如何使用 PHP 计算两个日期之间的差异?

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【中文标题】如何使用 PHP 计算两个日期之间的差异?【英文标题】:How to calculate the difference between two dates using PHP? 【发布时间】:2010-10-15 04:16:50 【问题描述】:

我有两个日期的表格:

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

现在我需要用下面的形式找出这两者之间的区别:

2 years, 3 months and 2 days

我如何在 php 中做到这一点?

【问题讨论】:

2 年 94 天。考虑到闰年来计算月份是有问题的。这需要有多准确? How do I calculate relative time?的可能重复 【参考方案1】:

使用这个功能

//function Diff between Dates
//////////////////////////////////////////////////////////////////////
//PARA: Date Should In YYYY-MM-DD Format
//RESULT FORMAT:
// '%y Year %m Month %d Day %h Hours %i Minute %s Seconds' =>  1 Year 3 Month 14 Day 11 Hours 49 Minute 36 Seconds
// '%y Year %m Month %d Day'                       =>  1 Year 3 Month 14 Days
// '%m Month %d Day'                                     =>  3 Month 14 Day
// '%d Day %h Hours'                                   =>  14 Day 11 Hours
// '%d Day'                                                 =>  14 Days
// '%h Hours %i Minute %s Seconds'         =>  11 Hours 49 Minute 36 Seconds
// '%i Minute %s Seconds'                           =>  49 Minute 36 Seconds
// '%h Hours                                          =>  11 Hours
// '%a Days                                                =>  468 Days
//////////////////////////////////////////////////////////////////////
function dateDifference($date_1 , $date_2 , $differenceFormat = '%a' )

    $datetime1 = date_create($date_1);
    $datetime2 = date_create($date_2);

    $interval = date_diff($datetime1, $datetime2);

    return $interval->format($differenceFormat);


只设置参数$differenceFormat根据你的需要 例如,我希望您的年龄与月份和天数之间的差异

dateDifference(date('Ym-d') , $date , '%y %m %d')

或其他格式

dateDifference(date('Ym-d') , $date , '%y-%m-%d')

【讨论】:

【参考方案2】:

我想提出一个稍微不同的观点,这似乎没有被提及。

您可以以声明的方式解决这个问题(就像任何其他问题一样)。重点是问什么你需要,而不是如何到达那里。

在这里,您需要与众不同。但那有什么区别呢?正如最受欢迎的答案中已经提到的那样,这是一个间隔。问题是如何得到它。无需显式调用diff() 方法,您只需按开始日期和结束日期(即按日期范围)创建一个间隔:

$startDate = '2007-03-24';
$endDate = '2009-06-26';
$range = new FromRange(new ISO8601DateTime($startDate), new ISO8601DateTime($endDate));

诸如闰年之类的所有复杂问题都已经处理好了。现在,当您有一个固定开始日期时间的间隔时,您可以获得一个人类可读的版本:

var_dump((new HumanReadable($range))->value());

它准确地输出你需要的东西。

如果您需要一些自定义格式,这也不是问题。您可以使用ISO8601Formatted 类,该类接受带有六个参数的可调用对象:年、月、日、小时、分钟和秒:

(new ISO8601Formatted(
    new FromRange(
        new ISO8601DateTime('2017-07-03T14:27:39+00:00'),
        new ISO8601DateTime('2018-07-05T14:27:39.235487+00:00')
    ),
    function (int $years, int $months, int $days, int $hours, int $minutes, int $seconds) 
        return $years >= 1 ? 'More than a year' : 'Less than a year';
    
))
    ->value();

它输出More than a year

有关此方法的更多信息,请查看quick start entry。

【讨论】:

【参考方案3】:

将此用于遗留代码 (PHP

您可以使用 strtotime() 将两个日期转换为 unix 时间,然后计算它们之间的秒数。由此很容易计算出不同的时间段。

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

$diff = abs(strtotime($date2) - strtotime($date1));

$years = floor($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));

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

编辑: 显然,这样做的首选方式如下 jurka 所述。我的代码通常仅在您没有 PHP 5.3 或更高版本时才推荐使用。

cmets 中的一些人指出,上面的代码只是一个近似值。我仍然相信对于大多数目的来说这很好,因为范围的使用更多地是为了提供已经过去或剩余多少时间的感觉,而不是提供精确度 - 如果你想这样做,只需输出日期。

尽管如此,我还是决定解决这些投诉。如果您确实需要一个确切的范围但无法访问 PHP 5.3,请使用下面的代码(它也应该在 PHP 4 中工作)。这是 PHP 在内部用于计算范围的代码的直接端口,但它不考虑夏令时。这意味着它最多关闭一个小时,但除此之外它应该是正确的。

