PHP strtotime 不带闰年
关于这个线程，我开发了部分解决方案：
function strtosecs($time,$now=null){
static $LEAPDIFF=86400;
$time=strtotime($time,$now);
return $time-(round((date('Y',$time)-1968)/4)*$LEAPDIFF);
}
该函数应该获取给定字符串的秒数而不检查闰年。
它计算 1970 年的闰年数 [(year-1986)/4]，将其乘以闰年与平年之间的秒数差（最终，它只是 1970 年的秒数）一天）。
最后，我简单地从计算的时间中删除所有多余的闰年秒。以下是输入/输出的一些示例：
// test code
echo strtosecs('+20 years',0).'=>'.(strtosecs('+20 years',0)/31536000);
echo strtosecs('+1 years',0).'=>'.(strtosecs('+1 years',0)/31536000);
// test output
630676800 => 19.998630136986
31471200 => 0.99794520547945
您可能会问为什么我要对输出进行除法？是为了测试一下； 31536000 是一年中的秒数，因此 19.99... 应该是 20，0.99... 应该是 1。 当然，我可以将其全部舍入并得到“正确”答案，但我担心不准确。
编辑1：因为它看起来并不明显，所以我的问题在于顽固性；你只要 20 年不问 PHP，它就会给你 19.99...，对吗？
编辑2：这一切似乎都可以归结为1968年的部分；
- 1970；在我尝试过的所有测试中发现它都是准确的。
- 1969；发现它这里使用（
...例如：（2008- 1969)/4 = 9.75...) 以及此处提到< /a>.第 2 年（+3 年）以后准确。 - 1968；如下所述，这是 UNIX 时间（1970 年）闰年的“零年”。 （对我来说）这听起来“正确”，但根本不准确。。
With regards to this thread, I've developed a partial solution:
function strtosecs($time,$now=null){
static $LEAPDIFF=86400;
$time=strtotime($time,$now);
return $time-(round((date('Y',$time)-1968)/4)*$LEAPDIFF);
}
The function is supposed to get the number of seconds given a string without checking leap-years.
It does this calculating the number of leap-years 1970 [(year-1986)/4], multiplying it by the difference in seconds between a leap-year and a normal year (which in the end, it's just the number of seconds in a day).
Finally, I simply remove all those excess leap-year seconds from the calculated time. Here's some examples of the inputs/outputs:
// test code
echo strtosecs('+20 years',0).'=>'.(strtosecs('+20 years',0)/31536000);
echo strtosecs('+1 years',0).'=>'.(strtosecs('+1 years',0)/31536000);
// test output
630676800 => 19.998630136986
31471200 => 0.99794520547945
You will probably ask why am I doing a division on the output? It's to test it out; 31536000 is the number of seconds in a year, so that 19.99... should be 20 and 0.99... should be a 1.
Sure, I could round it all and get "correct" answer, but I'm worried about the inaccuracies.
Edit1: Since it doesn't seem obvious, my problem is with inveteracies; you just don't ask PHP for 20 years and it gives you 19.99..., right?
Edit2: It all seems to boil down to the part about 1968;
- 1970; found it accurate in all tests I've tried.
- 1969; Found it used here (
...ex: (2008-1969)/4 = 9.75...) as well as mentioned here. Accurate after the 2nd year (+3 years) onwards. - 1968; as detailed below, this is "year zero" of leap years from unix time (1970). It sounds "right" (to me) but it isn't accurate, at all.
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评论（2）
这是否与使用 PHP 管理浮点数时固有的不准确性有关？
http://php.net/manual/en/language.types.float.php
Could this be related to the inherent inaccuracy experienced when using PHP to manage floating point numbers?
http://php.net/manual/en/language.types.float.php
您应该将计算中的1968（它从哪里来？）替换为unix时间的原点：1970，这样您将得到更准确的结果。编辑
您必须执行
intval来计算闰年的数量，它必须是整数：这将为您提供
+0 - 范围内的正确结果> +68，32位机器上unix时间结束You should replace1968in your calculation (where does it come from ?) by the origine of unix time :1970and you will get more accurate results.Edit
You have to do an
intvalto count the number of leapyears which must be an integer :This will give you correct results within the range
+0 -> +68, end of unix time on 32bit machine