给定周数计算星期几

【问题标题】：Calculating days of week given a week number给定周数计算星期几
【发布时间】：2008-10-09 08:42:11
【问题描述】：

给定一个星期数，例如date -u +%W，从星期一开始，那一周的天数如何计算？

第 40 周的示例 rfc-3339 输出：

2008-10-06
2008-10-07
2008-10-08
2008-10-09
2008-10-10
2008-10-11
2008-10-12

【问题讨论】：

    标签： php date


    【解决方案1】：

    PHP

    $week_number = 40;
$year = 2008;
for($day=1; $day<=7; $day++)
{
    echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n";
}

    下面的帖子是因为我是一个没有正确阅读问题的白痴，但是会从星期一开始得到一周内的日期，给定日期，而不是周数..

    在 PHP 中，改编自 this post 上的 PHP date manual page：

    function week_from_monday($date) {
    // Assuming $date is in format DD-MM-YYYY
    list($day, $month, $year) = explode("-", $_REQUEST["date"]);

    // Get the weekday of the given date
    $wkday = date('l',mktime('0','0','0', $month, $day, $year));

    switch($wkday) {
        case 'Monday': $numDaysToMon = 0; break;
        case 'Tuesday': $numDaysToMon = 1; break;
        case 'Wednesday': $numDaysToMon = 2; break;
        case 'Thursday': $numDaysToMon = 3; break;
        case 'Friday': $numDaysToMon = 4; break;
        case 'Saturday': $numDaysToMon = 5; break;
        case 'Sunday': $numDaysToMon = 6; break;   
    }

    // Timestamp of the monday for that week
    $monday = mktime('0','0','0', $month, $day-$numDaysToMon, $year);

    $seconds_in_a_day = 86400;

    // Get date for 7 days from Monday (inclusive)
    for($i=0; $i<7; $i++)
    {
        $dates[$i] = date('Y-m-d',$monday+($seconds_in_a_day*$i));
    }

    return $dates;
}

    week_from_monday('07-10-2008') 的输出给出：

    Array
(
    [0] => 2008-10-06
    [1] => 2008-10-07
    [2] => 2008-10-08
    [3] => 2008-10-09
    [4] => 2008-10-10
    [5] => 2008-10-11
    [6] => 2008-10-12
)

    【讨论】：

    • 该输入与指定不符。例如，输入将是“41”。
    • 谢谢，这里周一早上大脑的坏情况，没有正确阅读。用更短的代码 sn-p 更新了答案！
    • 愚蠢的错字，到目前为止还没有早上好！现已修复，感谢现场。
    • 我更喜欢这样： echo date('Y-m-d', strtotime(date("Y")."W".date("W").$day))."\n ";无论如何，非常感谢您的回答。
    • eeek，刚刚注意到：php -r "echo date("W");" 和 date -u +%W 41 != 40
    【解决方案2】：

    如果你有 Zend 框架，你可以使用 Zend_Date 类来做到这一点：

    require_once 'Zend/Date.php';

$date = new Zend_Date();
$date->setYear(2008)
     ->setWeek(40)
     ->setWeekDay(1);

$weekDates = array();

for ($day = 1; $day <= 7; $day++) {
    if ($day == 1) {
        // we're already at day 1
    }
    else {
        // get the next day in the week
        $date->addDay(1);
    }

    $weekDates[] = date('Y-m-d', $date->getTimestamp());
}

echo '<pre>';
print_r($weekDates);
echo '</pre>';

    【讨论】：

      【解决方案3】：

      自从发布了这个问题和接受的答案后，DateTime 课程让这件事变得更简单：-

      function daysInWeek($weekNum)
{
    $result = array();
    $datetime = new DateTime('00:00:00');
    $datetime->setISODate((int)$datetime->format('o'), $weekNum, 1);
    $interval = new DateInterval('P1D');
    $week = new DatePeriod($datetime, $interval, 6);

    foreach($week as $day){
        $result[] = $day->format('D d m Y H:i:s');
    }
    return $result;
}

var_dump(daysInWeek(24));

      这具有照顾闰年等的额外优势。

      See it working。包括艰难的第 1 周和第 53 周。

      【讨论】：

      • 不幸的是，这不起作用，因为 setISODate() 包含的时间部分等于今天的时间（即不是 0:00:00）。更改为 $result[] = $day->format('r');你会明白我在说什么。
      • @Dalin 我不明白这意味着它不起作用。 OP从未提及一天中的某个时间，只是想要这些日子。但是，对代码的一个小修改将所有时间重置为 00:00:00。
      【解决方案4】：

