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
我有一些简单的逻辑:
<?php
per_days_diff('2011-12-12','2011-12-29')
function per_days_diff($start_date, $end_date) {
$per_days = 0;
$noOfWeek = 0;
$noOfWeekEnd = 0;
$highSeason=array("7", "8");
$current_date = strtotime($start_date);
$current_date += (24 * 3600);
$end_date = strtotime($end_date);
$seassion = (in_array(date('m', $current_date), $highSeason))?"2":"1";
$noOfdays = array('');
while ($current_date <= $end_date) {
if ($current_date <= $end_date) {
$date = date('N', $current_date);
array_push($noOfdays,$date);
$current_date = strtotime('+1 day', $current_date);
}
}
$finalDays = array_shift($noOfdays);
//print_r($noOfdays);
$weekFirst = array("week"=>array(),"weekEnd"=>array());
for($i = 0; $i < count($noOfdays); $i++)
{
if ($noOfdays[$i] == 1)
{
//echo "This is week";
//echo "<br/>";
if($noOfdays[$i+6]==7)
{
$noOfWeek++;
$i=$i+6;
}
else
{
$per_days++;
}
//array_push($weekFirst["week"],$day);
}
else if($noOfdays[$i]==5)
{
//echo "This is weekend";
//echo "<br/>";
if($noOfdays[$i+2] ==7)
{
$noOfWeekEnd++;
$i = $i+2;
}
else
{
$per_days++;
}
//echo "After weekend value:- ".$i;
//echo "<br/>";
}
else
{
$per_days++;
}
}
/*echo $noOfWeek;
echo "<br/>";
echo $noOfWeekEnd;
echo "<br/>";
print_r($per_days);
echo "<br/>";
print_r($weekFirst);
*/
$duration = array("weeks"=>$noOfWeek, "weekends"=>$noOfWeekEnd, "perDay"=>$per_days, "seassion"=>$seassion);
return $duration;
?>
其他回答
简单的功能
function time_difference($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,$time_2,“天”);
将返回2年8个月2天
您还可以使用以下代码通过向上舍入分数来返回日期差异$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。
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);
$date1 = date_create('2007-03-24');
$date2 = date_create('2009-06-26');
$interval = date_diff($date1, $date2);
echo "difference : " . $interval->y . " years, " . $interval->m." months, ".$interval->d." days ";