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
我想带来一个稍微不同的视角，这似乎没有被提及。
你可以用声明的方式解决这个问题（就像任何其他问题一样）。重点是问你需要什么，而不是如何到达那里。
在这里，你需要与众不同。但这有什么不同？这是一个间隔，正如在最受欢迎的答案中所提到的。问题是如何获取它。您可以不显式调用diff（）方法，而是按开始日期和结束日期创建一个间隔，即按日期范围：
$startDate = '2007-03-24';
$endDate = '2009-06-26';
$range = new FromRange(new ISO8601DateTime($startDate), new ISO8601DateTime($endDate));
所有诸如闰年之类的复杂问题都已经解决了。现在，当您有一个固定开始日期时间的间隔时，您可以获得一个人类可读的版本：
var_dump((new HumanReadable($range))->value());
它输出的正是你所需要的。
如果您需要一些自定义格式，这也不是问题。您可以使用ISO8601格式化类，该类接受具有六个参数的调用：年、月、日、小时、分钟和秒：
(new ISO8601Formatted(
new FromRange(
new ISO8601DateTime('2017-07-03T14:27:39+00:00'),
new ISO8601DateTime('2018-07-05T14:27:39.235487+00:00')
),
function (int $years, int $months, int $days, int $hours, int $minutes, int $seconds) {
return $years >= 1 ? 'More than a year' : 'Less than a year';
}
))
->value();
它的产量超过一年。
有关此方法的更多信息，请查看快速入门条目。
其他回答
我有一些简单的逻辑：
<?php
per_days_diff('2011-12-12','2011-12-29')
function per_days_diff($start_date, $end_date) {
$per_days = 0;
$noOfWeek = 0;
$noOfWeekEnd = 0;
$highSeason=array("7", "8");
$current_date = strtotime($start_date);
$current_date += (24 * 3600);
$end_date = strtotime($end_date);
$seassion = (in_array(date('m', $current_date), $highSeason))?"2":"1";
$noOfdays = array('');
while ($current_date <= $end_date) {
if ($current_date <= $end_date) {
$date = date('N', $current_date);
array_push($noOfdays,$date);
$current_date = strtotime('+1 day', $current_date);
}
}
$finalDays = array_shift($noOfdays);
//print_r($noOfdays);
$weekFirst = array("week"=>array(),"weekEnd"=>array());
for($i = 0; $i < count($noOfdays); $i++)
{
if ($noOfdays[$i] == 1)
{
//echo "This is week";
//echo "<br/>";
if($noOfdays[$i+6]==7)
{
$noOfWeek++;
$i=$i+6;
}
else
{
$per_days++;
}
//array_push($weekFirst["week"],$day);
}
else if($noOfdays[$i]==5)
{
//echo "This is weekend";
//echo "<br/>";
if($noOfdays[$i+2] ==7)
{
$noOfWeekEnd++;
$i = $i+2;
}
else
{
$per_days++;
}
//echo "After weekend value:- ".$i;
//echo "<br/>";
}
else
{
$per_days++;
}
}
/*echo $noOfWeek;
echo "<br/>";
echo $noOfWeekEnd;
echo "<br/>";
print_r($per_days);
echo "<br/>";
print_r($weekFirst);
*/
$duration = array("weeks"=>$noOfWeek, "weekends"=>$noOfWeekEnd, "perDay"=>$per_days, "seassion"=>$seassion);
return $duration;
?>
前段时间，我编写了一个format_date函数，因为它提供了许多关于日期的选项：
function format_date($date, $type, $seperator="-")
{
if($date)
{
$day = date("j", strtotime($date));
$month = date("n", strtotime($date));
$year = date("Y", strtotime($date));
$hour = date("H", strtotime($date));
$min = date("i", strtotime($date));
$sec = date("s", strtotime($date));
switch($type)
{
case 0: $date = date("Y".$seperator."m".$seperator."d",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 1: $date = date("D, F j, Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 2: $date = date("d".$seperator."m".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 3: $date = date("d".$seperator."M".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 4: $date = date("d".$seperator."M".$seperator."Y h:i A",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 5: $date = date("m".$seperator."d".$seperator."Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 6: $date = date("M",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 7: $date = date("Y",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 8: $date = date("j",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 9: $date = date("n",mktime($hour, $min, $sec, $month, $day, $year)); break;
case 10:
$diff = abs(strtotime($date) - strtotime(date("Y-m-d h:i:s")));
$years = floor($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));
$date = $years . " years, " . $months . " months, " . $days . "days";
}
}
return($date);
}
使用示例：
echo time_diff_string('2013-05-01 00:22:35', 'now');
echo time_diff_string('2013-05-01 00:22:35', 'now', true);
输出：
4 months ago
4 months, 2 weeks, 3 days, 1 hour, 49 minutes, 15 seconds ago
功能：
function time_diff_string($from, $to, $full = false) {
$from = new DateTime($from);
$to = new DateTime($to);
$diff = $to->diff($from);
$diff->w = floor($diff->d / 7);
$diff->d -= $diff->w * 7;
$string = array(
'y' => 'year',
'm' => 'month',
'w' => 'week',
'd' => 'day',
'h' => 'hour',
'i' => 'minute',
's' => 'second',
);
foreach ($string as $k => &$v) {
if ($diff->$k) {
$v = $diff->$k . ' ' . $v . ($diff->$k > 1 ? 's' : '');
} else {
unset($string[$k]);
}
}
if (!$full) $string = array_slice($string, 0, 1);
return $string ? implode(', ', $string) . ' ago' : 'just now';
}
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, $days);
最终结果是：26年零个月零天
这就是我在2013年6月22日做生意的时间——哎呦！
使用此函数
//function Diff between Dates
//////////////////////////////////////////////////////////////////////
//PARA: Date Should In YYYY-MM-DD Format
//RESULT FORMAT:
// '%y Year %m Month %d Day %h Hours %i Minute %s Seconds' => 1 Year 3 Month 14 Day 11 Hours 49 Minute 36 Seconds
// '%y Year %m Month %d Day' => 1 Year 3 Month 14 Days
// '%m Month %d Day' => 3 Month 14 Day
// '%d Day %h Hours' => 14 Day 11 Hours
// '%d Day' => 14 Days
// '%h Hours %i Minute %s Seconds' => 11 Hours 49 Minute 36 Seconds
// '%i Minute %s Seconds' => 49 Minute 36 Seconds
// '%h Hours => 11 Hours
// '%a Days => 468 Days
//////////////////////////////////////////////////////////////////////
function dateDifference($date_1 , $date_2 , $differenceFormat = '%a' )
{
$datetime1 = date_create($date_1);
$datetime2 = date_create($date_2);
$interval = date_diff($datetime1, $datetime2);
return $interval->format($differenceFormat);
}
仅根据需要设置参数$differenceFormat例如，我希望你的年龄在到岁之间，以月和日为单位
dateDifference（日期（'Y-m-d'），$date，'%Y%m%d'）
或其他格式
dateDifference（日期（'Y-m-d'），$date，'%Y-%m-%d'）
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