      此计算在很大程度上取决于您居住的地方。例如，在欧洲，我们从星期一开始一周，在美国，星期日是一周的第一天。在英国，第 1 周是 1 月 1 日，其他国家/地区从包含一年中第一个星期四的那一周开始第 1 周。

      您可以在http://en.wikipedia.org/wiki/Week#Week_number找到更多一般信息

      【讨论】：

        【解决方案5】：

        此函数将给出找到 $date 的星期几的时间戳。如果 $date 没有给出，它假定“现在”。如果您更喜欢可读日期而不是时间戳，请将日期格式传递给第二个参数。如果您不是在星期一开始您的一周（幸运），请为第三个参数传入不同的日期。

        function week_dates($date = null, $format = null, $start = 'monday') {
  // is date given? if not, use current time...
  if(is_null($date)) $date = 'now';

  // get the timestamp of the day that started $date's week...
  $weekstart = strtotime('last '.$start, strtotime($date));

  // add 86400 to the timestamp for each day that follows it...
  for($i = 0; $i < 7; $i++) {
    $day = $weekstart + (86400 * $i);
    if(is_null($format)) $dates[$i] = $day;
    else $dates[$i] = date($format, $day);
  }

  return $dates;
}

        所以 week_dates() 应该返回类似...

        Array ( 
  [0] => 1234155600 
  [1] => 1234242000 
  [2] => 1234328400 
  [3] => 1234414800 
  [4] => 1234501200
  [5] => 1234587600
  [6] => 1234674000
)

        【讨论】：

        • 如果当天是星期一，这个函数不能正常工作！，这个修复它 if($currDay == 'Monday'){ $weekstart = strtotime("midnight",$date); }else{ $weekstart = strtotime('last '.$start, $date); }
        【解决方案6】： 
        $week_number = 40;
$year = 2008;

for($day=1; $day<=7; $day++)
{
    echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n";
}

        如果$week_number 小于 10，这将失败。

        //============Try this================//

$week_number = 40;
$year = 2008;

if($week_number < 10){
   $week_number = "0".$week_number;
}

for($day=1; $day<=7; $day++)
{
    echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n";
}

//==============================//

        【讨论】：

        • 这应该是公认的答案，因为它可以正常工作！
        【解决方案7】：

        另一个代码嘿嘿：

        public function getAllowedDays($year, $week) {
    $weekDaysArray = array();
    $dto = new \DateTime();
    $dto->setISODate($year, $week);

    for($i = 0; $i < 7; $i++) {
        array_push($weekDaysArray, $dto->format('Y-m-d'));
        $dto->modify("+1 days");
    }

    return $weekDaysArray;
}

        【讨论】：

          【解决方案8】：

          我发现此解决方案存在问题。 我必须将周数填零，否则它会中断。

          我的解决方案现在看起来像这样：

          $week_number = 40;
$year = 2008;
for($day=1; $day<=7; $day++)
{
    echo date('m/d/Y', strtotime($year."W".str_pad($week_number,2,'0',STR_PAD_LEFT).$day))."\n";
}

          【讨论】：

            【解决方案9】：

            对于那些在给定周数 (1-52) 的情况下寻找一周中的日子的人 从星期天开始然后这是我的小工作。考虑到检查周是否在正确的范围内，并填充值 1-9 以保持一切正常。

            $week = 2; $year = 2009;

$week = (($week >= 1) AND ($week <= 52))?($week-1):(1);

$dayrange  = array(7,1,2,3,4,5,6);

for($count=0; $count<=6; $count++) {
    $week = ($count == 1)?($week + 1): ($week);
    $week = str_pad($week,2,'0',STR_PAD_LEFT);
    echo date('d m Y', strtotime($year."W".$week.($dayrange[$count]))); }

            【讨论】：

            • 未能返回 $week #1 上一年的日期。
            【解决方案10】：

            我有同样的问题，只使用 strftime 而不是日期作为我的起点，即使用 %W 从 strftime 导出周数我想知道该周的日期范围 - 周一到周日（或者实际上是任何开始日）。对几个类似帖子的评论，特别是尝试了上述几种方法并没有让我得到我想要的解决方案。当然，我可能误解了一些东西，但我无法得到我想要的。

