我有一个随机日期的数组（不是来自MySQL）。我需要按周分组，从第一周开始到第五周为止，分别命名为Week1、Week2等。
我现在拥有的是：
$dates = array('2015-09-01','2015-09-05','2015-09-06','2015-09-15','2015-09-17');
date('W',strtotime('2015-09-01'));
但这个周数是一年中的周数（1-52），我需要的是一个月中的周数。例如，在2015年9月有5周:
$weekNumber = getWeekNumber('2015-09-01') //output 1;
$weekNumber = getWeekNumber('2015-09-17') //output 3;
我认为这个关系应该是真实的，并且会派上用场：
Week of the month = Week of the year - Week of the year of first day of month + 1
我们还需要确保正确处理上一年的“重叠”周-如果1月1日在第52或53周，则应将其计为第0周。同样，如果12月的某一天是下一年的第一周，它应该被计为第53周。(此前版本的答案未能正确处理此问题。)
<?php
function weekOfMonth($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
要获得以星期日为开头的周数，只需将date("W", ...)替换为strftime("%U", ...)。
/**
* Returns the number of week in a month for the specified date.
*
* @param string $date (can be in YYYY-MM-DD, DD-MM-YYYY or MM/DD/YYYY format)
* @return int
*/
function weekOfMonth($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;
}
$weekOfTheMonth = weekOfMonth("2023-10-19");
echo $weekOfTheMonth; // should print 3
$i = 1; $i <= $d; ++$i 改成 $i = 1; $i < $d; ++$i 可以得到完美的结果。 - Jazuly正确的方式是
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，似乎它能够正确地工作。如果有其他更好的方法，请分享一下。这就是我的做法。
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;
$weekCnt 要初始化为1？ - vidhiecho weekOfMonth('2017-10-01'); 它会输出 2。我认为它应该是 1。 - vidhifunction getWeekOfMonth(DateTime $date) {
$firstDayOfMonth = new DateTime($date->format('Y-m-1'));
return ceil(($firstDayOfMonth->format('N') + $date->format('j') - 1) / 7);
}
Goendg解决方案在2016-10-31失效了。
function weekOfMonth($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') // returns 3
2023-10-20返回4。 - undefined$currentWeek = ceil((date("d",strtotime($today_date)) - date("w",strtotime($today_date)) - 1) / 7) + 1;
算法:
日期 = '2018-08-08' => Y-m-d
2023-10-20为3，但是被接受的答案给出的是4。 - undefined根据月份中第一个星期的时间_t wday (0代表星期日，6代表星期六)的firstWday，此函数返回该月份内以星期日为基准的周数:
weekOfMonth = floor((dayOfMonth + firstWday - 1)/7) + 1
翻译成PHP：
function weekOfMonth($dateString) {
list($year, $month, $mday) = explode("-", $dateString);
$firstWday = date("w",strtotime("$year-$month-1"));
return floor(($mday + $firstWday - 1)/7) + 1;
}
2023-10-20为3，但是被接受的答案给出的是4。 - undefined$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($weeks_ar);
86400的技术往往容易出现问题。这个答案缺少了相关的教育性解释。 - undefined我的函数。主要思想：我们将计算从本月第一天到当前日期经过的周数。当前周数将是下一个周数。遵循规则：“一周从星期一开始”（对于基于星期日的类型，我们需要转换增加算法）。
function GetWeekNumberOfMonth ($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);