计算一周的周数

给出一周的数字,例如date -u +%W,你如何计算从周一开始的那一周的天数？

第40周的rfc-3339输出示例:

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

ConroyP.. 61

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,但会在星期一开始的一周内得到日期,给出日期,而不是周数.

在PHP中,改编自PHP日期手册页上的这篇文章:

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;
}

输出来自week_from_monday('07-10-2008'):

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
)

---

小智.. 7

如果你有Zend Framework,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 '
';
print_r($weekDates);
echo '
';

---

vascowhite.. 6

由于发布了这个问题和接受的答案,DateTime课程使这更容易: -

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.

看它工作.包括困难的第1周和第53周.

1> ConroyP..：

---

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,但会在星期一开始的一周内得到日期,给出日期,而不是周数.

在PHP中,改编自PHP日期手册页上的这篇文章:

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;
}

输出来自week_from_monday('07-10-2008'):

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
)

2> 小智..：

---

如果你有Zend Framework,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 '
';
print_r($weekDates);
echo '
';

3> vascowhite..：

---

由于发布了这个问题和接受的答案,DateTime课程使这更容易: -

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.

看它工作.包括困难的第1周和第53周.

---

@Dalin我不明白这意味着它不起作用.OP从来没有提到过一天的时间,只是想要的日子.但是,对代码的少量修改会将所有时间重置为00:00:00.