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
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, $days);
最终结果是：26年零个月零天
这就是我在2013年6月22日做生意的时间——哎呦！
其他回答
这是我的职责。所需PHP>=5.3.4。它使用DateTime类。非常快，很快，可以区分两个日期，甚至所谓的“开始时间”。
if(function_exists('grk_Datetime_Since') === FALSE){
function grk_Datetime_Since($From, $To='', $Prefix='', $Suffix=' ago', $Words=array()){
# Est-ce qu'on calcul jusqu'à un moment précis ? Probablement pas, on utilise maintenant
if(empty($To) === TRUE){
$To = time();
}
# On va s'assurer que $From est numérique
if(is_int($From) === FALSE){
$From = strtotime($From);
};
# On va s'assurer que $To est numérique
if(is_int($To) === FALSE){
$To = strtotime($To);
}
# On a une erreur ?
if($From === FALSE OR $From === -1 OR $To === FALSE OR $To === -1){
return FALSE;
}
# On va créer deux objets de date
$From = new DateTime(@date('Y-m-d H:i:s', $From), new DateTimeZone('GMT'));
$To = new DateTime(@date('Y-m-d H:i:s', $To), new DateTimeZone('GMT'));
# On va calculer la différence entre $From et $To
if(($Diff = $From->diff($To)) === FALSE){
return FALSE;
}
# On va merger le tableau des noms (par défaut, anglais)
$Words = array_merge(array(
'year' => 'year',
'years' => 'years',
'month' => 'month',
'months' => 'months',
'week' => 'week',
'weeks' => 'weeks',
'day' => 'day',
'days' => 'days',
'hour' => 'hour',
'hours' => 'hours',
'minute' => 'minute',
'minutes' => 'minutes',
'second' => 'second',
'seconds' => 'seconds'
), $Words);
# On va créer la chaîne maintenant
if($Diff->y > 1){
$Text = $Diff->y.' '.$Words['years'];
} elseif($Diff->y == 1){
$Text = '1 '.$Words['year'];
} elseif($Diff->m > 1){
$Text = $Diff->m.' '.$Words['months'];
} elseif($Diff->m == 1){
$Text = '1 '.$Words['month'];
} elseif($Diff->d > 7){
$Text = ceil($Diff->d/7).' '.$Words['weeks'];
} elseif($Diff->d == 7){
$Text = '1 '.$Words['week'];
} elseif($Diff->d > 1){
$Text = $Diff->d.' '.$Words['days'];
} elseif($Diff->d == 1){
$Text = '1 '.$Words['day'];
} elseif($Diff->h > 1){
$Text = $Diff->h.' '.$Words['hours'];
} elseif($Diff->h == 1){
$Text = '1 '.$Words['hour'];
} elseif($Diff->i > 1){
$Text = $Diff->i.' '.$Words['minutes'];
} elseif($Diff->i == 1){
$Text = '1 '.$Words['minute'];
} elseif($Diff->s > 1){
$Text = $Diff->s.' '.$Words['seconds'];
} else {
$Text = '1 '.$Words['second'];
}
return $Prefix.$Text.$Suffix;
}
}
我更喜欢使用date_create和date_diff对象。
代码：
$date1 = date_create("2007-03-24");
$date2 = date_create("2009-06-26");
$dateDifference = date_diff($date1, $date2)->format('%y years, %m months and %d days');
echo $dateDifference;
输出：
2 years, 3 months and 2 days
有关更多信息，请阅读PHP date_diff手册
根据手册date_diff是的别名日期时间：：diff（）
我不知道你是否在使用PHP框架，但很多PHP框架都有日期/时间库和助手来帮助你避免重新发明轮子。
例如，CodeIgniter具有timespan（）函数。只需输入两个Unix时间戳，就会自动生成如下结果：
1 Year, 10 Months, 2 Weeks, 5 Days, 10 Hours, 16 Minutes
http://codeigniter.com/user_guide/helpers/date_helper.html
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！
最好的做法是使用PHP的DateTime（和DateInterval）对象。每个日期都封装在DateTime对象中，然后可以在两者之间进行区别：
$first_date = new DateTime("2012-11-30 17:03:30");
$second_date = new DateTime("2012-12-21 00:00:00");
DateTime对象将接受strtotime（）的任何格式。如果需要更具体的日期格式，则可以使用DateTime:：createFromFormat（）创建DateTime对象。
两个对象实例化后，使用DateTime:：diff（）从另一个对象中减去一个对象。
$difference = $first_date->diff($second_date);
$difference现在保存一个包含差异信息的DateInterval对象。var_dump（）如下所示：
object(DateInterval)
public 'y' => int 0
public 'm' => int 0
public 'd' => int 20
public 'h' => int 6
public 'i' => int 56
public 's' => int 30
public 'invert' => int 0
public 'days' => int 20
要格式化DateInterval对象，我们需要检查每个值，如果值为0，则将其排除：
/**
* Format an interval to show all existing components.
* If the interval doesn't have a time component (years, months, etc)
* That component won't be displayed.
*
* @param DateInterval $interval The interval
*
* @return string Formatted interval string.
*/
function format_interval(DateInterval $interval) {
$result = "";
if ($interval->y) { $result .= $interval->format("%y years "); }
if ($interval->m) { $result .= $interval->format("%m months "); }
if ($interval->d) { $result .= $interval->format("%d days "); }
if ($interval->h) { $result .= $interval->format("%h hours "); }
if ($interval->i) { $result .= $interval->format("%i minutes "); }
if ($interval->s) { $result .= $interval->format("%s seconds "); }
return $result;
}
现在剩下的就是调用$differenceDateInterval对象上的函数：
echo format_interval($difference);
我们得到了正确的结果：
20天6小时56分30秒
用于实现目标的完整代码：
/**
* Format an interval to show all existing components.
* If the interval doesn't have a time component (years, months, etc)
* That component won't be displayed.
*
* @param DateInterval $interval The interval
*
* @return string Formatted interval string.
*/
function format_interval(DateInterval $interval) {
$result = "";
if ($interval->y) { $result .= $interval->format("%y years "); }
if ($interval->m) { $result .= $interval->format("%m months "); }
if ($interval->d) { $result .= $interval->format("%d days "); }
if ($interval->h) { $result .= $interval->format("%h hours "); }
if ($interval->i) { $result .= $interval->format("%i minutes "); }
if ($interval->s) { $result .= $interval->format("%s seconds "); }
return $result;
}
$first_date = new DateTime("2012-11-30 17:03:30");
$second_date = new DateTime("2012-12-21 00:00:00");
$difference = $first_date->diff($second_date);
echo format_interval($difference);
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