用PHP显示周数[duplicate]

我认为这种关系应该是真实的,而且会派上用场:

Week of the month = Week of the year - Week of the year of first day of month + 1

function weekOfMonth($date) {
    //Get the first day of the month.
    $firstOfMonth = strtotime(date("Y-m-01", $date));
    //Apply above formula.
    return intval(date("W", $date)) - intval(date("W", $firstOfMonth)) + 1;
}

要想得到从周日开始的周数,只需将两者都替换掉 date("W", ...) strftime("%U", ...) .

您可以使用下面的函数,完全注释:

/**
 * Returns the number of week in a month for the specified date.
 *
 * @param string $date
 * @return int
 */
function weekOfMonth($date) {
    // estract date parts
    list($y, $m, $d) = explode('-', date('Y-m-d', strtotime($date)));

    // current week, min 1
    $w = 1;

    // for each day since the start of the month
    for ($i = 1; $i <= $d; ++$i) {
        // if that day was a sunday and is not the first day of month
        if ($i > 1 && date('w', strtotime("$y-$m-$i")) == 0) {
            // increment current week
            ++$w;
        }
    }

    // now return
    return $w;
}

function weekOfMonth($date) {
    $firstOfMonth = date("Y-m-01", strtotime($date));
    return intval(date("W", strtotime($date))) - intval(date("W", strtotime($firstOfMonth)));
}

我自己创建了这个函数,似乎可以正常工作。如果其他人有更好的方法,请分享。。这是我所做的。

function weekOfMonth($qDate) {
    $dt = strtotime($qDate);
    $day  = date('j',$dt);
    $month = date('m',$dt);
    $year = date('Y',$dt);
    $totalDays = date('t',$dt);
    $weekCnt = 1;
    $retWeek = 0;
    for($i=1;$i<=$totalDays;$i++) {
        $curDay = date("N", mktime(0,0,0,$month,$i,$year));
        if($curDay==7) {
            if($i==$day) {
                $retWeek = $weekCnt+1;
            }
            $weekCnt++;
        } else {
            if($i==$day) {
                $retWeek = $weekCnt;
            }
        }
    }
    return $retWeek;
}

echo weekOfMonth('2015-09-08') // gives me 2;

function getWeekOfMonth(DateTime $date) {
    $firstDayOfMonth = new DateTime($date->format('Y-m-1'));

    return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}

Goendg solution 不适用于2016-10-31。

function weekOfMonth($strDate) {
  $dateArray = explode("-", $strDate);
  $date = new DateTime();
  $date->setDate($dateArray[0], $dateArray[1], $dateArray[2]);
  return floor((date_format($date, 'j') - 1) / 7) + 1;  
}

weekofmount('2015-09-17')//返回3

给定月1日的时间(0=星期日到6=星期六) firstWday ,返回该月内的(基于星期日的)周数:

weekOfMonth = floor((dayOfMonth + firstWday - 1)/7) + 1 

翻译成PHP:

function weekOfMonth($dateString) {
  list($year, $month, $mday) = explode("-", $dateString);
  $firstWday = date("w",strtotime("$year-$month-1"));
  return floor(($mday + $firstWday - 1)/7) + 1;
}

$currentWeek = ceil((date("d",strtotime($today_date)) - date("w",strtotime($today_date)) - 1) / 7) + 1;

算法:

日期='2018-08-08'=>Y-m-d

2. 找出一周中某一天减去1(一周中的天数)的数字表示,例如(3-1)
5. 除以7得到结果,然后计算结果的值
6. 结果加1,如ceil((08-3)-1)/7)+1=2

function GetWeekNumberOfMonth ($date){
    echo $date -> format('d.m.Y');
    //define current year, month and day in numeric
    $_year = $date -> format('Y');
    $_month = $date -> format('n');
    $_day = $date -> format('j');
    $_week = 0; //count of weeks passed
    for ($i = 1; $i < $_day; $i++){
        echo "\n\n-->";
        $_newDate = mktime(0,0,1, $_month, $i, $_year);
        echo "\n";
        echo date("d.m.Y", $_newDate);
        echo "-->";
        echo date("N", $_newDate);
        //on sunday increasing weeks passed count
        if (date("N", $_newDate) == 7){
            echo "New week";
            $_week += 1;
        }