<?php

/**
 * Calculate differences between two dates with precise semantics. Based on PHPs DateTime::diff()
 * implementation by Derick Rethans. Ported to PHP by Emil H, 2011-05-02. No rights reserved.
 * 
 * See here for original code:
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/tm2unixtime.c?revision=302890&view=markup
 * http://svn.php.net/viewvc/php/php-src/trunk/ext/date/lib/interval.c?revision=298973&view=markup
 */

function _date_range_limit($start, $end, $adj, $a, $b, $result)

    if ($result[$a] < $start) 
        $result[$b] -= intval(($start - $result[$a] - 1) / $adj) + 1;
        $result[$a] += $adj * intval(($start - $result[$a] - 1) / $adj + 1);
    

    if ($result[$a] >= $end) 
        $result[$b] += intval($result[$a] / $adj);
        $result[$a] -= $adj * intval($result[$a] / $adj);
    

    return $result;


function _date_range_limit_days($base, $result)

    $days_in_month_leap = array(31, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);
    $days_in_month = array(31, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31);

    _date_range_limit(1, 13, 12, "m", "y", &$base);

    $year = $base["y"];
    $month = $base["m"];

    if (!$result["invert"]) 
        while ($result["d"] < 0) 
            $month--;
            if ($month < 1) 
                $month += 12;
                $year--;
            

            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;
        
     else 
        while ($result["d"] < 0) 
            $leapyear = $year % 400 == 0 || ($year % 100 != 0 && $year % 4 == 0);
            $days = $leapyear ? $days_in_month_leap[$month] : $days_in_month[$month];

            $result["d"] += $days;
            $result["m"]--;

            $month++;
            if ($month > 12) 
                $month -= 12;
                $year++;
            
        
    

    return $result;


function _date_normalize($base, $result)

    $result = _date_range_limit(0, 60, 60, "s", "i", $result);
    $result = _date_range_limit(0, 60, 60, "i", "h", $result);
    $result = _date_range_limit(0, 24, 24, "h", "d", $result);
    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    $result = _date_range_limit_days(&$base, &$result);

    $result = _date_range_limit(0, 12, 12, "m", "y", $result);

    return $result;


/**
 * Accepts two unix timestamps.
 */
function _date_diff($one, $two)

    $invert = false;
    if ($one > $two) 
        list($one, $two) = array($two, $one);
        $invert = true;
    

    $key = array("y", "m", "d", "h", "i", "s");
    $a = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $one))));
    $b = array_combine($key, array_map("intval", explode(" ", date("Y m d H i s", $two))));

    $result = array();
    $result["y"] = $b["y"] - $a["y"];
    $result["m"] = $b["m"] - $a["m"];
    $result["d"] = $b["d"] - $a["d"];
    $result["h"] = $b["h"] - $a["h"];
    $result["i"] = $b["i"] - $a["i"];
    $result["s"] = $b["s"] - $a["s"];
    $result["invert"] = $invert ? 1 : 0;
    $result["days"] = intval(abs(($one - $two)/86400));

    if ($invert) 
        _date_normalize(&$a, &$result);
     else 
        _date_normalize(&$b, &$result);
    

    return $result;


$date = "1986-11-10 19:37:22";

print_r(_date_diff(strtotime($date), time()));
print_r(_date_diff(time(), strtotime($date)));

【讨论】:

如果你使用的是 DateTime 类,你可以使用 $date->format('U') 来获取 unix 时间戳 如果您必须处理夏季/冬季时间,这不是真的。在这种特殊情况下,当您调整夏季/冬季时间时,一天等于 23 或 25 小时。 嗯,闰年也可以这样。它也没有考虑到这一点。不过,我不相信你甚至想考虑到这一点,因为我们在这里讨论了一个范围。范围的语义与绝对日期的语义有些不同。 这个函数不正确。它适用于近似值,但不适用于精确范围。一方面,它假设一个月有 30 天,也就是说,2 月 1 日和 3 月 1 日之间的天数差异与 7 月 1 日至 8 月 1 日的天数差异相同(无论闰年如何)。 在 PHP 中,引用变量在函数签名中,而不是在调用中。将您所有的 &amp; 移至签名处。【参考方案4】:

使用date_diff() 尝试这个非常简单的答案,这是经过测试的。

$date1 = date_create("2017-11-27");
$date2 = date_create("2018-12-29");
$diff=date_diff($date1,$date2);
$months = $diff->format("%m months");
$years = $diff->format("%y years");
$days = $diff->format("%d days");

echo $years .' '.$months.' '.$days;

输出是:

1 years 1 months 2 days

【讨论】:

【参考方案5】:

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);

【讨论】:

【参考方案6】:

一个简单的函数

function time_difference($time_1, $time_2, $limit = null)


    $val_1 = new DateTime($time_1);
    $val_2 = new DateTime($time_2);

    $interval = $val_1->diff($val_2);

    $output = array(
        "year" => $interval->y,
        "month" => $interval->m,
        "day" => $interval->d,
        "hour" => $interval->h,
        "minute" => $interval->i,
        "second" => $interval->s
    );

