Question

假设我有一个初始日期，其年份早于当年的年份，我想每7天重复一次，但仅限于当年。

我如何找到当年的第一次出现？

我意识到我可以用这样的循环来做到这一点：

$reOccurringEvent =new DateTime('2013-12-01');
$interval = new DateInterval('P7D');
while($reOccurringEvent->format('Y') < date('Y') ){
    $reOccurringEvent->add($interval);
}
echo $reOccurringEvent->format('d m Y'); //05 01 2014

但是我觉得应该有一种更有效的方法来实现这一目标，而不是在日期中重复添加一个间隔（如果初始日期是几年前会发生很多次）。

我希望能够计算应该添加间隔的次数，并且只需要一次。

我在想这样的事情：

$date = new DateTime();
$diff = $date->diff($reOccurringEvent)->days%7;

但显然这不起作用，我无法弄清楚如何做到这一点的逻辑。

Answer 1

我认为，如果你每隔7天做一次，你可以找到你的初始日期的星期几，然后在星期几那天得到当年的第一个日期......

查找星期几：How to find the day of week from a date using PHP?

查找今年某一天的日期：Getting first weekday in a month with strtotime

把它放在一起：

$date=Date("2/8/2012");

//Get the day of week for the date in question
$dayOfWeek  = date('l', strtotime($date));
echo "The day of week for the given date is: $dayOfWeek <br>"; 

//Get the current year
$thisYear = date("Y");
echo "This year: $thisYear <br>"; 

//Create a date with the first occurence of the day of week of the given date for the current year
$firstOccurenceThisYear = date("m/d/y", strtotime("January " .$thisYear ." " . $dayOfWeek));
echo "The first interval of the year is: $firstOccurenceThisYear"; 

/*
Output:
This year: 2014 
The day of week for the given date is: Wednesday 
The first interval of the year is: 01/01/14
*/

Answer 2

更一般地说，算法是找到给定日期和去年最后一天之间的间隔数。然后将间隔乘以间隔数+ 1，得到当前年份的第一个间隔。

$date1="12/9/2013"; 
$ts1 = strtotime($date1);
$ts2 = strtotime("12/31/" . Date("Y")-1);

//get the number of seconds between the date and first of the year    
$seconds_diff = $ts2 - $ts1;
echo "$seconds_diff <br>";

//get the number of days
$dayDiff=$seconds_diff/86400;

//how many intervals?
$intervalDays = "10";

//get the number of intervals from start date to last day of last year
$numIntervals = floor($dayDiff/$intervalDays);
echo  $numIntervals."<br>";

//now the total intervals to get into the current year is one more interval, turn this into days
$totIntervals= ($numIntervals* $intervalDays)+$intervalDays;

//Date Time date in question
$theDt = new DateTime($date1);

//Add the intervals we calculated to the date in question, and we have the first date of the interval for the current year...
$theDt->add(new DateInterval('P' . $totIntervals. 'D'));

echo "The first date of the intreval is: " . $theDt->format('Y-m-d');

Answer 3

以下是@ Dan的第二个答案的略微修改版本，对我来说效果很好。

基准如下所示。

$date="1985-02-18";
$intervalDays = "5";

//original version
$benchMark = microtime(true);
$dt1 = new DateTime($date);
$interval = new DateInterval("P{$intervalDays}D");
while ($dt1->format('Y') < date('Y')) {
    $dt1->add($interval);
}
echo $dt1->format('d m Y') . '<br>';
echo microtime(true)-$benchMark.'<br>';


//new version
$benchMark = microtime(true);
$dt1 = new DateTime($date);
$dt2 = new DateTime("12/31/" . ((int) Date("Y") - 1));
$dayDiff = $dt1->diff($dt2)->days;
$numIntervals = floor($dayDiff / $intervalDays);
$totIntervals = ($numIntervals * $intervalDays) + $intervalDays;
$dt1->add(new DateInterval('P' . $totIntervals . 'D'));
echo $dt1->format('d m Y').'<br>';
echo microtime(true)-$benchMark.'<br>';
exit;

<强>输出

02 01 2014
0.0145111083984
02 01 2014
0.000123977661133

添加多个间隔，直到日期为当前年份

3 个答案: