从给定的日期查找以前的三个工作日 [英] Find three previous working days from a given date

问题描述

从给定的日期起，我需要找到三个工作日，省略周末和假期。这本身并不是一项艰巨的任务，但似乎我会这样做的方式过于复杂，所以我以为我会先征询你的意见。

为了使事情更有趣，让我们来做一个比赛。我提供300作为赏金，无论谁提出了符合这个规范的最短，最干净的解决方案：

• 编写一个函数从指定的日期返回三个前一个工作日


• 工作日被定义为不是星期六或星期天的任何一天，而不是假期


• 该函数知道给定日期的年份的假期，并可以考虑这些因素


• 该函数接受一个参数，日期在 Ymd 格式


• 该函数返回一个包含 Ymd 格式的三个日期的数组，从最旧到最新。
  

额外：

• 该功能还可以找到 next 三个工作日除了前三个

假日数组的一个例子：

  $ holidays = array（
'2010-01-01'，
'2010-01-06'，
'2010-04-02 '，
'2010-04-04'，
'2010-04-05'，
'2010-05-01'，
'2010-05-13' 
'2010-05-23'，
'2010-06-26'，
'2010-11-06'，
'2010-12-06'，
'2010-12-25'，
'2010-12-26'
）; 
  

请注意，在实际情况下，假期不是硬编码，但来自 get_holidays（$ year）函数。如果你愿意，你可以在你的答案中加入/使用。

当我提供赏金时，这意味着至少有三天之前我可以标记答复为接受（2天添加赏金，1天直到我可以接受）。

---

注意

如果您使用固定的日期长度（如86400秒）从一天跳到另一天，则会遇到夏令时问题。使用 strtotime（' - 1天'，$ timestamp）

这个问题的一个例子： p>

http://codepad.org/ uSYiIu5w

---

最终解决方案 p>

这是我最终使用的最终解决方案，改编自Keith Minkler使用 strtotime 的上周末。检测来自传递的计数的方向，如果为负值，向后搜索，向前搜索正数：

  function working_days（$ date，$ count）{
 
 $ working_days = array（）; 
 $ direction = $ count< 0？ 'last'：'next'; 
 $ holidays = get_holidays（date（Y，strtotime（$ date）））; 
 
 while（count（$ working_days）< abs（$ count））{
 $ date = date（Ymd，strtotime（$ direction weekday，strtotime（$ date） ））; 
 if（！in_array（$ date，$ holidays））{
 $ working_days [] = $ date; 
} 
} 
 
 sort（$ working_days）; 
 return $ working_days; 
} 
  

解决方案

如下所示：

  function last_working_days（$ date，$ backwards = true） 
 {
 $ holidays = get_holidays（date（Y，strtotime（$ date）））; 
 
 $ working_days = array（）; 
 
 do 
 {
 $ direction = $ backwards？ 'last'：'next'; 
 $ date = date（Y-m-d，strtotime（$ direction weekday，strtotime（$ date）））; 
 if（！in_array（$ date，$ holidays））
 {
 $ working_days [] = $ date; 
} 
} 
 while（count（$ working_days）< 3）; 
 
 return $ working_days; 
} 
  

I need to find three previous working days from a given date, omitting weekends and holidays. This isn't a hard task in itself, but it seems that the way I was going to do it would be overly complicated, so I thought I'd ask for your opinion first.

To make things more interesting, let's make this a contest. I'm offering 300 as a bounty to whoever comes up with the shortest, cleanest solution that adheres to this specification:

• Write a function that returns three previous working days from a given date
• Working day is defined as any day that is not saturday or sunday and isn't an holiday
• The function knows the holidays for the year of the given date and can take these into account
• The function accepts one parameter, the date, in Y-m-d format
• The function returns an array with three dates in Y-m-d format, sorted from oldest to newest.

Extra:

• The function can find also the next three working days in addition to the previous three

An example of the holidays array:

$holidays = array(
    '2010-01-01',
    '2010-01-06',
    '2010-04-02',
    '2010-04-04',
    '2010-04-05',
    '2010-05-01',
    '2010-05-13',
    '2010-05-23',
    '2010-06-26',
    '2010-11-06',
    '2010-12-06',
    '2010-12-25',
    '2010-12-26'
);

Note that in the real scenario, the holidays aren't hardcoded but come from get_holidays($year) function. You can include / use that in your answer if you wish.

As I'm offering a bounty, that means there will be at least three days before I can mark an answer as accepted (2 days to add a bounty, 1 day until I can accept).

---

Note

If you use a fixed day length such as 86400 seconds to jump from day to another, you'll run into problems with daylight savings time. Use strtotime('-1 day', $timestamp) instead.

An example of this problem:

http://codepad.org/uSYiIu5w

---

Final solution

Here's the final solution I ended up using, adapted from Keith Minkler's idea of using strtotime's last weekday. Detects the direction from the passed count, if negative, searches backwards, and forwards on positive:

function working_days($date, $count) {

    $working_days = array();
    $direction    = $count < 0 ? 'last' : 'next';
    $holidays     = get_holidays(date("Y", strtotime($date)));

    while(count($working_days) < abs($count)) {
        $date = date("Y-m-d", strtotime("$direction weekday", strtotime($date)));
        if(!in_array($date, $holidays)) {
            $working_days[] = $date;
        }
    }

    sort($working_days);
    return $working_days;
}

解决方案

You can use expressions like "last weekday" or "next thursday" in strtotime, such as this:

function last_working_days($date, $backwards = true)
{
    $holidays = get_holidays(date("Y", strtotime($date)));

    $working_days = array();

    do
    {
        $direction = $backwards ? 'last' : 'next';
        $date = date("Y-m-d", strtotime("$direction weekday", strtotime($date)));
        if (!in_array($date, $holidays))
        {
            $working_days[] = $date;
        }
    }
    while (count($working_days) < 3);

    return $working_days;
}

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