    $return = "";
    foreach ($output AS $key => $value) 

        if ($value == 1)
            $return .= $value . " " . $key . " ";
        elseif ($value >= 1)
            $return .= $value . " " . $key . "s ";

        if ($key == $limit)
            return trim($return);
    
    return trim($return);

使用喜欢

echo time_difference ($time_1, $time_2, "day");

会像2 years 8 months 2 days一样返回

【讨论】:

【参考方案7】:

我更喜欢使用 date_createdate_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

更多信息请阅读PHP date_diff manual

根据手册date_diff是一个别名 DateTime::diff()

【讨论】:

【参考方案8】:
$date = '2012.11.13';
$dateOfReturn = '2017.10.31';

$substract = str_replace('.', '-', $date);

$substract2 = str_replace('.', '-', $dateOfReturn);



$date1 = $substract;
$date2 = $substract2;

$ts1 = strtotime($date1);
$ts2 = strtotime($date2);

$year1 = date('Y', $ts1);
$year2 = date('Y', $ts2);

$month1 = date('m', $ts1);
$month2 = date('m', $ts2);

echo $diff = (($year2 - $year1) * 12) + ($month2 - $month1);

【讨论】:

【参考方案9】:

对于 php 版本 >=5.3 :创建两个日期对象,然后使用 date_diff() 函数。它将返回 php DateInterval 对象。 see documentation

$date1=date_create("2007-03-24");
$date2=date_create("2009-06-26");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");

【讨论】:

【参考方案10】:

你总是可以使用下面的函数,它可以返回以年和月为单位的年龄(即1年4个月)

function getAge($dob, $age_at_date)
  
    $d1 = new DateTime($dob);
    $d2 = new DateTime($age_at_date);
    $age = $d2->diff($d1);
    $years = $age->y;
    $months = $age->m;

    return $years.'.'.months;

或者如果你想在当前日期计算年龄,你可以使用

function getAge($dob)
  
    $d1 = new DateTime($dob);
    $d2 = new DateTime(date());
    $age = $d2->diff($d1);
    $years = $age->y;
    $months = $age->m;

    return $years.'.'.months;

【讨论】:

【参考方案11】:

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 manual

来自手册:

从 PHP 5.2.2 开始,可以使用比较运算符来比较 DateTime 对象。

$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)

【讨论】:

+1 DateTime 可以正确处理闰年和时区,而且书架上有一本好书:phparch.com/books/… 有没有一种方法可以给出两个日期时间之间的总秒数? (不加组件,即) @Panique $interval->days 和 $interval->d 是不同的度量。您上面的评论是正确的“显示总天数(不像上面那样分为年,月和日)” @potatoe 你可能想要$date2-&gt;format('U') - $date1-&gt;format('U') 请注意,在某些 PHP 版本的 Windows 上,DateInterval 的天数属性(始终为 6015)存在一个错误:bugs.php.net/bug.php?id=51184(请参阅那里的 cmets 以获取修复/解决方法)【参考方案12】:

很简单:

    <?php
        $date1 = date_create("2007-03-24");
        echo "Start date: ".$date1->format("Y-m-d")."<br>";
        $date2 = date_create("2009-06-26");
        echo "End date: ".$date2->format("Y-m-d")."<br>";
        $diff = date_diff($date1,$date2);
        echo "Difference between start date and end date: ".$diff->format("%y years, %m months and %d days")."<br>";
    ?>

详情请查看以下链接:

PHP: date_diff - Manual

请注意,它适用于 PHP 5.3.0 或更高版本。

【讨论】:

【参考方案13】:

对此我有一些简单的逻辑:

<?php
    per_days_diff('2011-12-12','2011-12-29')
    function per_days_diff($start_date, $end_date) 
        $per_days = 0;
        $noOfWeek = 0;
        $noOfWeekEnd = 0;
        $highSeason=array("7", "8");

        $current_date = strtotime($start_date);
        $current_date += (24 * 3600);
        $end_date = strtotime($end_date);

        $seassion = (in_array(date('m', $current_date), $highSeason))?"2":"1";

        $noOfdays = array('');

        while ($current_date <= $end_date) 
            if ($current_date <= $end_date) 
                $date = date('N', $current_date);
                array_push($noOfdays,$date);
                $current_date = strtotime('+1 day', $current_date);
            
        

        $finalDays = array_shift($noOfdays);
        //print_r($noOfdays);
        $weekFirst = array("week"=>array(),"weekEnd"=>array());
        for($i = 0; $i < count($noOfdays); $i++)
        
            if ($noOfdays[$i] == 1)
            
                //echo "This is week";
                //echo "<br/>";
                if($noOfdays[$i+6]==7)
                
                    $noOfWeek++;
                    $i=$i+6;
                
                else
                
                    $per_days++;
                
                //array_push($weekFirst["week"],$day);
            
            else if($noOfdays[$i]==5)
            
                //echo "This is weekend";
                //echo "<br/>";
                if($noOfdays[$i+2] ==7)
                
                    $noOfWeekEnd++;
                    $i = $i+2;
                
                else
                
                    $per_days++;
                
