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
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, $days);
}
其他回答
当PHP 5.3(分别为date_diff())不可用时,我使用了我编写的以下函数:
function dateDifference($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, $days);
}
我在下面的页面上找到了您的文章,其中包含了许多PHP日期时间计算的参考。
使用PHP计算两个日期(和时间)之间的差异。下一页提供了一系列不同的方法(共7种),用于使用PHP执行日期/时间计算,以确定两个日期之间的时间差(小时、弹药)、天、月或年。
请参阅PHP日期时间–计算两个日期之间差值的7种方法。
我不知道你是否在使用PHP框架,但很多PHP框架都有日期/时间库和助手来帮助你避免重新发明轮子。
例如,CodeIgniter具有timespan()函数。只需输入两个Unix时间戳,就会自动生成如下结果:
1 Year, 10 Months, 2 Weeks, 5 Days, 10 Hours, 16 Minutes
http://codeigniter.com/user_guide/helpers/date_helper.html
查看以下链接。这是迄今为止我找到的最好的答案
function dateDiff ($d1, $d2) {
// Return the number of days between the two dates:
return round(abs(strtotime($d1) - strtotime($d2))/86400);
} // end function dateDiff
当你通过日期参数。函数使用PHP ABS()绝对值始终返回正数作为两者之间的天数日期。请记住,两个日期之间的天数不是包括两个日期。因此,如果您正在寻找天数由输入日期之间的所有日期表示,您需要向该函数的结果添加一(1)。例如,差异(由上述函数返回)2013-02-09和2013-02-14之间的值为5。但天数或日期范围2013-02-09-2013-02-14表示的日期为6。
http://www.bizinfosys.com/php/date-difference.html
前段时间,我编写了一个format_date函数,因为它提供了许多关于日期的选项:
function format_date($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);
}