如何使用PHP找到两个日期之间的天数?
当前回答
这是我的改进版本,显示1年2月25天(s),如果通过第二个参数。
class App_Sandbox_String_Util {
/**
* Usage: App_Sandbox_String_Util::getDateDiff();
* @param int $your_date timestamp
* @param bool $hr human readable. e.g. 1 year(s) 2 day(s)
* @see http://stackoverflow.com/questions/2040560/finding-the-number-of-days-between-two-dates
* @see http://qSandbox.com
*/
static public function getDateDiff($your_date, $hr = 0) {
$now = time(); // or your date as well
$datediff = $now - $your_date;
$days = floor( $datediff / ( 3600 * 24 ) );
$label = '';
if ($hr) {
if ($days >= 365) { // over a year
$years = floor($days / 365);
$label .= $years . ' Year(s)';
$days -= 365 * $years;
}
if ($days) {
$months = floor( $days / 30 );
$label .= ' ' . $months . ' Month(s)';
$days -= 30 * $months;
}
if ($days) {
$label .= ' ' . $days . ' day(s)';
}
} else {
$label = $days;
}
return $label;
}
}
其他回答
看看所有的答案,我写了一个通用函数,适用于所有的PHP版本。
if(!function_exists('date_between')) :
function date_between($date_start, $date_end)
{
if(!$date_start || !$date_end) return 0;
if( class_exists('DateTime') )
{
$date_start = new DateTime( $date_start );
$date_end = new DateTime( $date_end );
return $date_end->diff($date_start)->format('%a');
}
else
{
return abs( round( ( strtotime($date_start) - strtotime($date_end) ) / 86400 ) );
}
}
endif;
一般来说,我使用“DateTime”来查找两个日期之间的天数。但如果出于某种原因,一些服务器设置没有启用'DateTime',它将使用'strtotime()'简单(但不安全)计算。
从PHP 5.3及以上版本开始,添加了新的日期/时间函数来获得不同:
$datetime1 = new DateTime("2010-06-20");
$datetime2 = new DateTime("2011-06-22");
$difference = $datetime1->diff($datetime2);
echo 'Difference: '.$difference->y.' years, '
.$difference->m.' months, '
.$difference->d.' days';
print_r($difference);
结果如下:
Difference: 1 years, 0 months, 2 days
DateInterval Object
(
[y] => 1
[m] => 0
[d] => 2
[h] => 0
[i] => 0
[s] => 0
[invert] => 0
[days] => 367
)
希望能有所帮助!
使用这个简单的函数。声明函数
<?php
function dateDiff($firstDate,$secondDate){
$firstDate = strtotime($firstDate);
$secondDate = strtotime($secondDate);
$datediff = $firstDate - $secondDate;
$output = round($datediff / (60 * 60 * 24));
return $output;
}
?>
像这样调用这个函数
<?php
echo dateDiff("2018-01-01","2018-12-31");
// OR
$firstDate = "2018-01-01";
$secondDate = "2018-01-01";
echo dateDiff($firstDate,$secondDate);
?>
// Change this to the day in the future
$day = 15;
// Change this to the month in the future
$month = 11;
// Change this to the year in the future
$year = 2012;
// $days is the number of days between now and the date in the future
$days = (int)((mktime (0,0,0,$month,$day,$year) - time(void))/86400);
echo "There are $days days until $day/$month/$year";
这个工作!
$start = strtotime('2010-01-25');
$end = strtotime('2010-02-20');
$days_between = ceil(abs($end - $start) / 86400);
推荐文章
- 如何将XML转换成PHP数组?
- 如何将对象转换为数组?
- 如何将python datetime转换为字符串,具有可读格式的日期?
- 从IP地址获取位置
- 在c#中创建一个特定时区的DateTime
- 获取数组值的键名
- HTTPS和SSL3_GET_SERVER_CERTIFICATE:证书验证失败,CA is OK
- PHP -获取bool值,当为false时返回false
- 在foreach中通过引用传递
- 如何触发命令行PHP脚本的XDebug分析器?
- 如何找出如果你使用HTTPS没有$_SERVER['HTTPS']
- 更好的方法检查变量为null或空字符串?
- 当使用Composer的开发/生产开关时,如何正确部署?
- 自动删除Laravel (Eloquent ORM)中的相关行
- 如何减去X天从一个日期对象在Java?