                //echo "After weekend value:- ".$i;
                //echo "<br/>";
            
            else
            
                $per_days++;
            
        

        /*echo $noOfWeek;
          echo "<br/>";
          echo $noOfWeekEnd;
          echo "<br/>";
          print_r($per_days);
          echo "<br/>";
          print_r($weekFirst);
        */

        $duration = array("weeks"=>$noOfWeek, "weekends"=>$noOfWeekEnd, "perDay"=>$per_days, "seassion"=>$seassion);
        return $duration;
      ?>

【讨论】:

示例代码的末尾似乎缺少一些东西(结束大括号和“?>”?)。 “简单”逻辑。这些至少是 40 行纯代码。【参考方案14】:

前段时间我写了一个format_date 函数,因为它提供了许多关于你想要约会方式的选项

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);
    

【讨论】:

这个答案和 khaldonno 的答案一样错误。它假设(案例 10)一年有 365 天(每四年有 366 天(公历的 100 年 / 400 年规则除外)),并且一个月有 30 天(大约是 30.42 天在非闰年)。即使有更好的常数,它也只是平均正确,不一定对任何两个特定日期都是正确的。【参考方案15】:

看看下面的链接。这是迄今为止我找到的最佳答案.. :)

function dateDiff ($d1, $d2) 

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

 // end function dateDiff

无论您在哪个日期更早或更晚 日期参数。该函数使用 PHP ABS() 绝对值 总是返回一个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。所以如果你正在寻找天数 由输入日期之间(包括输入日期)之间的所有日期表示, 您需要在此函数的结果中添加一 (1)。

例如,差异(由上述函数返回) 在 2013-02-09 和 2013-02-14 之间是 5。但是天数还是 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

【讨论】:

问题要求的差异是年数、月数和天数,而不是总天数。 很棒的人,为我在几天内得到了差异,谢谢【参考方案16】:
<?php
    $today = strtotime("2011-02-03 00:00:00");
    $myBirthDate = strtotime("1964-10-30 00:00:00");
    printf("Days since my birthday: ", ($today - $myBirthDate)/60/60/24);
?>

【讨论】:

问题要求的差异是。这会将差异输出为总天数。【参考方案17】:

我在下一页找到了您的文章,其中包含许多关于 PHP 日期时间计算的参考资料。

使用 PHP 计算两个日期(和时间)之间的差异。以下页面提供了一系列不同的方法(总共 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(小时、分钟)、天、月或年的差异。

见PHP Date Time – 7 Methods to Calculate the Difference between 2 dates。

【讨论】:

【参考方案18】:

“如果”日期存储在mysql中,我发现在数据库级别进行差异计算更容易......然后根据Day,Hour,Min,Sec输出,适当地解析和显示结果.. .

mysql> select firstName, convert_tz(loginDate, '+00:00', '-04:00') as loginDate, TIMESTAMPDIFF(DAY, loginDate, now()) as 'Day', TIMESTAMPDIFF(HOUR, loginDate, now())+4 as 'Hour', TIMESTAMPDIFF(MINUTE, loginDate, now())+(60*4) as 'Min', TIMESTAMPDIFF(SECOND, loginDate, now())+(60*60*4) as 'Sec' from User_ where userId != '10158' AND userId != '10198' group by emailAddress order by loginDate desc;
 +-----------+---------------------+------+------+------+--------+
 | firstName | loginDate           | Day  | Hour | Min  | Sec    |
 +-----------+---------------------+------+------+------+--------+
 | Peter     | 2014-03-30 18:54:40 |    0 |    4 |  244 |  14644 |
 | Keith     | 2014-03-30 18:54:11 |    0 |    4 |  244 |  14673 |
 | Andres    | 2014-03-28 09:20:10 |    2 |   61 | 3698 | 221914 |
 | Nadeem    | 2014-03-26 09:33:43 |    4 |  109 | 6565 | 393901 |
 +-----------+---------------------+------+------+------+--------+
 4 rows in set (0.00 sec)

【讨论】:

【参考方案19】:

由于每个人都在发布代码示例,这里是另一个版本。

我想要一个函数来显示从秒到年的差异(只有一个单位)。对于超过 1 天的时间段,我希望它在午夜翻转(从周三上午 9 点看到的周一上午 10 点是 2 天前,而不是 1 天)。对于超过一个月的时间段,我希望翻转在当月的同一天(包括 30/31 天的月份和闰年)。

这是我想出的:

/**
 * Returns how long ago something happened in the past, showing it
 * as n seconds / minutes / hours / days / weeks / months / years ago.
 *
 * For periods over a day, it rolls over at midnight (so doesn't depend
 * on current time of day), and it correctly accounts for month-lengths
 * and leap-years (months and years rollover on current day of month).
 *
 * $param string $timestamp in DateTime format
 * $return string description of interval
 */
function ago($timestamp)

    $then = date_create($timestamp);

