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?
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()) ? $datediff_q->fetchColumn(0) : false;
此处有更多信息http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_datediff
其他回答
使用此函数
//function Diff between Dates
//////////////////////////////////////////////////////////////////////
//PARA: Date Should In YYYY-MM-DD Format
//RESULT FORMAT:
// '%y Year %m Month %d Day %h Hours %i Minute %s Seconds' => 1 Year 3 Month 14 Day 11 Hours 49 Minute 36 Seconds
// '%y Year %m Month %d Day' => 1 Year 3 Month 14 Days
// '%m Month %d Day' => 3 Month 14 Day
// '%d Day %h Hours' => 14 Day 11 Hours
// '%d Day' => 14 Days
// '%h Hours %i Minute %s Seconds' => 11 Hours 49 Minute 36 Seconds
// '%i Minute %s Seconds' => 49 Minute 36 Seconds
// '%h Hours => 11 Hours
// '%a Days => 468 Days
//////////////////////////////////////////////////////////////////////
function dateDifference($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);
}
仅根据需要设置参数$differenceFormat例如,我希望你的年龄在到岁之间,以月和日为单位
dateDifference(日期('Y-m-d'),$date,'%Y%m%d')
或其他格式
dateDifference(日期('Y-m-d'),$date,'%Y-%m-%d')
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)
<?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);
?>
由于每个人都在发布代码示例,这里有另一个版本。
我想要一个函数来显示从秒到年的差异(仅一个单位)。对于超过1天的时段,我希望它在午夜滚动(周一上午10点到周三上午9点是2天前,而不是1天前)。对于超过一个月的时间段,我希望滚动在当月的同一天(包括30/31天的月份和闰年)。
这就是我想到的:
/**
* 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';
}
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);