    }
    return $_week + 1; // as we are counting only passed weeks the current one would be on one higher
}

$date = new DateTime("2019-04-08");
echo "\n\nResult: ". GetWeekNumberOfMonth($date);

// self::DAYS_IN_WEEK = 7;
function getWeeksNumberOfMonth(): int
{
    $currentDate            = new \DateTime();
    $dayNumberInMonth       = (int) $currentDate->format('j');
    $dayNumberInWeek        = (int) $currentDate->format('N');
    $dayNumberToLastSunday  = $dayNumberInMonth - $dayNumberInWeek;
    $daysCountInFirstWeek   = $dayNumberToLastSunday % self::DAYS_IN_WEEK;
    $weeksCountToLastSunday = ($dayNumberToLastSunday - $daysCountInFirstWeek) / self::DAYS_IN_WEEK;

    $weeks = [];
    array_push($weeks, $daysCountInFirstWeek);
    for ($i = 0; $i < $weeksCountToLastSunday; $i++) {
        array_push($weeks, self::DAYS_IN_WEEK);
    }
    array_push($weeks, $dayNumberInWeek);

    if (array_sum($weeks) !== $dayNumberInMonth) {
        throw new Exception('Logic is not valid');
    }

    return count($weeks);
}

短变型:

(int) (new \DateTime())->format('W') - (int) (new \DateTime('first day of this month'))->format('W') + 1;

有很多解决方案,但这里有一个我的解决方案,在大多数情况下都很有效。

function current_week ($date = NULL) {
    if($date) {
        if(is_numeric($date) && ctype_digit($date) && strtotime(date('Y-m-d H:i:s',$date)) === (int)$date)
            $unix_timestamp = $date;
        else
            $unix_timestamp = strtotime($date);
    } else $unix_timestamp = time();

    return (ceil((date('d', $unix_timestamp) - date('w', $unix_timestamp) - 1) / 7) + 1);
}

time()

享受吧!

$datetime0 = date_create("1970-01-01");
$datetime1 = date_create(date("Y-m-d",mktime(0,0,0,$m,"01",$Y)));
$datetime2 = date_create(date("Y-m-d",mktime(0,0,0,$m,$d,$Y)));

$interval1 = date_diff($datetime0, $datetime1);
$daysdiff1= $interval1->format('%a');

$interval2 = date_diff($datetime0, $datetime2);
$daysdiff2= $interval2->format('%a');

$week1=round($daysdiff1/7);
$week2=round($daysdiff2/7);

$WeekOfMonth=$week2-$week1+1;

$date = new DateTime('first Monday of this month');
$thisMonth = $date->format('m');
$mondays_arr = [];

// Get all the Mondays in the current month and store in array
while ($date->format('m') === $thisMonth) {
    //echo $date->format('Y-m-d'), "\n";
    $mondays_arr[] = $date->format('d');
    $date->modify('next Monday');
}

// Get the day of the week (1-7 from monday to sunday)
$day_of_week = date('N') - 1;

// Get the day of month (1 to 31) 
$current_week_monday_date = date('j') - $day_of_week;

/*$day_of_week = date('N',mktime(0, 0, 0, 2, 11, 2020)) - 1;
$current_week_monday_date = date('j',mktime(0, 0, 0, 2, 11, 2020)) - $day_of_week;*/

$week_no = array_search($current_week_monday_date,$mondays_arr) + 1;
echo "Week No: ". $week_no;

//It's easy, no need to use php function
//Let's say your date is 2017-07-02

$Date = explode("-","2017-07-02");
$DateNo = $Date[2];
$WeekNo = $DateNo / 7; // devide it with 7
if(is_float($WeekNo) == true)
{
   $WeekNo = ceil($WeekNo); //So answer will be 1
}  

//If value is not float then ,you got your answer directly