    // for anything over 1 day, make it rollover on midnight
    $today = date_create('tomorrow'); // ie end of today
    $diff = date_diff($then, $today);

    if ($diff->y > 0) return $diff->y.' year'.($diff->y>1?'s':'').' ago';
    if ($diff->m > 0) return $diff->m.' month'.($diff->m>1?'s':'').' ago';
    $diffW = floor($diff->d / 7);
    if ($diffW > 0) return $diffW.' week'.($diffW>1?'s':'').' ago';
    if ($diff->d > 1) return $diff->d.' day'.($diff->d>1?'s':'').' ago';

    // for anything less than 1 day, base it off 'now'
    $now = date_create();
    $diff = date_diff($then, $now);

    if ($diff->d > 0) return 'yesterday';
    if ($diff->h > 0) return $diff->h.' hour'.($diff->h>1?'s':'').' ago';
    if ($diff->i > 0) return $diff->i.' minute'.($diff->i>1?'s':'').' ago';
    return $diff->s.' second'.($diff->s==1?'':'s').' ago';

【讨论】:

【参考方案20】:

我在 PHP 5.2 中遇到了同样的问题,并用 MySQL 解决了它。可能不是您正在寻找的内容,但这可以解决问题并返回天数:

$datediff_q = $dbh->prepare("SELECT DATEDIFF(:date2, :date1)");
$datediff_q->bindValue(':date1', '2007-03-24', PDO::PARAM_STR);
$datediff_q->bindValue(':date2', '2009-06-26', PDO::PARAM_STR);
$datediff = ($datediff_q->execute()) ? $datediff_q->fetchColumn(0) : false;

更多信息在这里http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff

【讨论】:

【参考方案21】:

这将尝试检测是否给出了时间戳,并将返回未来的日期/时间作为负值:

<?php

function time_diff($start, $end = NULL, $convert_to_timestamp = FALSE) 
  // If $convert_to_timestamp is not explicitly set to TRUE,
  // check to see if it was accidental:
  if ($convert_to_timestamp || !is_numeric($start)) 
    // If $convert_to_timestamp is TRUE, convert to timestamp:
    $timestamp_start = strtotime($start);
  
  else 
    // Otherwise, leave it as a timestamp:
    $timestamp_start = $start;
  
  // Same as above, but make sure $end has actually been overridden with a non-null,
  // non-empty, non-numeric value:
  if (!is_null($end) && (!empty($end) && !is_numeric($end))) 
    $timestamp_end = strtotime($end);
  
  else 
    // If $end is NULL or empty and non-numeric value, assume the end time desired
    // is the current time (useful for age, etc):
    $timestamp_end = time();
  
  // Regardless, set the start and end times to an integer:
  $start_time = (int) $timestamp_start;
  $end_time = (int) $timestamp_end;

  // Assign these values as the params for $then and $now:
  $start_time_var = 'start_time';
  $end_time_var = 'end_time';
  // Use this to determine if the output is positive (time passed) or negative (future):
  $pos_neg = 1;

  // If the end time is at a later time than the start time, do the opposite:
  if ($end_time <= $start_time) 
    $start_time_var = 'end_time';
    $end_time_var = 'start_time';
    $pos_neg = -1;
  

  // Convert everything to the proper format, and do some math:
  $then = new DateTime(date('Y-m-d H:i:s', $$start_time_var));
  $now = new DateTime(date('Y-m-d H:i:s', $$end_time_var));

  $years_then = $then->format('Y');
  $years_now = $now->format('Y');
  $years = $years_now - $years_then;

  $months_then = $then->format('m');
  $months_now = $now->format('m');
  $months = $months_now - $months_then;

  $days_then = $then->format('d');
  $days_now = $now->format('d');
  $days = $days_now - $days_then;

  $hours_then = $then->format('H');
  $hours_now = $now->format('H');
  $hours = $hours_now - $hours_then;

  $minutes_then = $then->format('i');
  $minutes_now = $now->format('i');
  $minutes = $minutes_now - $minutes_then;

  $seconds_then = $then->format('s');
  $seconds_now = $now->format('s');
  $seconds = $seconds_now - $seconds_then;

  if ($seconds < 0) 
    $minutes -= 1;
    $seconds += 60;
  
  if ($minutes < 0) 
    $hours -= 1;
    $minutes += 60;
  
  if ($hours < 0) 
    $days -= 1;
    $hours += 24;
  
  $months_last = $months_now - 1;
  if ($months_now == 1) 
    $years_now -= 1;
    $months_last = 12;
  

  // "Thirty days hath September, April, June, and November" ;)
  if ($months_last == 9 || $months_last == 4 || $months_last == 6 || $months_last == 11) 
    $days_last_month = 30;
  
  else if ($months_last == 2) 
    // Factor in leap years:
    if (($years_now % 4) == 0) 
      $days_last_month = 29;
    
    else 
      $days_last_month = 28;
    
  
  else 
    $days_last_month = 31;
  
  if ($days < 0) 
    $months -= 1;
    $days += $days_last_month;
  
  if ($months < 0) 
    $years -= 1;
    $months += 12;
  

