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
最好的做法是使用PHP的DateTime(和DateInterval)对象。每个日期都封装在DateTime对象中,然后可以在两者之间进行区别:
$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()的任何格式。如果需要更具体的日期格式,则可以使用DateTime::createFromFormat()创建DateTime对象。
两个对象实例化后,使用DateTime::diff()从另一个对象中减去一个对象。
$difference = $first_date->diff($second_date);
$difference现在保存一个包含差异信息的DateInterval对象。var_dump()如下所示:
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,如果值为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 $result;
}
现在剩下的就是调用$differenceDateInterval对象上的函数:
echo format_interval($difference);
我们得到了正确的结果:
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($difference);
其他回答
查看以下链接。这是迄今为止我找到的最好的答案
function dateDiff ($d1, $d2) {
// Return the number of days between the two dates:
return round(abs(strtotime($d1) - strtotime($d2))/86400);
} // end function dateDiff
当你通过日期参数。函数使用PHP ABS()绝对值始终返回正数作为两者之间的天数日期。请记住,两个日期之间的天数不是包括两个日期。因此,如果您正在寻找天数由输入日期之间的所有日期表示,您需要向该函数的结果添加一(1)。例如,差异(由上述函数返回)2013-02-09和2013-02-14之间的值为5。但天数或日期范围2013-02-09-2013-02-14表示的日期为6。
http://www.bizinfosys.com/php/date-difference.html
我不知道你是否在使用PHP框架,但很多PHP框架都有日期/时间库和助手来帮助你避免重新发明轮子。
例如,CodeIgniter具有timespan()函数。只需输入两个Unix时间戳,就会自动生成如下结果:
1 Year, 10 Months, 2 Weeks, 5 Days, 10 Hours, 16 Minutes
http://codeigniter.com/user_guide/helpers/date_helper.html
当PHP 5.3(分别为date_diff())不可用时,我使用了我编写的以下函数:
function dateDifference($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);
}
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);
$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);