            因此，我想分享我的解决方案。

            我的第一个想法是，鉴于 strftime %W 的描述是：

            当年的周数，从第一个星期一开始 第一周的第一天

            如果我确定了每年的第一个星期一是什么，我可以计算一个日期范围数组，其索引等于 %W 的值。此后我可以使用 strftime 调用该函数。

            所以这里是：

            功能：

            <?php

/*
 *  function to establish scope of week given a week of the year value returned from strftime %W
 */

// note strftime %W reports 1/1/YYYY as wk 00 unless 1/1/YYYY is a monday when it reports wk 01
// note strtotime Monday [last, this, next] week - runs sun - sat

function date_Range_For_Week($W,$Y){

// where $W = %W returned from strftime
//       $Y = %Y returned from strftime

    // establish 1st day of 1/1/YYYY

    $first_Day_Of_Year = mktime(0,0,0,1,1,$Y);

    // establish the first monday of year after 1/1/YYYY    

    $first_Monday_Of_Year = strtotime("Monday this week",(mktime(0,0,0,1,1,$Y)));   

    // Check for week 00 advance first monday if found
    // We could use strtotime "Monday next week" or add 604800 seconds to find next monday
    // I have decided to avoid any potential strtotime overhead and do the arthimetic

    if (strftime("%W",$first_Monday_Of_Year) != "01"){
        $first_Monday_Of_Year += (60 * 60 * 24 * 7);
    }

    // create array to ranges for the year. Note 52 wks is the norm but it is possible to have 54 weeks
    // in a given yr therefore allow for this in array index

    $week_Start = array();
    $week_End = array();        

    for($i=0;$i<=53;$i++){

        if ($i == 0){   
            if ($first_Day_Of_Year != $first_Monday_Of_Year){
                $week_Start[$i] = $first_Day_Of_Year;
                $week_End[$i] = $first_Monday_Of_Year - (60 * 60 * 24 * 1);
            } else {
                // %W returns no week 00
                $week_Start[$i] = 0;
                $week_End[$i] = 0;                              
            }
            $current_Monday = $first_Monday_Of_Year;
        } else {
            $week_Start[$i] = $current_Monday;
            $week_End[$i] = $current_Monday + (60 * 60 * 24 * 6);
            // find next monday
            $current_Monday += (60 * 60 * 24 * 7);
            // test for end of year
            if (strftime("%W",$current_Monday) == "01"){ $i = 999; };
        }
    };

    $result = array("start" => strftime("%a on %d, %b, %Y", $week_Start[$W]), "end" => strftime("%a on %d, %b, %Y", $week_End[$W]));

    return $result;

    }   

?>

            例子：

            // usage example

//assume we wish to find the date range of a week for a given date July 12th 2011

$Y = strftime("%Y",mktime(0,0,0,7,12,2011));
$W = strftime("%W",mktime(0,0,0,7,12,2011));

// use dynamic array variable to check if we have range if so get result if not run function

$date_Range = date_Range . "$Y";

isset(${$date_Range}) ? null : ${$date_Range} = date_Range_For_Week($W, $Y);

echo "Date sought: " . strftime(" was %a on %b %d, %Y, %X time zone: %Z",mktime(0,0,0,7,12,2011)) . "<br/>";
echo "start of week " . $W . " is " . ${$date_Range}["start"] . "<br/>";
echo "end of week " . $W . " is " . ${$date_Range}["end"];

            输出：

            > Date sought: was Tue on Jul 12, 2011, 00:00:00 time zone: GMT Daylight
> Time start of week 28 is Mon on 11, Jul, 2011 end of week 28 is Sun on
> 17, Jul, 2011

            我已经对此进行了几年的测试，包括 2018 年，即 2018 年 1 月 1 日 = 星期一的下一年。到目前为止，似乎提供了正确的日期范围。

            所以我希望这会有所帮助。

            问候

            【讨论】：

              【解决方案11】：

              另一种解决方案：

              //$date Date in week
//$start Week start (out)
//$end Week end (out)

function week_bounds($date, &$start, &$end) {
    $date = strtotime($date);
    $start = $date;
    while( date('w', $start)>1 ) {
        $start -= 86400;
    }
    $end = date('Y-m-d', $start + (6*86400) );
    $start = date('Y-m-d', $start);
}