  // Finally, multiply each value by either 1 (in which case it will stay the same),
  // or by -1 (in which case it will become negative, for future dates).
  // Note: 0 * 1 == 0 * -1 == 0
  $out = new stdClass;
  $out->years = (int) $years * $pos_neg;
  $out->months = (int) $months * $pos_neg;
  $out->days = (int) $days * $pos_neg;
  $out->hours = (int) $hours * $pos_neg;
  $out->minutes = (int) $minutes * $pos_neg;
  $out->seconds = (int) $seconds * $pos_neg;
  return $out;

示例用法:

<?php
  $birthday = 'June 2, 1971';
  $check_age_for_this_date = 'June 3, 1999 8:53pm';
  $age = time_diff($birthday, $check_age_for_this_date)->years;
  print $age;// 28

或者:

<?php
  $christmas_2020 = 'December 25, 2020';
  $countdown = time_diff($christmas_2020);
  print_r($countdown);

【讨论】:

【参考方案22】:

使用示例:

echo time_diff_string('2013-05-01 00:22:35', 'now');
echo time_diff_string('2013-05-01 00:22:35', 'now', true);

输出:

4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago

功能:

function time_diff_string($from, $to, $full = false) 
    $from = new DateTime($from);
    $to = new DateTime($to);
    $diff = $to->diff($from);

    $diff->w = floor($diff->d / 7);
    $diff->d -= $diff->w * 7;

    $string = array(
        'y' => 'year',
        'm' => 'month',
        'w' => 'week',
        'd' => 'day',
        'h' => 'hour',
        'i' => 'minute',
        's' => 'second',
    );
    foreach ($string as $k => &$v) 
        if ($diff->$k) 
            $v = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
         else 
            unset($string[$k]);
        
    

    if (!$full) $string = array_slice($string, 0, 1);
    return $string ? implode(', ', $string) . ' ago' : 'just now';

【讨论】:

如果我想确定差异是否大于 30 分钟,我该怎么办? @OfirAttia:您在 SO 上有很多类似的问题,请使用搜索。 Simple demo【参考方案23】:

这是我的功能。需要 PHP >= 5.3.4。它使用 DateTime 类。非常快,很快,可以区分两个日期,甚至是所谓的“时间”。

if(function_exists('grk_Datetime_Since') === FALSE)
    function grk_Datetime_Since($From, $To='', $Prefix='', $Suffix=' ago', $Words=array())
        #   Est-ce qu'on calcul jusqu'à un moment précis ? Probablement pas, on utilise maintenant
        if(empty($To) === TRUE)
            $To = time();
        

        #   On va s'assurer que $From est numérique
        if(is_int($From) === FALSE)
            $From = strtotime($From);
        ;

        #   On va s'assurer que $To est numérique
        if(is_int($To) === FALSE)
            $To = strtotime($To);
        

        #   On a une erreur ?
        if($From === FALSE OR $From === -1 OR $To === FALSE OR $To === -1)
            return FALSE;
        

        #   On va créer deux objets de date
        $From = new DateTime(@date('Y-m-d H:i:s', $From), new DateTimeZone('GMT'));
        $To   = new DateTime(@date('Y-m-d H:i:s', $To), new DateTimeZone('GMT'));

        #   On va calculer la différence entre $From et $To
        if(($Diff = $From->diff($To)) === FALSE)
            return FALSE;
        

        #   On va merger le tableau des noms (par défaut, anglais)
        $Words = array_merge(array(
            'year'      => 'year',
            'years'     => 'years',
            'month'     => 'month',
            'months'    => 'months',
            'week'      => 'week',
            'weeks'     => 'weeks',
            'day'       => 'day',
            'days'      => 'days',
            'hour'      => 'hour',
            'hours'     => 'hours',
            'minute'    => 'minute',
            'minutes'   => 'minutes',
            'second'    => 'second',
            'seconds'   => 'seconds'
        ), $Words);

        #   On va créer la chaîne maintenant
        if($Diff->y > 1)
            $Text = $Diff->y.' '.$Words['years'];
         elseif($Diff->y == 1)
            $Text = '1 '.$Words['year'];
         elseif($Diff->m > 1)
            $Text = $Diff->m.' '.$Words['months'];
         elseif($Diff->m == 1)
            $Text = '1 '.$Words['month'];
         elseif($Diff->d > 7)
            $Text = ceil($Diff->d/7).' '.$Words['weeks'];
         elseif($Diff->d == 7)
            $Text = '1 '.$Words['week'];
         elseif($Diff->d > 1)
            $Text = $Diff->d.' '.$Words['days'];
         elseif($Diff->d == 1)
            $Text = '1 '.$Words['day'];
         elseif($Diff->h > 1)
            $Text = $Diff->h.' '.$Words['hours'];
         elseif($Diff->h == 1)
            $Text = '1 '.$Words['hour'];
         elseif($Diff->i > 1)
            $Text = $Diff->i.' '.$Words['minutes'];
         elseif($Diff->i == 1)
            $Text = '1 '.$Words['minute'];
         elseif($Diff->s > 1)
            $Text = $Diff->s.' '.$Words['seconds'];
         else 
            $Text = '1 '.$Words['second'];
        

        return $Prefix.$Text.$Suffix;
    

