and 2 days

如何在PHP中执行此操作?


当前回答

我更喜欢使用date_create和date_diff对象。

代码:

$date1 = date_create("2007-03-24");
$date2 = date_create("2009-06-26");

$dateDifference = date_diff($date1, $date2)->format('%y years, %m months and %d days');

echo $dateDifference;

输出:

2 years, 3 months and 2 days

有关更多信息,请阅读PHP date_diff手册

根据手册date_diff是的别名日期时间::diff()

其他回答

您可以使用

getdate()

函数,该函数返回包含所提供日期/时间的所有元素的数组:

$diff = abs($endDate - $startDate);
$my_t=getdate($diff);
print("$my_t[year] years, $my_t[month] months and $my_t[mday] days");

如果开始和结束日期为字符串格式,则使用

$startDate = strtotime($startDateStr);
$endDate = strtotime($endDateStr);

在上述代码之前

我建议使用DateTime和DateInterval对象。

$date1 = new DateTime("2007-03-24");
$date2 = new DateTime("2009-06-26");
$interval = $date1->diff($date2);
echo "difference " . $interval->y . " years, " . $interval->m." months, ".$interval->d." days "; 

// shows the total amount of days (not divided into years, months and days like above)
echo "difference " . $interval->days . " days ";

阅读更多php DateTime::diff手册

根据手册:

从PHP 5.2.2开始,DateTime对象可以使用比较运算符进行比较。

$date1 = new DateTime("now");
$date2 = new DateTime("tomorrow");

var_dump($date1 == $date2); // bool(false)
var_dump($date1 < $date2);  // bool(true)
var_dump($date1 > $date2);  // bool(false)
function showTime($time){

    $start      = strtotime($time);
    $end        = strtotime(date("Y-m-d H:i:s"));
    $minutes    = ($end - $start)/60;


    // years 
    if(($minutes / (60*24*365)) > 1){
        $years = floor($minutes/(60*24*365));
        return "From $years year( s ) ago";
    }


    // monthes 
    if(($minutes / (60*24*30)) > 1){
        $monthes = floor($minutes/(60*24*30));
        return "From $monthes monthe( s ) ago";
    }


    // days 
    if(($minutes / (60*24)) > 1){
        $days = floor($minutes/(60*24));
        return "From $days day( s ) ago";
    }

    // hours 
    if(($minutes / 60) > 1){
        $hours = floor($minutes/60);
        return "From $hours hour( s ) ago";
    }

    // minutes 
    if($minutes > 1){
        $minutes = floor($minutes);
        return "From $minutes minute( s ) ago";
    }
}

echo showTime('2022-05-05 21:33:00');

前段时间,我编写了一个format_date函数,因为它提供了许多关于日期的选项:

function format_date($date, $type, $seperator="-")
{
    if($date)
    {
        $day = date("j", strtotime($date));
        $month = date("n", strtotime($date));
        $year = date("Y", strtotime($date));
        $hour = date("H", strtotime($date));
        $min = date("i", strtotime($date));
        $sec = date("s", strtotime($date));

        switch($type)
        {
            case 0:  $date = date("Y".$seperator."m".$seperator."d",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 1:  $date = date("D, F j, Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 2:  $date = date("d".$seperator."m".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 3:  $date = date("d".$seperator."M".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 4:  $date = date("d".$seperator."M".$seperator."Y h:i A",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 5:  $date = date("m".$seperator."d".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 6:  $date = date("M",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 7:  $date = date("Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 8:  $date = date("j",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 9:  $date = date("n",mktime($hour, $min, $sec, $month, $day, $year)); break;
            case 10: 
                     $diff = abs(strtotime($date) - strtotime(date("Y-m-d h:i:s"))); 
                     $years = floor($diff / (365*60*60*24));
                     $months = floor(($diff - $years * 365*60*60*24) / (30*60*60*24));
                     $days = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24)/ (60*60*24));
                     $date = $years . " years, " . $months . " months, " . $days . "days";
        }
    }
    return($date);
}    

我在下面的页面上找到了您的文章,其中包含了许多PHP日期时间计算的参考。

使用PHP计算两个日期(和时间)之间的差异。下一页提供了一系列不同的方法(共7种),用于使用PHP执行日期/时间计算,以确定两个日期之间的时间差(小时、弹药)、天、月或年。

请参阅PHP日期时间–计算两个日期之间差值的7种方法。