              例子：

              week_bounds("2014/02/10", $start, $end);
echo $start."<br>".$end;

              输出：

              2014-02-10
2014-02-16

              【讨论】：

                【解决方案12】： 
                $year      = 2016; //enter the year
$wk_number = 46;   //enter the weak nr

$start = new DateTime($year.'-01-01 00:00:00');
$end   = new DateTime($year.'-12-31 00:00:00');

$start_date = $start->format('Y-m-d H:i:s');

$output[0]= $start;    
$end   = $end->format('U');    
$x = 1;

//create array full of data objects
for($i=0;;$i++){
    if($i == intval(date('z',$end)) || $i === 365){
        break;
    }
    $a = new DateTime($start_date);
    $b = $a->modify('+1 day');
    $output[$x]= $a;        
    $start_date = $b->format('Y-m-d H:i:s');
    $x++;
}    

//create a object to use
for($i=0;$i<count($output);$i++){
    if(intval ($output[$i]->format('W')) === $wk_number){
        $output_[$output[$i]->format('N')]        = $output[$i];
    }
}

$dayNumberOfWeek = 1; //enter the desired day in 1 = Mon -> 7 = Sun

echo '<pre>';
print_r($output_[$dayNumberOfWeek]->format('Y-m-d'));
echo '</pre>';

                在 php 中用作 date() 对象 date php

                【讨论】：

                  【解决方案13】：

                  基于周数和天数的周期

                  一些国家（如斯堪的纳维亚国家和德国）使用周数，作为预订假期、会议等的实用方式。此功能可以根据周数开始日期和周期长度（以天为单位）发送有关周期的短信。 

                  function MakePeriod($year,$Week,$StartDay,$NumberOfDays, $lan='DK'){
    //Please note that start dates in january of week 53 must be entered as "the year before"
    switch($lan){
    case "NO":
        $WeekDays=['mandag','tirsdag','onsdag','torsdag','fredag','lørdag','søndag'];
        $the=" den ";
        $weekName="Uke ";
        $dateformat="j/n Y";
    break;      
    case "DK":
        $WeekDays=['mandag','tirsdag','onsdag','torsdag','fredag','lørdag','søndag'];
        $the=" den ";
        $weekName="Uge ";
        $dateformat="j/n Y";
    break;
    case "SV":
        $WeekDays=['måndag','tisdag','onsdag','torsdag','fredag','lördag','söndag'];
        $the=" den ";
        $weekName="Vecka ";
        $dateformat="j/n Y";
    break;
    case "GE":
        $WeekDays=['Montag','Dienstag','Mittwoch','Donnerstag','Freitag','Samstag','Sonntag'];
        $the=" die ";
        $weekName="Woche ";
        $dateformat="j/n Y";
    break;
    case "EN":
    case "US":  
        $WeekDays=['Monday','Tuesday','Wednesday','Thursday','Friday','Saturday','Sunday'];
        $the=" the ";
        $weekName="Week ";
        $dateformat="n/j/Y";
    break;  
    }   
    $EndDay= (($StartDay-1+$NumberOfDays) % 7)+1;
    $ExtraDays= $NumberOfDays % 7;
    $FirstWeek=$Week;
    $LastWeek=$Week;    
    $NumberOfWeeks=floor($NumberOfDays / 7) ;
    $LastWeek=$Week+$NumberOfWeeks;

    if($StartDay+$ExtraDays>7){
        $LastWeek++;
    }       

    if($FirstWeek<10) $FirstWeek='0'.$FirstWeek;
    if($LastWeek<10) $LastWeek='0'.$LastWeek;

    
    $date1 = date( $dateformat, strtotime($year."W".$FirstWeek.$StartDay) ); // First day of week

    $date2 = date( $dateformat, strtotime($year."W".$LastWeek.$EndDay) ); // Last day of week

    if($LastWeek>53){
        $LastWeek=$LastWeek-53;
        $year++;
        if($LastWeek<10) $LastWeek='0'.$LastWeek;
        $date2 = date( $dateformat, strtotime($year."W".$LastWeek.$EndDay) );
    }
    $EndDayName=$WeekDays[$EndDay-1];
    $StartDayName=$WeekDays[$StartDay-1];
    $retval= " $weekName $Week $StartDayName  $the $date1 - $EndDayName $the $date2 ";
    return $retval;     
    
