用php获取下一个15日和/或第30天的日期？ [英] Get the date of the next 15th and/or 30th day with php?

问题描述

我需要获得今后几个月的即将到来的 15日和 30日的日期。 （如果二月份在范围内，则必须是28/29）。

I require to get the dates of the upcoming 15th and 30ths of the next months as for the current date. (If February is within range it must be 28/29th of course).

我可以使用mktime / strtotime还是使用其他方法？

Can I do this using mktime/strtotime or maybe using another method?

我有这个，但当然这只是得到这个和下个月的最后一天。我需要即将到来的15和30日。

I got this but of course this is for only get the last day of this and next month. I need the upcoming 15ths and 30ths instead.

$cuota1 = date('t-m-Y', strtotime('+15 days'));
    return $cuota1;

$cuota2 = date('t-m-Y', strtotime('+30 days'));
    return $cuota2;

$cuota3 = date('t-m-Y', strtotime('+45 days'));
    return $cuota3;

提前感谢

推荐答案

希望我有你的想法，可能这个解决方案是相当长的，但看起来像是防弹：

Hope I've got your idea right, and may be this solution is quite long, but looks like it's bullet-proof:

$numOfDays = date('t', $todayStamp);
$base = strtotime('+'.$numOfDays.' days', strtotime(date('Y-m-01', $todayStamp)));
$day15 = date('Y-m-15', $base);
$day30 = date('Y-m-'.min(date('t', $base), 30), $base);

其中 $ todayStamp 通常是 time（），但为了调试的目的，它可以是任意日期的 strtotime（）。例如，让我们来看下一个闰年的困难情况：

where $todayStamp is generally the value of time(), but for debug purposes it can be strtotime() of an arbitrary date. For example, let's take "difficult" case of the next leap year:

$today = '2016-01-30';
$todayStamp = strtotime($today);
$numOfDays = date('t', $todayStamp);
$base = strtotime('+'.$numOfDays.' days', strtotime(date('Y-m-01', $todayStamp)));
$day15 = date('Y-m-15', $base);
$day30 = date('Y-m-'.min(date('t', $base), 30), $base);
var_dump($day15);
var_dump($day30);

输出是

string(10) "2016-02-15"
string(10) "2016-02-29"

而对于20162-29，输出将为

and for 2016-02-29 the output will be

string(10) "2016-03-15"
string(10) "2016-03-30"

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