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
这是可运行的代码
$date1 = date_create('2007-03-24');
$date2 = date_create('2009-06-26');
$diff1 = date_diff($date1,$date2);
$daysdiff = $diff1->format("%R%a");
$daysdiff = abs($daysdiff);
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。
一便士一英镑:我刚刚回顾了几个解决方案,所有这些方案都使用floor()提供了一个复杂的解决方案,然后四舍五入到26年12个月零2天的解决方案中,原本应该是25年11个月零20天!!!!
这是我对这个问题的看法:可能不优雅,可能编码不好,但如果不计算LEAP年份,则提供了更接近答案的答案,显然闰年可以编码为,但在这种情况下-正如其他人所说,也许您可以提供以下答案:我已经包含了所有测试条件和print_r,以便您可以更清楚地看到结果的构造:在这里,
//设置输入日期/变量::
$ISOstartDate = "1987-06-22";
$ISOtodaysDate = "2013-06-22";
//我们需要将ISO yyyy-mm-dd格式分解为yyyy-mm-d格式,如下所示:
$yDate[]=爆炸('-',$ISOstartDate);print_r($yDate);
$zDate[]=爆炸('-',$ISOtodaysDate);print_r($zDate);
// Lets Sort of the Years!
// Lets Sort out the difference in YEARS between startDate and todaysDate ::
$years = $zDate[0][0] - $yDate[0][0];
// We need to collaborate if the month = month = 0, is before or after the Years Anniversary ie 11 months 22 days or 0 months 10 days...
if ($months == 0 and $zDate[0][1] > $ydate[0][1]) {
$years = $years -1;
}
// TEST result
echo "\nCurrent years => ".$years;
// Lets Sort out the difference in MONTHS between startDate and todaysDate ::
$months = $zDate[0][1] - $yDate[0][1];
// TEST result
echo "\nCurrent months => ".$months;
// Now how many DAYS has there been - this assumes that there is NO LEAP years, so the calculation is APPROXIMATE not 100%
// Lets cross reference the startDates Month = how many days are there in each month IF m-m = 0 which is a years anniversary
// We will use a switch to check the number of days between each month so we can calculate days before and after the years anniversary
switch ($yDate[0][1]){
case 01: $monthDays = '31'; break; // Jan
case 02: $monthDays = '28'; break; // Feb
case 03: $monthDays = '31'; break; // Mar
case 04: $monthDays = '30'; break; // Apr
case 05: $monthDays = '31'; break; // May
case 06: $monthDays = '30'; break; // Jun
case 07: $monthDays = '31'; break; // Jul
case 08: $monthDays = '31'; break; // Aug
case 09: $monthDays = '30'; break; // Sept
case 10: $monthDays = '31'; break; // Oct
case 11: $monthDays = '30'; break; // Nov
case 12: $monthDays = '31'; break; // Dec
};
// TEST return
echo "\nDays in start month ".$yDate[0][1]." => ".$monthDays;
// Lets correct the problem with 0 Months - is it 11 months + days, or 0 months +days???
$days = $zDate[0][2] - $yDate[0][2] +$monthDays;
echo "\nCurrent days => ".$days."\n";
// Lets now Correct the months to being either 11 or 0 Months, depending upon being + or - the years Anniversary date
// At the same time build in error correction for Anniversary dates not being 1yr 0m 31d... see if ($days == $monthDays )
if($days < $monthDays && $months == 0)
{
$months = 11; // If Before the years anniversary date
}
else {
$months = 0; // If After the years anniversary date
$years = $years+1; // Add +1 to year
$days = $days-$monthDays; // Need to correct days to how many days after anniversary date
};
// Day correction for Anniversary dates
if ($days == $monthDays ) // if todays date = the Anniversary DATE! set days to ZERO
{
$days = 0; // days set toZERO so 1 years 0 months 0 days
};
echo "\nTherefore, the number of years/ months/ days/ \nbetween start and todays date::\n\n";
printf("%d years, %d months, %d days\n", $years, $months, $days);
最终结果是:26年零个月零天
这就是我在2013年6月22日做生意的时间——哎呦!
我建议使用DateTime和DateInterval对象。
$date1 = new DateTime("2007-03-24");
$date2 = new DateTime("2009-06-26");
$interval = $date1->diff($date2);
echo "difference " . $interval->y . " years, " . $interval->m." months, ".$interval->d." days ";
// shows the total amount of days (not divided into years, months and days like above)
echo "difference " . $interval->days . " days ";
阅读更多php DateTime::diff手册
根据手册:
从PHP 5.2.2开始,DateTime对象可以使用比较运算符进行比较。
$date1 = new DateTime("now");
$date2 = new DateTime("tomorrow");
var_dump($date1 == $date2); // bool(false)
var_dump($date1 < $date2); // bool(true)
var_dump($date1 > $date2); // bool(false)
DateInterval很好,但它有几个注意事项:
仅适用于PHP 5.3+(但这真的不再是一个好借口)仅支持年、月、日、小时、分钟和秒(无周)它计算上述所有+天的差异(你不能只计算月的差异)
为了克服这个问题,我编写了以下代码(由@enobrev答案改进而来):
function date_dif($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);
}