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

我不知道你是否在使用PHP框架,但很多PHP框架都有日期/时间库和助手来帮助你避免重新发明轮子。

例如,CodeIgniter具有timespan()函数。只需输入两个Unix时间戳,就会自动生成如下结果:

1 Year, 10 Months, 2 Weeks, 5 Days, 10 Hours, 16 Minutes

http://codeigniter.com/user_guide/helpers/date_helper.html

其他回答

查看小时、分钟和秒。。

$date1 = "2008-11-01 22:45:00"; 

$date2 = "2009-12-04 13:44:01"; 

$diff = abs(strtotime($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));

$hours   = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24)/ (60*60)); 

$minuts  = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60)/ 60); 

$seconds = floor(($diff - $years * 365*60*60*24 - $months*30*60*60*24 - $days*60*60*24 - $hours*60*60 - $minuts*60)); 

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

您还可以使用以下代码通过向上舍入分数来返回日期差异$date1=$duedate;//指定到期日echo$date2=日期(“Y-m-d”);//当前日期$ts1=字符串时间($date1);$ts2=字符串时间($date2);$seconds_diff=$ts1-$ts2;echo$datediff=ceil(($seconds_diff/3600)/24);//天内返回

如果您使用php的floor方法而不是ceil,它将返回舍入分数。请检查此处的差异,有时,如果您的临时服务器时区与现场站点时区不同,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。

<?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);
?>

简单的功能

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);
}

像这样使用

回波时间差($time_1,$time_2,“天”);

将返回2年8个月2天

我在PHP5.2中遇到了同样的问题,并用MySQL解决了这个问题。可能并不是你想要的,但这会奏效,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;

此处有更多信息http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff