开始我很抱歉我的英语不好，我希望你能理解我的问题并找到解决方案......
我的问题是我有两个时期
the first period: dateStart1-dateEnd1
the secondperiod: dateStart2-dateEnd2
对于第一对夫妇，频率 = 2 ：
dte=dateStar1;dateEnd1>dte;dte+2week
对于第二个，频率 = 3 ：
dte=dateStar2;dateEnd2>dte;dte+3week
示例：
first period 2016-04-04 -> 2016-05-09
frequence 2 weeks 2016-04-04 , 2016-04-18 , 2016-05-02
the second : 2016-04-11 -> 2016-05-09
frequence 3 weeks 2016-04-11, 2016-05-02
the two periods overlaps in 2016-05-02
我的问题是如何知道两个时期或日期重叠的最小周数？谢谢
可能有更好的算法，但我唯一能想到的是检查一段时间内所有可用的周数。
$result = [];
$current = clone $dateStart1;
while ($current < $dateEnd1) {
// Calculate the difference betweenthe current date and dateStart2
$diff = $current->diff($dateStart2);
// Check if the difference is a multiple of the frequency, i.e. if the number of days is congruent to 7 times the frequency
if ($diff->days % ($frequency2 * 7) == 0) {
// If the anwser is yes, adds the current date to the list of overlapping dates.
$result[] = clone $current;
}
// Get the next date possible for period 1 by adding $frequency1 weeks to the current date, and then try again till $dateEnd1 is reached.
$current = $current->add(new 'DateInterval('PT'.$frequency1.'W'););
}
return $result;
我还没有完全测试它，但这至少可以帮助您上轨。