[英]How can i use date_diff function with string values
我有2个字符串值,例如“ 24/10/2015”和“ 23/10/2015”,它们是动态值。 我需要这两个值之间的夜数,所以我试图使用date_diff但我无法管理它。 我尝试了strtotime,date_create_from_format等,但是没有用。 有什么建议吗?
示例代码:
$checkout = $_COOKIE['cout'];
$checkin = $_COOKIE['cin'];
$nights = date_diff(strtotime($checkout),strtotime($checkin));
这应该为您工作:
只需使用DateTime::createFromFormat()
创建一个DateTime
对象,然后可以使用diff()
获得区别,如下所示:
<?php
$dateOne = "24/10/2015";
$dateTwo = "23/10/2015";
$dateOne = DateTime::createFromFormat("d/m/Y", $dateOne);
$dateTwo = DateTime::createFromFormat("d/m/Y", $dateTwo);
$interval = $dateOne->diff($dateTwo);
echo $interval->format("%d " . ($interval->d > 1 || $interval->d == 0?"nights":"night"));
?>
输出:
1 night
这是解决方案。
$checkout = $_COOKIE['cout'];
$checkin = $_COOKIE['cin'];
$datediff = strtotime($checkout) - strtotime($checkin);
$night = floor($datediff/(60*60*24));
可以在php中这样使用:
$date1=date_create("2013-03-15");
$date2=date_create("2013-12-12");
echo date_diff($date1,$date2);
如果您没有使用正确格式的日期,则使用phps date()函数
Try this one to solve your problem
/****Code Start*****/
<?php
$checkout = $_COOKIE['cout'];
$checkin = $_COOKIE['cin'];
$diff = abs(strtotime($checkout) - strtotime($checkin));
$years = floor($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));
printf("%d years, %d months, %d days\n", $years, $months, $days);
?>
/****Code End*****/
尝试这个
$checkout = date_create("2015-10-12");
$checkin = date_create("2015-10-05");
$nights = date_diff($checkout,$checkin);
print_r($nights);
您将获得像这样的对象数组,
DateInterval Object
(
[y] => 0
[m] => 0
[d] => 7
[h] => 0
[i] => 0
[s] => 0
[weekday] => 0
[weekday_behavior] => 0
[first_last_day_of] => 0
[invert] => 1
[days] => 7
[special_type] => 0
[special_amount] => 0
[have_weekday_relative] => 0
[have_special_relative] => 0
)
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