【讨论】:

【参考方案24】:
$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 ";

【讨论】:

【参考方案25】:

一分钱,一磅: 我刚刚回顾了几个解决方案,所有解决方案都使用 floor() 提供了一个复杂的解决方案,然后四舍五入到 26 年 12 个月和 2 天的解决方案,本来应该是 25 年、11 个月和 20 天!!!!

这是我的这个问题的版本: 可能不优雅,可能没有很好的编码,但如果你不计算闰年,可以更接近答案,显然闰年可以编码到这个,但在这种情况下 - 正如其他人所说,也许你可以提供这个答案:: 我已经包含了所有 TEST 条件和 print_r,以便您可以更清楚地看到结果的构造: 来了,

// 设置你的输入日期/变量::

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

//我们需要将ISO yyyy-mm-dd格式EXPLODE为yyyy mm dd,如下::

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

$zDate[ ] = explode('-', $ISOtodaysDate); print_r ($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0个月0天

这就是我在 2013 年 6 月 22 日的营业时间 - 哎呀!

【讨论】:

【参考方案26】:

您还可以使用以下代码通过向上取整返回日期差异 $date1 = $duedate; // 分配截止日期 echo $date2 = date("Y-m-d"); // 当前的日期 $ts1 = strtotime($date1); $ts2 = strtotime($date2); $seconds_diff = $ts1 - $ts2; 回声 $datediff = ceil(($seconds_diff/3600)/24); // 返回天数

如果您使用 php 的 floor 方法而不是 ceil ,它将返回向下的整数部分。请在此处检查差异,有时如果您的登台服务器时区与实时站点时区不同,在这种情况下您可能会得到不同的结果,因此请相应地更改条件。

【讨论】:

【参考方案27】:

最好的做法是使用 PHP 的 DateTime(和 DateInterval)对象。每个日期都封装在一个DateTime对象中,然后就可以做两者的区别了:

$first_date = new DateTime("2012-11-30 17:03:30");
$second_date = new DateTime("2012-12-21 00:00:00");

DateTime 对象将接受strtotime() 将接受的任何格式。如果需要更具体的日期格式,可以使用 DateTime::createFromFormat() 来创建 DateTime 对象。

两个对象都被实例化后,你用DateTime::diff()从另一个中减去一个。

$difference = $first_date->diff($second_date);

$difference 现在拥有一个带有差异信息的DateInterval 对象。 var_dump() 看起来像这样:

object(DateInterval)
  public 'y' => int 0
  public 'm' => int 0
  public 'd' => int 20
  public 'h' => int 6
  public 'i' => int 56
  public 's' => int 30
  public 'invert' => int 0
  public 'days' => int 20

要格式化 DateInterval 对象,我们需要检查每个值,如果为 0 则将其排除:

/**
 * Format an interval to show all existing components.
 * If the interval doesn't have a time component (years, months, etc)
 * That component won't be displayed.
 *
 * @param DateInterval $interval The interval
 *
 * @return string Formatted interval string.
 */
function format_interval(DateInterval $interval) 
    $result = "";
    if ($interval->y)  $result .= $interval->format("%y years "); 
    if ($interval->m)  $result .= $interval->format("%m months "); 
    if ($interval->d)  $result .= $interval->format("%d days "); 
    if ($interval->h)  $result .= $interval->format("%h hours "); 
    if ($interval->i)  $result .= $interval->format("%i minutes "); 
    if ($interval->s)  $result .= $interval->format("%s seconds "); 

    return $result;

现在剩下的就是在 $difference DateInterval 对象上调用我们的函数:

echo format_interval($difference);

我们得到正确的结果:

20 天 6 小时 56 分 30 秒

用于实现目标的完整代码:

/**
 * Format an interval to show all existing components.
 * If the interval doesn't have a time component (years, months, etc)
 * That component won't be displayed.
 *
 * @param DateInterval $interval The interval
 *
 * @return string Formatted interval string.
 */
function format_interval(DateInterval $interval) 
    $result = "";
    if ($interval->y)  $result .= $interval->format("%y years "); 
    if ($interval->m)  $result .= $interval->format("%m months "); 
    if ($interval->d)  $result .= $interval->format("%d days "); 
    if ($interval->h)  $result .= $interval->format("%h hours "); 
    if ($interval->i)  $result .= $interval->format("%i minutes "); 
    if ($interval->s)  $result .= $interval->format("%s seconds "); 

    return $result;


$first_date = new DateTime("2012-11-30 17:03:30");
$second_date = new DateTime("2012-12-21 00:00:00");

