从PHP中的日期获取月份中的周数？

Question

我有一个随机日期数组（不是来自 MySQL）。我需要按周将它们分组为第 1 周、第 2 周等等，直到第 5 周。

我拥有的是这样的：

$dates = array('2015-09-01','2015-09-05','2015-09-06','2015-09-15','2015-09-17');

我需要的是一个通过提供日期来获取月份周数的函数。

我知道我可以通过做 date('W',strtotime('2015-09-01')); 但是这个周数是年份（1-52）之间的数字，但我只需要当月的周数，例如2015 年 9 月有 5 周：

• 第 1 周 = 第 1 到第 5
• 第 2 周 = 第 6 至 12 日
• 第 3 周 = 13 日至 19 日
• 第 4 周 = 20 日至 26 日
• 第 5 周 = 27 日至 30 日

我应该能够通过提供日期来获得第 1 周 例如

$weekNumber = getWeekNumber('2015-09-01') //output 1;
$weekNumber = getWeekNumber('2015-09-17') //output 3;

Answer 1

我觉得这种关系应该是真的，并且派上用场：

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

我们还需要确保正确处理与上一年“重叠”的周 - 如果 1 月 1 日在第 52 周或第 53 周，则应计为第 0 周。以类似的方式，如果 12 月的一天是在明年的第一周，应该算作53。（这个答案的以前版本没有正确做到这一点。）

<?php

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

function weekOfYear($date) {
    $weekOfYear = intval(date("W", $date));
    if (date('n', $date) == "1" && $weekOfYear > 51) {
        // It's the last week of the previos year.
        return 0;
    }
    else if (date('n', $date) == "12" && $weekOfYear == 1) {
        // It's the first week of the next year.
        return 53;
    }
    else {
        // It's a "normal" week.
        return $weekOfYear;
    }
}

// A few test cases.
echo weekOfMonth(strtotime("2020-04-12")) . " "; // 2
echo weekOfMonth(strtotime("2020-12-31")) . " "; // 5
echo weekOfMonth(strtotime("2020-01-02")) . " "; // 1
echo weekOfMonth(strtotime("2021-01-28")) . " "; // 5
echo weekOfMonth(strtotime("2018-12-31")) . " "; // 6

要获得从星期日开始的星期，只需将 date("W", ...) 替换为 strftime("%U", ...)。

Answer 2

你可以使用下面的函数，完整注释：

/**
 * 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;
}

Answer 3

正确的方法是

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

Answer 4

我自己创建了这个函数，它似乎可以正常工作。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;

Answer 5

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。

Answer 6

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;  
}

weekOfMonth ('2015-09-17') // 返回 3

Answer 7

给定firstWday 中每月第一天的 time_t wday（0=星期日到 6=星期六），这将返回该月内（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;
}

Answer 8

您还可以使用这个简单的公式来查找月份中的星期

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

算法：

日期 = '2018-08-08' => Y-m-d

1. 找出一个月中的哪一天，例如。 08
2. 找出星期几的数字表示减去 1（星期几），例如。 (3-1)
3. 取差并存储在结果中
4. 从结果中减去 1
5. 将其除以 7 得到结果，并将结果的值设为上限
6. 将结果加 1，例如。 ceil(( 08 - 3 ) - 1 ) / 7) + 1 = 2

Answer 9

我的功能。主要思想：我们将计算从该月的第一个日期到当前日期的周数。当前的周数将是下一个周数。适用于规则：“星期从星期一开始”（对于基于星期日的类型，我们需要转换递增算法）

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);

Answer 10

$month = 6;
$year = 2021;           
$week = date("W", strtotime($year . "-" . $month ."-01"));

$str='';
$str .= date("d-m-Y", strtotime($year . "-" . $month ."-01")) ."to";
$unix = strtotime($year."W".$week ."+1 week");
while(date("m", $unix) == $month){
 $str .= date("d-m-Y", $unix-86400) . "|";
 $str .= date("d-m-Y", $unix) ."to"; 
 $unix = $unix + (86400*7);
}
$str .= date("d-m-Y", strtotime("last day of ".$year . "-" . $month));

$weeks_ar = explode('|',$str);
echo '<pre>'; print_r($weeks_ar);

工作正常。

Answer 11

    // Current week of the month starts with Sunday
 
    $first_day_of_the_week = 'Sunday';
    $start_of_the_week1    = strtotime("Last $first_day_of_the_week");     
    
    if (strtolower(date('l')) === strtolower($first_day_of_the_week)) {
         $start_of_the_week1 = strtotime('today');
     }
     $end_of_the_week1   = $start_of_the_week1 + (60 * 60 * 24 * 7) - 1;

