真的需要帮助解决这个问题。
我有一系列offerDays-$offerDays = array(1,6);（平均周一和周六）；
例如，输入日期为2014年9月3日星期三。下一个优惠日期是星期六2014/09/06
问题：如何在php中确定最近的第二天？类似的东西
$offerDays = array(1,6);
$inputDate = '2014/09/03'; // filled date, not actual date
$offerDate = findDate($inputDate, $offerDays); // returns 2014/09/06
function findDate($inputDate, $offerDays) {
return 'nearest next date to $inputDate defined by $offerDays';
}
这里可以有一个offerdays的变量数组。
$offerDays = array(1,6);
$inputDate = '2014/09/04'; // filled date, not actual date
$offerDate = findDate($inputDate, $offerDays);
function findDate($inputDate, $offerDays) {
$date = DateTime::createFromFormat('Y/m/d', $inputDate);
$num = $date->format('w');
$min = 10; //initialize minimum days
foreach($offerDays as $o){ //loop through all the offerdays to find the minimum difference
$dif = $o - $num;
if($dif>0 && $dif < $min){
$min = $dif ;
}
}
// if minimum days is not changed then get the first offerday from next week
if($min == 10){
$min = 6 - $num + min($offerDays);
}
//add the days till next offerday
$add = new DateInterval('P'.$min.'D');
$nextOfferDay = $date->add($add)->format('Y/m/d');
return $nextOfferDay;
}
$currentDate = date('Y-m-d');
$frequencyDays = [2, 4, 5];
function nextNearestDate($currentDate, $frequencyDays) {
$weekDay = date('w', strtotime($currentDate));
$min = 10; // Some random number>7
foreach ($frequencyDays as $day) {
$dif = $day - $weekDay;
if ($dif == 0) {
$min = $dif;
break;
}else if ($dif > 0) {
$positiveDay[] = $dif;
}else if ($dif < 0) {
$negativeDay[] = $dif;
}
}
if ($min != 0) {
if (isset($positiveDay) && count($positiveDay)) {
$min = min($positiveDay);
} else {
$min = min($negativeDay);
$min = 7 + $min;
}
}
return $newDay = date('Y-m-d', strtotime('+' . $min . ' days', strtotime($currentDate)));
}
$date_str="2014-09-03";
if(date('l' , $date_str) != 'Saturday')
{
//Your Code Here
}
简单版本
<?php
/**
* Return the next date that falls on one of the specified weekdays.
*
* @param DateTimeInterface $inputDate The search begins on the day after this
* @param array $weekdays An array of permitted weekdays
* represented by an integer (0=Sunday)
*
* @return DateTimeInterface|null The first date that meets the criteria
*/
function findDate(DateTimeInterface $inputDate, array $weekdays)
{
$weekdays = array_flip($weekdays);
foreach (new DatePeriod($inputDate, new DateInterval('P1D'), 7, DatePeriod::EXCLUDE_START_DATE) as $possibleDate)
{
if (isset($weekdays[$possibleDate->format('w')]) === true)
{
return $possibleDate;
}
}
}
// Example:
$weekdays = [1, 6];
$inputDate = '2014/09/03';
$offerDate = findDate(new DateTimeImmutable($inputDate), $weekdays);
echo $offerDate->format('Y-m-d'), "'n";
扩展版本
如果你想通过建立一个稍微抽象一点的基础来解决这个问题，它可能看起来像：
<?php
function WeekDayIterator(DateTimeInterface $inputDate, array $weekdays)
{
$weekdays = array_intersect($weekdays, range(0,6));
if (!$weekdays)
{
return;
}
$possibleDate = new DateTimeImmutable('@' . $inputDate->getTimeStamp());
$oneDay = new DateInterval('P1D');
$weekdays = array_flip($weekdays);
while (true)
{
$possibleDate = $possibleDate->add($oneDay);
if (isset($weekdays[$possibleDate->format('w')]) === true)
{
yield $possibleDate;
}
}
}
function findDate(DateTimeInterface $inputDate, array $weekdays)
{
return WeekDayIterator($inputDate, $weekdays)->current();
}
// Example:
$weekdays = [1, 6];
$inputDate = '2014/09/03';
$offerDate = findDate(new DateTimeImmutable($inputDate), $weekdays);
echo $offerDate->format('Y-m-d'), "'n";
foreach (new LimitIterator(WeekDayIterator(new DateTimeImmutable($inputDate), $weekdays), 0, 10) as $date)
{
echo $date->format('Y-m-d (w)'), "'n";
}
这个较长的版本为您提供了一个迭代器，它将在所有有效日期上无限循环。findDate方法现在只是一种隐藏实现细节的方便方法。
但这里的优点是，您现在可以解决类似的问题，而不必编写更多的管道。例如，您可以使用LimitIterator计算接下来的10个日期。（也许你想提供可用日期的下拉列表等）
第二个版本在很大程度上依赖于5.5，但第一个版本不是（如果将DateTimeImmutable替换为DateTime）。