}

                  测试：

                  $Year=2021;
$Week=22;   
$StartDay=4;    
$NumberOfDays=3;
$Period=MakePeriod($Year,$Week,$StartDay,$NumberOfDays,"DK");
echo $Period;

                  Uge 22 torsdag den 3/6 2021 - søndag den 6/6 2021

                  【讨论】：

                    【解决方案14】： 
                        <?php
    $iWeeksAgo = 5;// need weeks ago
    $sWeekDayStartOn = 0;// 0 - Sunday, 1 - Monday, 2 - Tuesday
    $aWeeksDetails = getWeekDetails($iWeeksAgo, $sWeekDayStartOn);

    print_r($aWeeksDetails);
    die('end of line of getWeekDetails ');

    function getWeekDetails($iWeeksAgo, $sWeekDayStartOn){
        $date = new DateTime();
        $sCurrentDate = $date->format('W, Y-m-d, w');
        #echo 'Current Date (Week of the year, YYYY-MM-DD, day of week ): ' . $sCurrentDate . "\n";

        $iWeekOfTheYear = $date->format('W');// Week of the Year i.e. 19-Feb-2014 = 08
        $iDayOfWeek = $date->format('w');// day of week for the current month i.e. 19-Feb-2014 = 4
        $iDayOfMonth = $date->format('d'); // date of the month i.e. 19-Feb-2014 = 19

        $iNoDaysAdd = 6;// number of days adding to get last date of the week i.e. 19-Feb-2014  + 6 days = 25-Feb-2014

        $date->sub(new DateInterval("P{$iDayOfWeek}D"));// getting start date of the week
        $sStartDateOfWeek = $date->format('Y-m-d');// getting start date of the week

        $date->add(new DateInterval("P{$iNoDaysAdd}D"));// getting end date of the week
        $sEndDateOfWeek = $date->format('Y-m-d');// getting end date of the week

        $iWeekOfTheYearWeek = (string) $date->format('YW');//week of the year
        $iWeekOfTheYearWeekWithPeriod = (string) $date->format('Y-W');//week of the year with year

        //To check uncomment
        #echo "Start Date / End Date of Current week($iWeekOfTheYearWeek), week with - ($iWeekOfTheYearWeekWithPeriod) : " . $sStartDateOfWeek . ',' . $sEndDateOfWeek . "\n";

        $iDaysAgo = ($iWeeksAgo*7) + $iNoDaysAdd + $sWeekDayStartOn;// getting 4 weeks ago i.e. no. of days to substract

        $date->sub(new DateInterval("P{$iDaysAgo}D"));// getting 4 weeks ago i.e. no. of days to substract
        $sStartDateOfWeekAgo = $date->format('Y-m-d');// getting 4 weeks ago start date i.e. 19-Jan-2014

        $date->add(new DateInterval("P{$iNoDaysAdd}D")); // getting 4 weeks ago end date i.e. 25-Jan-2014
        $sEndDateOfWeekAgo = $date->format('Y-m-d');// getting 4 weeks ago start date i.e. 25-Jan-2014

        $iProccessedWeekAgoOfTheYear = (string) $date->format('YW');//ago week of the year
        $iProccessedWeekOfTheYearWeekAgo = (string) $date->format('YW');//ago week of the year with year
        $iProccessedWeekOfTheYearWeekWithPeriodAgo = (string) $date->format('Y-W');//ago week of the year with year

        //To check uncomment
        #echo "Start Date / End Date of week($iProccessedWeekOfTheYearWeekAgo), week with - ($iProccessedWeekOfTheYearWeekWithPeriodAgo) ago: " . $sStartDateOfWeekAgo . ',' . $sEndDateOfWeekAgo . "\n";

        $aWeeksDetails = array ('weeksago' => $iWeeksAgo, 'currentweek' => $iWeekOfTheYear, 'currentdate' => $sCurrentDate, 'startdateofcurrentweek' => $sStartDateOfWeek,  'enddateofcurrentweek' => $sEndDateOfWeek,
                                'weekagoyearweek' => $iProccessedWeekAgoOfTheYear, 'startdateofagoweek' => $sStartDateOfWeekAgo,  'enddateofagoweek' => $sEndDateOfWeekAgo);

        return $aWeeksDetails;
    }
?>

                    【讨论】：

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