// Get the date format
print date('Y-m-d', $start_of_the_week1) . ' 00:00:00';
print date('Y-m-d', $end_of_the_week1) . ' 23:59:59';

Answer 12

// 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);
}

短变体：

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

Answer 13

有很多解决方案，但这是我的一个解决方案，在大多数情况下运行良好。

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);
}

如果您没有传递任何值，它接受 unix 时间戳、正常日期或从 time() 返回当前周。

享受吧！

Answer 14

我知道这是一篇旧帖子，但我有个主意！

$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;

Answer 15

$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;

Answer 16

这个函数如何利用 PHP 的相对日期？ 此函数假定星期在星期六结束。但这很容易改变。

function get_weekNumMonth($date) {
    $CI = &get_instance();
    $strtotimedate = strtotime($date);
    $firstweekEnd = date('j', strtotime("FIRST SATURDAY OF " . date("F", $strtotimedate) . " " . date("Y", $strtotimedate)));
    $cutoff = date('j', strtotime($date));
    $weekcount = 1;
    while ($cutoff > $firstweekEnd) {
        $weekcount++;
        $firstweekEnd += 7; // move to next week
    }
    return $weekcount;
}

Answer 17

此函数返回当前月份的整数周数。周总是从星期一开始，计数总是从 1 开始。

function weekOfmonth(DateTime $date)
{
  $dayFirstMonday = date_create('first monday of '.$date->format('F Y'))->format('j');
  return (int)(($date->format('j') - $dayFirstMonday +7)/7) + ($dayFirstMonday == 1 ? 0 : 1);
}

使用示例

echo weekOfmonth(new DateTime("2020-04-12"));  //2

以@Anders 接受的解决方案作为参考，对 1900 年至 2038 年的所有日子进行测试：

//reference functions
//integer $date (Timestamp) 
function weekOfMonthAnders($date) {
    //Get the first day of the month.
    $firstOfMonth = strtotime(date("Y-m-01", $date));
    //Apply above formula.
    return weekOfYear($date) - weekOfYear($firstOfMonth) + 1;
}

function weekOfYear($date) {
    $weekOfYear = intval(date("W", $date));
    if (date('n', $date) == "1" && $weekOfYear > 51) {
        // It's the last week of the previos year.
        return 0;
    }
    else if (date('n', $date) == "12" && $weekOfYear == 1) {
        // It's the first week of the next year.
        return 53;
    }
    else {
        // It's a "normal" week.
        return $weekOfYear;
    }
}

//this function
function weekOfmonth(DateTime $date)
{
  $dayFirstMonday = date_create('first monday of '.$date->format('F Y'))->format('j');
  return (int)(($date->format('j') - $dayFirstMonday +7)/7) + ($dayFirstMonday == 1 ? 0 : 1);
}

$dt = date_create('1900-01-01');
$end = date_create('2038-01-02');
$countOk = 0;
$countError = 0;
for(;$dt < $end; $dt->modify('+1 Day')){
    $ts = $dt->getTimestamp();
    if(weekOfmonth($dt) === weekOfMonthAnders($ts)){
       ++$countOk; 
    }
    else {
       ++$countError;
    }
}
echo $countOk.' compare ok, '.$countError.' errors';

结果：50405 比较正常，0 个错误

Answer 18

我采用了视觉方法（就像我们在现实世界中的做法一样）。我没有使用公式或其他什么，而是通过可视化文字日历然后将日期放入多维数组来解决它（或者至少我认为我做到了）。第一个维度对应于星期。

我希望有人可以检查它是否经得起您的测试。或者用不同的方法帮助别人。

# date in this format 2021-08-03
# week_start is either Sunday or Monday
function getWeekOfMonth($date, $week_start = "Sunday"){

    list($year, $month, $day) = explode("-", $date);

    $dates = array();
    $current_week = 1;

    $new_week_signal = $week_start == "Sunday" ? 6 : 0;

    for($i = 1; $i <= date("t", strtotime($date)); $i++){
        $current_date = strtotime("{$year}-{$month}-".$i);
        $dates[$current_week][] = $i;
        if(date('w', $current_date) == $new_week_signal){
            $current_week++;
        }
    }

    foreach($dates as $week => $days){
        if(in_array(intval($day), $days)){
            return $week;
       }
    }

    return false;

}

Answer 19

//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