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
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
其他回答
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, $seconds);
我在下面的页面上找到了您的文章,其中包含了许多PHP日期时间计算的参考。
使用PHP计算两个日期(和时间)之间的差异。下一页提供了一系列不同的方法(共7种),用于使用PHP执行日期/时间计算,以确定两个日期之间的时间差(小时、弹药)、天、月或年。
请参阅PHP日期时间–计算两个日期之间差值的7种方法。
DateInterval很好,但它有几个注意事项:
仅适用于PHP 5.3+(但这真的不再是一个好借口)仅支持年、月、日、小时、分钟和秒(无周)它计算上述所有+天的差异(你不能只计算月的差异)
为了克服这个问题,我编写了以下代码(由@enobrev答案改进而来):
function date_dif($since, $until, $keys = 'year|month|week|day|hour|minute|second')
{
$date = array_map('strtotime', array($since, $until));
if ((count($date = array_filter($date, 'is_int')) == 2) && (sort($date) === true))
{
$result = array_fill_keys(explode('|', $keys), 0);
foreach (preg_grep('~^(?:year|month)~i', $result) as $key => $value)
{
while ($date[1] >= strtotime(sprintf('+%u %s', $value + 1, $key), $date[0]))
{
++$value;
}
$date[0] = strtotime(sprintf('+%u %s', $result[$key] = $value, $key), $date[0]);
}
foreach (preg_grep('~^(?:year|month)~i', $result, PREG_GREP_INVERT) as $key => $value)
{
if (($value = intval(abs($date[0] - $date[1]) / strtotime(sprintf('%u %s', 1, $key), 0))) > 0)
{
$date[0] = strtotime(sprintf('+%u %s', $result[$key] = $value, $key), $date[0]);
}
}
return $result;
}
return false;
}
它运行两个循环;第一个算法通过暴力强制处理相对间隔(年和月),第二个算法通过简单的算法计算额外的绝对间隔(因此速度更快):
echo humanize(date_dif('2007-03-24', '2009-07-31', 'second')); // 74300400 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'minute|second')); // 1238400 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'hour|minute|second')); // 20640 hours, 0 minutes, 0 seconds
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|day')); // 2 years, 129 days
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week')); // 2 years, 18 weeks
echo humanize(date_dif('2007-03-24', '2009-07-31', 'year|week|day')); // 2 years, 18 weeks, 3 days
echo humanize(date_dif('2007-03-24', '2009-07-31')); // 2 years, 4 months, 1 week, 0 days, 0 hours, 0 minutes, 0 seconds
function humanize($array)
{
$result = array();
foreach ($array as $key => $value)
{
$result[$key] = $value . ' ' . $key;
if ($value != 1)
{
$result[$key] .= 's';
}
}
return implode(', ', $result);
}
// 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;
}
由于每个人都在发布代码示例,这里有另一个版本。
我想要一个函数来显示从秒到年的差异(仅一个单位)。对于超过1天的时段,我希望它在午夜滚动(周一上午10点到周三上午9点是2天前,而不是1天前)。对于超过一个月的时间段,我希望滚动在当月的同一天(包括30/31天的月份和闰年)。
这就是我想到的:
/**
* Returns how long ago something happened in the past, showing it
* as n seconds / minutes / hours / days / weeks / months / years ago.
*
* For periods over a day, it rolls over at midnight (so doesn't depend
* on current time of day), and it correctly accounts for month-lengths
* and leap-years (months and years rollover on current day of month).
*
* $param string $timestamp in DateTime format
* $return string description of interval
*/
function ago($timestamp)
{
$then = date_create($timestamp);
// for anything over 1 day, make it rollover on midnight
$today = date_create('tomorrow'); // ie end of today
$diff = date_diff($then, $today);
if ($diff->y > 0) return $diff->y.' year'.($diff->y>1?'s':'').' ago';
if ($diff->m > 0) return $diff->m.' month'.($diff->m>1?'s':'').' ago';
$diffW = floor($diff->d / 7);
if ($diffW > 0) return $diffW.' week'.($diffW>1?'s':'').' ago';
if ($diff->d > 1) return $diff->d.' day'.($diff->d>1?'s':'').' ago';
// for anything less than 1 day, base it off 'now'
$now = date_create();
$diff = date_diff($then, $now);
if ($diff->d > 0) return 'yesterday';
if ($diff->h > 0) return $diff->h.' hour'.($diff->h>1?'s':'').' ago';
if ($diff->i > 0) return $diff->i.' minute'.($diff->i>1?'s':'').' ago';
return $diff->s.' second'.($diff->s==1?'':'s').' ago';
}