Php 我试图精确地确定两个（可能非常遥远的）日期之间的十进制年数，这有什么错？

当前代码/示例（请仔细阅读日期/时间）：

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:01'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-03 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-01 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"

预期/期望输出：

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:01'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-03 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-01 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"

实际输出：

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:01'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-03 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-01 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"

分析：

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-02 15:00:01'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-03 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

$a = date_diff(date_create('1980-01-02 15:00:00'), date_create('2020-01-01 15:00:00'));
$average_number_of_days_for_the_relevant_years = bcdiv($a->days, $a->y);
var_dump(bcdiv($a->days, $average_number_of_days_for_the_relevant_years));

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.001000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999"

string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "40.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"
string(102) "39.000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000"

无论我尝试了什么（在过去和今天提问之前，我尝试了无数的事情），PHP似乎从未让我计算两个日期之间的准确“十进制年份”。它要么迫使我用“365”除法，这将导致日期跨度越长，数字越不准确；要么像我上面所做的那样，只给我“整数”：有人被认为正好是39岁，直到它切换到40岁为止

我也用

（float）

s尝试过这一点。同样的基本问题。我在绝望中使用了

bcdiv（）

，认为可能是浮点数把它和舍入之类的东西搞砸了

在这一点上，我真的不知道该怎么办。所有在线“解决方案”都大错特错。我看过很多代码，它们都做得不正确。遗憾的是，

date\u diff

对象只提供了“零散”的数据和总天数，但对于年数没有很好的十进制/浮点数。我不知道他们为什么会排除最有用的数据，但他们确实排除了，而我就在这里，在这个问题上浪费了我的生命。。。这很讽刺，因为这是关于计算年龄的

我再也睡不着了，除非这件事最终得到妥善准确的解决，不管日期相隔多远。这种事情让我发疯，对我来说，有一种需求是典型的，而现有的机制无法满足这种需求，尽管在我看来，这是迄今为止最常见的期望输出

---

能不能请一个真正聪明的人帮我一劳永逸地解决这个问题？（我对另一个不准确的解决方案不感兴趣，例如，如果你比较相隔千年的日期倍，它就会分崩离析。）

不可能确定两个日期之间年份的精确小数差

可以给出几年的分数作为说明。例如，3年零6个月可以写成3.5年

humanYearDiff（）函数提供了这样一个值以供说明。结果不准确，因此不适合进一步计算。为了简化问题，该月也按31天计算

function humanYearDiff(DateTime $start,Datetime $end){
  $diff = $start->diff($end);
  $days = $diff->d + $diff->h/24 + $diff->i/1440 + $diff->s/86400;
  $years = $diff->y + $diff->m/12.0 + $days/372;
  return $diff->invert ? -$years : $years;
}

一个小测试：

$testData = [
  ['1980-02-01 15:00','2020-02-01 15:00'],
  ['1980-02-01 15:00','2020-08-01 15:00'],
  ['1980-02-01 15:00','2020-05-01 15:00'],
  ['1980-02-01 15:00','2020-02-01 15:01'],
  ['1980-02-01 15:00','2020-02-01 14:59'],
  ['today','today +6 month'],
];

foreach($testData as $dateStartEnd){
  $floatYear = humanYearDiff(date_create($dateStartEnd[0]),date_create($dateStartEnd[1]));
  echo $dateStartEnd[0].' -> '.$dateStartEnd[1].' = '.$floatYear." Years<br>\n";
}

---

什么是“十进制年”有确切的定义吗？特别是闰年应该如何考虑？我想，闰年的开始日期或结束日期是在2月28日之前还是之后，都必须加以考虑？@CBroe这两个日期之间一年平均长度的百分比。只有当单位是固定数量时，单位的分数才有意义。这适用于小时、分钟和秒。年和月不是固定的。它的一小部分没有意义。