从日期 PHP 获取周数（在一年中）
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Get week number (in the year) from a date PHP
提问by AJFMEDIA
I want to take a date and work out its week number.
我想约会并计算出它的周数。
So far, I have the following. It is returning 24 when it should be 42.
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;
?>
Is it wrong and a coincidence that the digits are reversed? Or am I nearly there?
数字颠倒是错误还是巧合？还是我快到了？
回答by Madara's Ghost
Today, using PHP's DateTimeobjects is better:
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";
It's because in mktime(), it goes like this:
这是因为在mktime()，它是这样的：
mktime(hour, minute, second, month, day, year);
Hence, your order is wrong.
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;
?>
回答by pbond
$date_string = "2012-10-18";
echo "Weeknummer: " . date("W", strtotime($date_string));
回答by cjtech
Use PHP's date function
http://php.net/manual/en/function.date.php
使用 PHP 的日期函数
http://php.net/manual/en/function.date.php
date("W", $yourdate)
回答by Andrew Myers
Just as a suggestion:
就像一个建议：
<?php echo date("W", strtotime("2012-10-18")); ?>
Might be a little simpler than all that lot.
可能比所有这些都简单一点。
Other things you could do:
你可以做的其他事情：
<?php echo date("Weeknumber: W", strtotime("2012-10-18 01:00:00")); ?>
<?php echo date("Weeknumber: W", strtotime($MY_DATE)); ?>
回答by user3133144
This get today date then tell the week number for the week
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";
?>
回答by FreudianSlip
<?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;
?>
You had the params to mktime wrong - needs to be Month/Day/Year, not Day/Month/Year
你对 mktime 的参数有误 - 需要是月/日/年，而不是日/月/年
回答by Marcel Bachus
I have tried to solve this question for years now, I thought I found a shorter solution but had to come back again to the long story. This function gives back the right ISO week notation:
多年来我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 周表示法：
/**
* 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+$weeknumber;
}
I found out that my short solution missed the 2018-12-31 as it gave back 1801 instead of 1901. So I had to put in this long version which is correct.
我发现我的简短解决方案错过了 2018-12-31，因为它返回了 1801 而不是 1901。所以我不得不输入这个正确的长版本。
回答by Kjeld
To get the week number for a date in North America I do like this:
要获取北美日期的周数，我这样做：
function week_number($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));
and get:
并得到：
Tue 2022-12-27: 53
Tue 2022-12-27: 53
回答by Bhushan D
To get Correct Week Count for Date 2018-12-31 Please use below Code
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'));
}
回答by majid rigi zadeh
for get week number in jalai calendaryou can use this:
要在jalai 日历中获取周数，您可以使用：
$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;
result:
结果：
15
week number change in saturday
周六周数变化
