php 在php中查找2个unix时间戳之间的天数
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Finding days between 2 unix timestamps in php
提问by dotty
Hay, i have a database holding events. There are 2 fields 'start' and 'end', these contain timestamps. When an admin enters these dates, they only have the ability to set the day,month,year. So we are only dealing with stamps containing days,months,years, not hours,minutes,seconds (hours,minutes and seconds are set to 0,0,0).
嘿，我有一个保存事件的数据库。有 2 个字段“开始”和“结束”，它们包含时间戳。当管理员输入这些日期时，他们只能设置日、月、年。所以我们只处理包含天、月、年的邮票，而不是小时、分钟、秒（小时、分钟和秒设置为 0,0,0）。
I have an event with the start time as 1262304000 and the end time as 1262908800. These convert to Jan 1 2010 and Jan 8 2010. How would i get all the days between these timestamps? I want to be able to return Jan 2 2010 (1262390400), Jan 3 2010 (1262476800) .. all the way to the end stamp. These events could cross over into different months, say May 28 to June 14.
我有一个事件，开始时间为 1262304000，结束时间为 1262908800。这些转换为 2010 年 1 月 1 日和 2010 年 1 月 8 日。我如何获得这些时间戳之间的所有天数？我希望能够返回 2010 年 1 月 2 日 (1262390400)、2010 年 1 月 3 日 (1262476800) .. 一直到结束戳。这些事件可能跨越不同的月份，比如 5 月 28 日至 6 月 14 日。
Any ideas how to do this?
任何想法如何做到这一点？
回答by Vincent Savard
You just have to calculate the number of seconds between the two dates, then divide to get days :
您只需要计算两个日期之间的秒数，然后除以得到天数：
$numDays = abs($smallestTimestamp - $biggestTimestamp)/60/60/24;
Then, you can use a for loop to retrieve the dates :
然后，您可以使用 for 循环来检索日期：
$numDays = abs($smallestTimestamp - $biggestTimestamp)/60/60/24;
for ($i = 1; $i < $numDays; $i++) {
echo date('Y m d', strtotime("+{$i} day", $smallestTimestamp)) . '<br />';
}
Again, if you don't know which timestamp is the smallest, you can use the min() function (second argument in strtotime).
同样，如果您不知道哪个时间戳最小，您可以使用 min() 函数（strtotime 中的第二个参数）。
回答by Etienne Marais
I think that a quick workaround for this is to subtract the amount of a days worth of seconds from the end_stamp until you get to the start_tag.
我认为对此的快速解决方法是从 end_stamp 中减去几天的秒数，直到到达 start_tag。
//1 day = 86400 seconds
I would build an array of the days to use later.
我会建立一个天数数组以供以后使用。
EDIT (example)
编辑（示例）
$difference = 86400;
$days = array();
while ( $start_time < $end_time )
{
$days[] = date('M j Y', $end_time);
$end_time -= $difference;
}
This should cover any time frame even if its over a bunch of months.
这应该涵盖任何时间范围，即使它超过几个月。
回答by Valentin Flachsel
Try this:
尝试这个：
while($date_start <= $date_end) {
echo date('M d Y', $date_start) . '<br>';
$date_start = $date_start + 86400;
}
Hope this helps !
希望这可以帮助 ！
回答by zod
$d1=mktime(22,0,0,1,1,2007);
$d2=mktime(0,0,0,1,2,2007);
echo "Hours difference = ".floor(($d2-$d1)/3600) . "<br>";
echo "Minutes difference = ".floor(($d2-$d1)/60) . "<br>";
echo "Seconds difference = " .($d2-$d1). "<br>";
echo "Month difference = ".floor(($d2-$d1)/2628000) . "<br>";
echo "Days difference = ".floor(($d2-$d1)/86400) . "<br>";
echo "Year difference = ".floor(($d2-$d1)/31536000) . "<br>";
http://www.plus2net.com/php_tutorial/date-diff.php
http://www.plus2net.com/php_tutorial/date-diff.php
回答by Halil ?zgür
$daysInBetween = range($startTs, $endTs, 86400);
$secondDay = date('M d Y', $daysInBetween[1]);
/*
$thirdDay = date('M d Y', $daysInBetween[2]);
...
*/
Note that the range()function is inclusive.
请注意，该range()函数是包含性的。
回答by sandy
**This is a very simple code for find days hours minutes and seconds in php**
$dbDate = strtotime("".$yourbdDate.""); // Database date
$endDate = time(); // current time
$diff = $endDate - $dbDate; /// diffrence
$days = floor($diff/86400); /// number of days
$hours = floor(($diff-$days*86400)/(60 * 60)); //// number of hours
$min = floor(($diff-($days*86400+$hours*3600))/60);///// numbers of minute
$second = $diff - ($days*86400+$hours*3600+$min*60); //// secondes
if($days > 0) echo $days." Days ago";
elseif($hours > 0) echo $hours." Hours ago";
elseif($min > 0) echo $min." Minute ago";
else echo "Just second ago";
回答by netcoder
Something like this?
像这样的东西？
$day = $start;
while ($day < $end) {
$day += 86400;
echo $day.' '.date('Y-m-d', $day).PHP_EOL;
}
By the way, 1262304000 is Dec 31, not Jan 1.
顺便说一下，1262304000 是 12 月 31 日，而不是 1 月 1 日。
回答by bkilinc
get the difference of two dates and divide it by 86400. abs(($date1 - $date2) / 86400) will produce the needed result
获取两个日期的差值并将其除以 86400。 abs(($date1 - $date2) / 86400) 将产生所需的结果
