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
$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 - $month1);
其他回答
当PHP 5.3(分别为date_diff())不可用时,我使用了我编写的以下函数:
function dateDifference($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);
}
// 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 $year;
}
function showTime($time){
$start = strtotime($time);
$end = strtotime(date("Y-m-d H:i:s"));
$minutes = ($end - $start)/60;
// years
if(($minutes / (60*24*365)) > 1){
$years = floor($minutes/(60*24*365));
return "From $years year( s ) ago";
}
// monthes
if(($minutes / (60*24*30)) > 1){
$monthes = floor($minutes/(60*24*30));
return "From $monthes monthe( s ) ago";
}
// days
if(($minutes / (60*24)) > 1){
$days = floor($minutes/(60*24));
return "From $days day( s ) ago";
}
// hours
if(($minutes / 60) > 1){
$hours = floor($minutes/60);
return "From $hours hour( s ) ago";
}
// minutes
if($minutes > 1){
$minutes = floor($minutes);
return "From $minutes minute( s ) ago";
}
}
echo showTime('2022-05-05 21:33:00');
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phpDateInterval对象。请参阅文档
$date1=date_create("2007-03-24");
$date2=date_create("2009-06-26");
$diff=date_diff($date1,$date2);
echo $diff->format("%R%a days");
<?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);
?>