85

我想约会并计算出它的周数。

到目前为止，我有以下内容。当它应该是 42 时，它返回 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;
?>

数字颠倒是错误的巧合吗？还是我快到了？

4

18 回答 18

193

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";

这是因为在 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;
?>
于 2012-03-05T13:51:57.970 回答
61 
$date_string = "2012-10-18";
echo "Weeknummer: " . date("W", strtotime($date_string));
于 2012-03-05T13:52:23.247 回答
20

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

date("W", $yourdate)
于 2012-03-05T13:50:08.847 回答
12

这得到今天的日期，然后告诉一周的周数

<?php
 $date=date("W");
 echo $date." Week Number";
 ?>
于 2013-12-24T18:01:58.127 回答
8

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点。

你可以做的其他事情：

<?php echo date("Weeknumber: W", strtotime("2012-10-18 01:00:00")); ?>
<?php echo date("Weeknumber: W", strtotime($MY_DATE)); ?>
于 2014-03-10T10:25:02.807 回答
2

使用IntlGregorianCalendar类怎么样？

要求：在开始使用之前，请IntlGregorianCalendar确保在服务器上安装了libicuor 。pecl/intl所以在 CLI 上运行：

php -m

如果您intl在[PHP Modules]列表中看到，那么您可以使用IntlGregorianCalendar.

DateTime 与 IntlGregorianCalendar： IntlGregorianCalendar不是更好DateTime。但好处IntlGregorianCalendar是它会给你周数作为int.

例子：

$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
于 2020-10-22T13:20:08.607 回答
2

当您需要年和周时变得更加困难。
尝试找出哪一周是 01.01.2017。
（这是 2016 年的第 52 周，从 Mon 26.12.2016 - Sun 01.01.2017）。

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
ISO-8601:1988 给定年份的周数，从一年的第一周开始，至少有 4 个工作日，星期一是开始一周中的。（01 到 53）

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

无法使用 date() 或 DateTime 找到相应的解决方案。
至少不是没有像“+1day, last Monday”这样的解决方案。

于 2020-11-02T08:18:38.013 回答
2

我多年来一直试图解决这个问题，我以为我找到了一个更短的解决方案，但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;
}

我发现我的简短解决方案错过了 2018-12-31，因为它返回了 1801 而不是 1901。所以我不得不输入这个正确的长版本。

于 2018-02-20T06:27:23.547 回答
1 
<?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;
?>

您的 mktime 参数错误 - 需要是月/日/年，而不是日/月/年

于 2012-03-05T13:51:35.070 回答
1

当一年有 53 周（如 2020 年）时，上面给出的大多数示例都会产生问题。所以每四年你会经历一周的差异。此代码不会：

$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
于 2020-04-24T11:09:02.947 回答
1

要获取北美日期的星期数，我这样做：

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));

并得到：

Tue 2022-12-27: 53

于 2018-08-01T05:36:45.507 回答
0

规则是一年的第一周是包含该年第一个星期四的那一周。

我个人使用 Zend_Date 进行这种计算，获得今天的星期就这么简单。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
于 2012-03-05T13:47:45.243 回答
0

您的代码将起作用，但您需要翻转第 4 和第 5 个参数。

我会这样做

$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}.";

此外，在一周不查看该代码后，您的变量名称会让您感到困惑，您应该考虑阅读http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/

于 2012-03-05T13:48:32.090 回答
0

要获取jalai 日历中的周数，您可以使用以下命令：

$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：

15

周六周数变化

于 2020-04-10T08:11:00.417 回答
0

要获得日期 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'));
}
于 2019-03-08T11:32:48.930 回答
0

很简单就一行：

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

例如，您还可以像我需要的图表一样，减去以得到前一周，例如：

<?php $date=date("W"); echo "Week " . ($date - 1); ?>
于 2021-12-01T14:20:20.670 回答
-1

试试这个解决方案

date( 'W', strtotime( "2017-01-01 + 1 day" ) );
于 2017-02-27T11:07:43.003 回答
-1 
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.
于 2015-12-08T08:18:31.253 回答