从一个日期 PHP 中获取周数(在一年中)

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The rule is that the first week of a year is the week that contains the first Thursday of the year.

I personally use Zend_Date for this kind of calculation and to get the week for today is this simple. They have a lot of other useful functions if you work with dates.

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

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Your code will work but you need to flip the 4th and the 5th argument.

I would do it this way

$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}.";

Also, your variable names will be confusing to you after a week of not looking at that code, you should consider reading http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/

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Use PHP's date function
http://php.net/manual/en/function.date.php

date("W", $yourdate)

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<?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

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最佳答案

Today, using PHP's DateTime objects is better:

<?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:

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

Hence, your order is wrong.

<?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;
?>

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

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This get today date then tell the week number for the week

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

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Just as a suggestion:

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

Might be a little simpler than all that lot.

Other things you could do:

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

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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.

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try this solution

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

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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:

/**
* 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.

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To get the week number for a date in North America I do like this:

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:

Tue 2022-12-27: 53

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To get Correct Week Count for Date 2018-12-31 Please use below Code

$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'));
}

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for get week number in jalai calendar you can use this:

$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:

week number change in saturday

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The most of the above given examples create a problem when a year has 53 weeks (like 2020). So every fourth year you will experience a week difference. This code does not:

$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

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How about using the IntlGregorianCalendar class?

Requirements: Before you start to use IntlGregorianCalendar make sure that libicu or pecl/intl is installed on the Server. So run on the CLI:

php -m

If you see intl in the [PHP Modules] list, then you can use IntlGregorianCalendar.

DateTime vs IntlGregorianCalendar: IntlGregorianCalendar is not better then DateTime. But the good thing about IntlGregorianCalendar is that it will give you the week number as an int.

Example:

$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

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Becomes more difficult when you need year and week.
Try to find out which week is 01.01.2017.
(It is the 52nd week of 2016, which is from Mon 26.12.2016 - Sun 01.01.2017).

After a longer search I found

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

Result: 2016-52

https://www.php.net/manual/de/function.strftime.php
ISO-8601:1988 week number of the given year, starting with the first week of the year with at least 4 weekdays, with Monday being the start of the week. (01 through 53)

The equivalent in mysql is DATE_FORMAT(date, '%x-%v') https://www.w3schools.com/sql/func_mysql_date_format.asp
Week where Monday is the first day of the week (01 to 53).

Could not find a corresponding solution with DateTime.
At least not without solutions like "+1day, last monday".

Edit: since strftime is now deprecated, maybe you can also use date. Didn't verify it though.

date('o-W',strtotime("2017-01-01"));

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Very simple Just one line:

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

You can also, for example like I needed for a graph, subtract to get the previous week like:

<?php $date=date("W"); echo "Week " . ($date - 1); ?>