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');

查看以下链接。这是迄今为止我找到的最好的答案

function dateDiff ($d1, $d2) {

    // Return the number of days between the two dates:    
    return round(abs(strtotime($d1) - strtotime($d2))/86400);

} // end function dateDiff

当你通过日期参数。函数使用PHP ABS（）绝对值始终返回正数作为两者之间的天数日期。请记住，两个日期之间的天数不是包括两个日期。因此，如果您正在寻找天数由输入日期之间的所有日期表示，您需要向该函数的结果添加一（1）。例如，差异（由上述函数返回）2013-02-09和2013-02-14之间的值为5。但天数或日期范围2013-02-09-2013-02-14表示的日期为6。

http://www.bizinfosys.com/php/date-difference.html

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)

$date1 = date_create('2007-03-24');
$date2 = date_create('2009-06-26');
$interval = date_diff($date1, $date2);
echo "difference : " . $interval->y . " years, " . $interval->m." months, ".$interval->d." days ";

这是可运行的代码

$date1 = date_create('2007-03-24');
$date2 = date_create('2009-06-26');
$diff1 = date_diff($date1,$date2);
$daysdiff = $diff1->format("%R%a");
$daysdiff = abs($daysdiff);

一便士一英镑：我刚刚回顾了几个解决方案，所有这些方案都使用floor（）提供了一个复杂的解决方案，然后四舍五入到26年12个月零2天的解决方案中，原本应该是25年11个月零20天！！！！

这是我对这个问题的看法：可能不优雅，可能编码不好，但如果不计算LEAP年份，则提供了更接近答案的答案，显然闰年可以编码为，但在这种情况下-正如其他人所说，也许您可以提供以下答案：我已经包含了所有测试条件和print_r，以便您可以更清楚地看到结果的构造：在这里，

//设置输入日期/变量：：

$ISOstartDate   = "1987-06-22";
$ISOtodaysDate = "2013-06-22";

//我们需要将ISO yyyy-mm-dd格式分解为yyyy-mm-d格式，如下所示：

$yDate[]=爆炸（'-'，$ISOstartDate）；print_r（$yDate）；

$zDate[]=爆炸（'-'，$ISOtodaysDate）；print_r（$zDate）；

// Lets Sort of the Years!
// Lets Sort out the difference in YEARS between startDate and todaysDate ::
$years = $zDate[0][0] - $yDate[0][0];

// We need to collaborate if the month = month = 0, is before or after the Years Anniversary ie 11 months 22 days or 0 months 10 days...
if ($months == 0 and $zDate[0][1] > $ydate[0][1]) {
    $years = $years -1;
}
// TEST result
echo "\nCurrent years => ".$years;

// Lets Sort out the difference in MONTHS between startDate and todaysDate ::
$months = $zDate[0][1] - $yDate[0][1];

// TEST result
echo "\nCurrent months => ".$months;

// Now how many DAYS has there been - this assumes that there is NO LEAP years, so the calculation is APPROXIMATE not 100%
// Lets cross reference the startDates Month = how many days are there in each month IF m-m = 0 which is a years anniversary
// We will use a switch to check the number of days between each month so we can calculate days before and after the years anniversary

switch ($yDate[0][1]){
    case 01:    $monthDays = '31';  break;  // Jan
    case 02:    $monthDays = '28';  break;  // Feb
    case 03:    $monthDays = '31';  break;  // Mar
    case 04:    $monthDays = '30';  break;  // Apr
    case 05:    $monthDays = '31';  break;  // May
    case 06:    $monthDays = '30';  break;  // Jun
    case 07:    $monthDays = '31';  break;  // Jul
    case 08:    $monthDays = '31';  break;  // Aug
    case 09:    $monthDays = '30';  break;  // Sept
    case 10:    $monthDays = '31';  break;  // Oct
    case 11:    $monthDays = '30';  break;  // Nov
    case 12:    $monthDays = '31';  break;  // Dec
};
// TEST return
echo "\nDays in start month ".$yDate[0][1]." => ".$monthDays;


// Lets correct the problem with 0 Months - is it 11 months + days, or 0 months +days???

$days = $zDate[0][2] - $yDate[0][2] +$monthDays;
echo "\nCurrent days => ".$days."\n";

// Lets now Correct the months to being either 11 or 0 Months, depending upon being + or - the years Anniversary date 
// At the same time build in error correction for Anniversary dates not being 1yr 0m 31d... see if ($days == $monthDays )
if($days < $monthDays && $months == 0)
    {
    $months = 11;       // If Before the years anniversary date
    }
else    {
    $months = 0;        // If After the years anniversary date
    $years = $years+1;  // Add +1 to year
    $days = $days-$monthDays;   // Need to correct days to how many days after anniversary date
    };
// Day correction for Anniversary dates
if ($days == $monthDays )   // if todays date = the Anniversary DATE! set days to ZERO
    {
    $days = 0;          // days set toZERO so 1 years 0 months 0 days
    };

    echo "\nTherefore, the number of years/ months/ days/ \nbetween start and todays date::\n\n";

    printf("%d years, %d months, %d days\n", $years, $months, $days);

最终结果是：26年零个月零天

这就是我在2013年6月22日做生意的时间——哎呦！