$difference = $first_date->diff($second_date);

echo format_interval($difference);

【讨论】:

DateTime() 不是一个函数,它是一个对象,它从 PHP 5.2 开始就存在了。确保您的服务器支持它。 @SecondRikudo DateTime::Diff 需要 PHP 5.3.0 我们有问题,将 first_date 换成 second_date 并且得到相同的结果?为什么不说 0 天 0 小时 0 分钟 0 秒或只有 0。例如:2012-11-30 17:03:30 - 2012-12-21 00:00:00 和 2012-12-21 00:00:00 - 2012-11-30 17:03:30 得到相同的结果。 因为 diff 为您提供了两次之间的差异。无论哪个日期晚,差值都不为 0。 这是一个真的很好的答案,因为它提供了一个清晰的函数,可以从代码库中的任何位置调用,而无需大量时间计算。其他答案允许您即时放弃解决症状而不是解决问题的回显计算......我添加的唯一元素(几乎所有其他帖子都没有涵盖这一点)是 $interval 元素的复数如果超过 1。【参考方案28】:

DateInterval 很棒,但有几点需要注意:

    仅适用于 PHP 5.3+(但这真的不再是一个好借口) 仅支持年、月、日、小时、分钟和秒(无周) 它计算上述所有 + 天的差异(您无法仅获得月份的差异)

为了克服这个问题,我编写了以下代码(从 @enobrev answer 改进):

function date_dif($since, $until, $keys = 'year|month|week|day|hour|minute|second')

    $date = array_map('strtotime', array($since, $until));

    if ((count($date = array_filter($date, 'is_int')) == 2) && (sort($date) === true))
    
        $result = array_fill_keys(explode('|', $keys), 0);

        foreach (preg_grep('~^(?:year|month)~i', $result) as $key => $value)
        
            while ($date[1] >= strtotime(sprintf('+%u %s', $value + 1, $key), $date[0]))
            
                ++$value;
            

            $date[0] = strtotime(sprintf('+%u %s', $result[$key] = $value, $key), $date[0]);
        

        foreach (preg_grep('~^(?:year|month)~i', $result, PREG_GREP_INVERT) as $key => $value)
        
            if (($value = intval(abs($date[0] - $date[1]) / strtotime(sprintf('%u %s', 1, $key), 0))) > 0)
            
                $date[0] = strtotime(sprintf('+%u %s', $result[$key] = $value, $key), $date[0]);
            
        

        return $result;
    

    return false;

它运行两个循环;第一个通过暴力破解处理相对间隔(年和月),第二个通过简单的算术计算额外的绝对间隔(所以它更快):

echo humanize(date_dif('2007-03-24', '2009-07-31', 'second')); // 74300400 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'minute|second')); // 1238400 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'hour|minute|second')); // 20640 hours, 0 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|day')); // 2 years, 129 days
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week')); // 2 years, 18 weeks
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week|day')); // 2 years, 18 weeks, 3 days
echo humanize(date_dif('2007-03-24', '2009-07-31')); // 2 years, 4 months, 1 week, 0 days, 0 hours, 0 minutes, 0 seconds

function humanize($array)

    $result = array();

    foreach ($array as $key => $value)
    
        $result[$key] = $value . ' ' . $key;

        if ($value != 1)
        
            $result[$key] .= 's';
        
    

    return implode(', ', $result);

【讨论】:

@PeterMortensen:应该可以,但我不保证。设置您的时区并尝试一下。【参考方案29】:
// If you just want to see the year difference then use this function.
// Using the logic I've created you may also create month and day difference
// which I did not provide here so you may have the efforts to use your brain.
// :)
$date1='2009-01-01';
$date2='2010-01-01';
echo getYearDifference ($date1,$date2);
function getYearDifference($date1=strtotime($date1),$date2=strtotime($date2))
    $year = 0;
    while($date2 > $date1 = strtotime('+1 year', $date1))
        ++$year;
    
    return $year;

【讨论】:

"strtotime('+1 year', $date1)" 是否考虑闰年?【参考方案30】:

当 PHP 5.3(分别为 date_diff())不可用时,我正在使用我编写的以下函数:

        function dateDifference($startDate, $endDate)
        
            $startDate = strtotime($startDate);
            $endDate = strtotime($endDate);
            if ($startDate === false || $startDate < 0 || $endDate === false || $endDate < 0 || $startDate > $endDate)
                return false;

            $years = date('Y', $endDate) - date('Y', $startDate);

            $endMonth = date('m', $endDate);
            $startMonth = date('m', $startDate);

            // Calculate months
            $months = $endMonth - $startMonth;
            if ($months <= 0)  
                $months += 12;
                $years--;
            
            if ($years < 0)
                return false;

            // Calculate the days
            $measure = ($months == 1) ? 'month' : 'months';
            $days = $endDate - strtotime('+' . $months . ' ' . $measure, $startDate);
            $days = date('z', $days);   

            return array($years, $months, $days);
        

【讨论】:

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