从日期PHP获取星期数（一年中的数字）

Question 1

我想确定一个日期并计算出其星期数。

到目前为止，我有以下几点。它应该是42时返回24。

<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[2], $duedt[1],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>

数字颠倒是巧合吗？还是我快到了？

Question 2

今天，使用PHP的DateTime对象更好：

<?php
$ddate = "2012-10-18";
$date = new DateTime($ddate);
$week = $date->format("W");
echo "Weeknummer: $week";

这是因为in中mktime()，它是这样的：

mktime(hour, minute, second, month, day, year);

因此，您的订单是错误的。

<?php
$ddate = "2012-10-18";
$duedt = explode("-", $ddate);
$date  = mktime(0, 0, 0, $duedt[1], $duedt[2], $duedt[0]);
$week  = (int)date('W', $date);
echo "Weeknummer: " . $week;
?>

Question 3

$date_string = "2012-10-18";
echo "Weeknummer: " . date("W", strtotime($date_string));

Question 4

使用PHP的日期函数
http://php.net/manual/en/function.date.php

date("W", $yourdate)

Question 5

该日期为今天，然后告诉该周的星期数

<?php
 $date=date("W");
 echo $date." Week Number";
 ?>

Question 6

就像一个建议：

<?php echo date("W", strtotime("2012-10-18")); ?>

可能比所有这些都简单一些。

您可以做的其他事情：

<?php echo date("Weeknumber: W", strtotime("2012-10-18 01:00:00")); ?>
<?php echo date("Weeknumber: W", strtotime($MY_DATE)); ?>

Question 7

<?php
$ddate = "2012-10-18";
$duedt = explode("-",$ddate);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2],$duedt[0]);
$week = (int)date('W', $date);
echo "Weeknummer: ".$week;
?>

您将参数设置为mktime错误-必须为月/日/年，而不是日/月/年

Question 8

我尝试解决这个问题已有多年了，我以为我找到了一个更短的解决方案，但是不得不再次回到长篇大论。此函数返回正确的ISO周符号：

/**
 * calcweek("2018-12-31") => 1901
 * This function calculates the production weeknumber according to the start on 
 * monday and with at least 4 days in the new year. Given that the $date has
 * the following format Y-m-d then the outcome is and integer.
 *
 * @author M.S.B. Bachus
 *
 * @param date-notation PHP "Y-m-d" showing the data as yyyy-mm-dd
 * @return integer
 **/
function calcweek($date) {
  // 1. Convert input to $year, $month, $day
  $dateset      = strtotime($date);
  $year         = date("Y", $dateset);
  $month        = date("m", $dateset);
  $day          = date("d", $dateset);

  $referenceday = getdate(mktime(0,0,0, $month, $day, $year));
  $jan1day      = getdate(mktime(0,0,0,1,1,$referenceday[year]));

  // 2. check if $year is a  leapyear
  if ( ($year%4==0 && $year%100!=0) || $year%400==0) {
    $leapyear = true;
  } else {
    $leapyear = false;
  }

  // 3. check if $year-1 is a  leapyear
  if ( (($year-1)%4==0 && ($year-1)%100!=0) || ($year-1)%400==0 ) {
    $leapyearprev = true;
  } else {
    $leapyearprev = false;
  }

  // 4. find the dayofyearnumber for y m d
  $mnth = array(0, 31, 59, 90, 120, 151, 181, 212, 243, 273, 304, 334);
  $dayofyearnumber = $day + $mnth[$month-1];
  if ( $leapyear && $month > 2 ) { $dayofyearnumber++; }

  // 5. find the jan1weekday for y (monday=1, sunday=7)
  $yy = ($year-1)%100;
  $c  = ($year-1) - $yy;
  $g  = $yy + intval($yy/4);
  $jan1weekday = 1+((((intval($c/100)%4)*5)+$g)%7);

  // 6. find the weekday for y m d
  $h = $dayofyearnumber + ($jan1weekday-1);
  $weekday = 1+(($h-1)%7);

  // 7. find if y m d falls in yearnumber y-1, weeknumber 52 or 53
  $foundweeknum = false;
  if ( $dayofyearnumber <= (8-$jan1weekday) && $jan1weekday > 4 ) {
    $yearnumber = $year - 1;
    if ( $jan1weekday = 5 || ( $jan1weekday = 6 && $leapyearprev )) {
      $weeknumber = 53;
    } else {
      $weeknumber = 52;
    }
    $foundweeknum = true;
  } else {
    $yearnumber = $year;
  }

  // 8. find if y m d falls in yearnumber y+1, weeknumber 1
  if ( $yearnumber == $year && !$foundweeknum) {
    if ( $leapyear ) {
      $i = 366;
    } else {
      $i = 365;
    }
    if ( ($i - $dayofyearnumber) < (4 - $weekday) ) {
      $yearnumber = $year + 1;
      $weeknumber = 1;
      $foundweeknum = true;
    }
  }

  // 9. find if y m d falls in yearnumber y, weeknumber 1 through 53
  if ( $yearnumber == $year && !$foundweeknum ) {
    $j = $dayofyearnumber + (7 - $weekday) + ($jan1weekday - 1);
    $weeknumber = intval( $j/7 );
    if ( $jan1weekday > 4 ) { $weeknumber--; }
  }

  // 10. output iso week number (YYWW)
  return ($yearnumber-2000)*100+$weeknumber;
}

我发现我的短期解决方案错过了2018-12-31，因为它提供了1801而不是1901。所以我不得不输入这个正确的较长版本。

Question 9

要获得北美某个日期的星期数，我会这样：

function week_number($n)
{
    $w = date('w', $n);
    return 1 + date('z', $n + (6 - $w) * 24 * 3600) / 7;
}

$n = strtotime('2022-12-27');
printf("%s: %d\n", date('D Y-m-d', $n), week_number($n));

并获得：

Tue 2022-12-27: 53

Question 10

您的代码可以使用，但是您需要翻转第4个和第5个参数。

我会这样

$date_string = "2012-10-18";
$date_int = strtotime($date_string);
$date_date = date($date_int);
$week_number = date('W', $date_date);
echo "Weeknumber: {$week_number}.";

另外，一个星期不看该代码后，您的变量名将使您感到困惑，您应该考虑阅读http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/

Question 11

规则是，一年的第一周是包含该年的第一个星期四的那一周。

我个人使用Zend_Date进行这种计算，因此获得今天的星期很简单。如果您使用日期，它们还有许多其他有用的功能。

$now = Zend_Date::now();
$week = $now->get(Zend_Date::WEEK);
// 10

Question 12

要获取日期2018-12-31的正确周数，请使用以下代码

$day_count = date('N',strtotime('2018-12-31'));
$week_count = date('W',strtotime('2018-12-31'));    


if($week_count=='01' && date('m',strtotime('2018-12-31'))==12){
    $yr_count = date('y',strtotime('2018-12-31')) + 1;
}else{
    $yr_count = date('y',strtotime('2018-12-31'));
}

Question 13

获取星期数 jalai日历中的您可以使用以下方法：

$weeknumber = date("W"); //number week in year
$dayweek = date("w"); //number day in week
if ($dayweek == "6")
{
    $weeknumberint = (int)$weeknumber;
    $date2int++; 
    $weeknumber = (string)$date2int;
}

echo $date2;

结果：

星期六的周数变化

Question 14

上面给出的大多数示例在一年有53周（例如2020年）时都会产生问题。因此，每四年，您将经历一周的差异。此代码不：

$thisYear = "2020";
$thisDate = "2020-04-24"; //or any other custom date
$weeknr = date("W", strtotime($thisDate)); //when you want the weeknumber of a specific week, or just enter the weeknumber yourself

$tempDatum = new DateTime();
$tempDatum->setISODate($thisYear, $weeknr);
$tempDatum_start = $tempDatum->format('Y-m-d');
$tempDatum->setISODate($thisYear, $weeknr, 7);
$tempDatum_end = $tempDatum->format('Y-m-d');

echo $tempDatum_start //will output the date of monday
echo $tempDatum_end // will output the date of sunday

Question 15

如何使用IntlGregorianCalendar课程？

要求：开始使用之前，请IntlGregorianCalendar确保已在服务器上安装libicu或pecl/intl。因此，在CLI上运行：

php -m

如果您intl在[PHP Modules]列表中看到，则可以使用IntlGregorianCalendar。

DateTime与IntlGregorianCalendar： IntlGregorianCalendar当时还不错DateTime。但是关于的好处IntlGregorianCalendar是它可以给你星期的数字int。

例：

$dateTime = new DateTime('21-09-2020 09:00:00');
echo $dateTime->format("W"); // string '39'

$intlCalendar = IntlCalendar::fromDateTime ('21-09-2020 09:00:00');
echo $intlCalendar->get(IntlCalendar::FIELD_WEEK_OF_YEAR); // integer 39

Question 16

当您需要一年和一周时会变得更加困难。
尝试找出哪个星期是2017年1月1日。
（这是2016年的第52周，从2016年12月26日星期一开始-2017年1月1日星期日）。

经过更长的搜索，我发现

strftime('%G-%V',strtotime("2017-01-01"))

结果：2016-52

https://www.php.net/manual/de/function.strftime.php
给定年份的ISO-8601：1988周号，从每年的第一周开始，至少有4个工作日，从星期一开始一周中的。（01至53）

mysql中的等效项是DATE_FORMAT（date，'％x-％v'） https://www.w3schools.com/sql/func_mysql_date_format.asp
Week，其中星期一是一周的第一天（01到53）。

使用date（）或DateTime找不到相应的解决方案。
至少并非没有诸如“ + 1day，last monday”之类的解决方案。

Question 17

function last_monday($date) 
{
    if (!is_numeric($date))
        $date = strtotime($date);
    if (date('w', $date) == 1)
        return $date;
    else
        return date('Y-m-d',strtotime('last monday',$date));
}
$date = '2021-01-04';  //Enter custom date
$year = date('Y',strtotime($date));
$date1 = new DateTime($date);
$ldate = last_monday($year."-01-01");
$date2 = new DateTime($ldate);
$diff = $date2->diff($date1)->format("%a");
$diff = $diff/7;
$week = intval($diff) + 1;
echo $week;
//Returns 2.

Question 18

试试这个解决方案

date( 'W', strtotime( "2017-01-01 + 1 day" ) );