blocks|key|417549|text|$date_string+=+"2012-10-18";
echo+"Weeknummer:+"+.+date("W",+strtotime($date_string));|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|417550|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>$date_string = "2012-10-18";
echo "Weeknummer: " . date("W", strtotime($date_string));
</code></pre>

blocks|key|632589|text|使用PHP的date函数|type|unstyled|depth|inlineStyleRanges|entityRanges|data|632590|http://php.net/manual/en/function.date.php|offset|length|632591|date("W",+$yourdate)|code-block|syntax|javascript|632592|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|16|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|S|8|@]|9|@$D|T|E|U|1|V]]|A|$]]|$1|F|3|G|5|H|7|W|8|@]|9|@]|A|$I|J]]|$1|K|3|-4|5|6|7|X|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|C]]]]

Use PHP's date function 
<a href="http://php.net/manual/en/function.date.php" rel="noreferrer">http://php.net/manual/en/function.date.php</a>

<pre><code>date("W", $yourdate)
</code></pre>

blocks|key|417585|text|get+today+date，然后告诉您该周的周编号|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417586|<?php
+$date=date("W");
+echo+$date."+Week+Number";
+?>|code-block|syntax|javascript|417587|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

<pre><code>&lt;?php
 $date=date("W");
 echo $date." Week Number";
 ?&gt;
</code></pre>

blocks|key|417604|text|只是作为一个建议：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417605|<?php+echo+date("W",+strtotime("2012-10-18"));+?>|code-block|syntax|javascript|417606|可能比所有这些都要简单一点。|417607|你可以做的其他事情：|417608|<?php+echo+date("Weeknumber:+W",+strtotime("2012-10-18+01:00:00"));+?>
<?php+echo+date("Weeknumber:+W",+strtotime($MY_DATE));+?>|417609|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Just as a suggestion:

<pre><code>&lt;?php echo date("W", strtotime("2012-10-18")); ?&gt;
</code></pre>

Might be a little simpler than all that lot.

Other things you could do: 

<pre><code>&lt;?php echo date("Weeknumber: W", strtotime("2012-10-18 01:00:00")); ?&gt;
&lt;?php echo date("Weeknumber: W", strtotime($MY_DATE)); ?&gt;
</code></pre>

blocks|key|632764|text|多年来，我一直试图解决这个问题，我想我找到了一个更短的解决方案，但不得不回到长篇大论上来。此函数返回正确的ISO周表示法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|632765|/**
+*+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%254==0+&&+$year%25100!=0)+%7C%7C+$year%25400==0)+{
++++$leapyear+=+true;
++}+else+{
++++$leapyear+=+false;
++}

++//+3.+check+if+$year-1+is+a++leapyear
++if+(+(($year-1)%254==0+&&+($year-1)%25100!=0)+%7C%7C+($year-1)%25400==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+=+$day+%2B+$mnth[$month-1];
++if+(+$leapyear+&&+$month+>+2+)+{+$dayofyearnumber%2B%2B;+}

++//+5.+find+the+jan1weekday+for+y+(monday=1,+sunday=7)
++$yy+=+($year-1)%25100;
++$c++=+($year-1)+-+$yy;
++$g++=+$yy+%2B+intval($yy/4);
++$jan1weekday+=+1%2B((((intval($c/100)%254)*5)%2B$g)%257);

++//+6.+find+the+weekday+for+y+m+d
++$h+=+$dayofyearnumber+%2B+($jan1weekday-1);
++$weekday+=+1%2B(($h-1)%257);

++//+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+(+$jan1weekday+=+5+%7C%7C+(+$jan1weekday+=+6+&&+$leapyearprev+))+{
++++++$weeknumber+=+53;
++++}+else+{
++++++$weeknumber+=+52;
++++}
++++$foundweeknum+=+true;
++}+else+{
++++$yearnumber+=+$year;
++}

++//+8.+find+if+y+m+d+falls+in+yearnumber+y%2B1,+weeknumber+1
++if+(+$yearnumber+==+$year+&&+!$foundweeknum)+{
++++if+(+$leapyear+)+{
++++++$i+=+366;
++++}+else+{
++++++$i+=+365;
++++}
++++if+(+($i+-+$dayofyearnumber)+<+(4+-+$weekday)+)+{
++++++$yearnumber+=+$year+%2B+1;
++++++$weeknumber+=+1;
++++++$foundweeknum+=+true;
++++}
++}

++//+9.+find+if+y+m+d+falls+in+yearnumber+y,+weeknumber+1+through+53
++if+(+$yearnumber+==+$year+&&+!$foundweeknum+)+{
++++$j+=+$dayofyearnumber+%2B+(7+-+$weekday)+%2B+($jan1weekday+-+1);
++++$weeknumber+=+intval(+$j/7+);
++++if+(+$jan1weekday+>+4+)+{+$weeknumber--;+}
++}

++//+10.+output+iso+week+number+(YYWW)
++return+($yearnumber-2000)*100%2B$weeknumber;
}|code-block|syntax|javascript|632766|我发现我的短线解决方案错过了2018-12-31，因为它返回了1801而不是1901。所以我不得不放入这个长版本，它是正确的。|632767|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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:

<pre><code>/**
 * calcweek("2018-12-31") =&gt; 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 &amp;&amp; $year%100!=0) || $year%400==0) {
 $leapyear = true;
 } else {
 $leapyear = false;
 }

 // 3. check if $year-1 is a leapyear
 if ( (($year-1)%4==0 &amp;&amp; ($year-1)%100!=0) || ($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 &amp;&amp; $month &gt; 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 ( $dayofyearnumber &lt;= (8-$jan1weekday) &amp;&amp; $jan1weekday &gt; 4 ) {
 $yearnumber = $year - 1;
 if ( $jan1weekday = 5 || ( $jan1weekday = 6 &amp;&amp; $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 == $year &amp;&amp; !$foundweeknum) {
 if ( $leapyear ) {
 $i = 366;
 } else {
 $i = 365;
 }
 if ( ($i - $dayofyearnumber) &lt; (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 == $year &amp;&amp; !$foundweeknum ) {
 $j = $dayofyearnumber + (7 - $weekday) + ($jan1weekday - 1);
 $weeknumber = intval( $j/7 );
 if ( $jan1weekday &gt; 4 ) { $weeknumber--; }
 }

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

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.

blocks|key|632633|text|<?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;
?>|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|632634|您的mktime参数错误-需要是月/日/年，而不是日/月/年|unstyled|632635|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>&lt;?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;
?&gt;
</code></pre>

You had the params to mktime wrong - needs to be Month/Day/Year, not Day/Month/Year

blocks|key|632799|text|要获取北美日期的周编号，我需要这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|632800|function+week_number($n)
{
++++$w+=+date('w',+$n);
++++return+1+%2B+date('z',+$n+%2B+(6+-+$w)+*+24+*+3600)+/+7;
}

$n+=+strtotime('2022-12-27');
printf("%25s:+%25d\n",+date('D+Y-m-d',+$n),+week_number($n));|code-block|syntax|javascript|632801|并获取：|632802|Tue+2022-12-27:+53|offset|length|style|CODE|632803|entityMap^0|0|0|0|0|I|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

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

<pre><code>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));
</code></pre>

and get:

<code>Tue 2022-12-27: 53</code>

blocks|key|633001|text|当一年有53周(比如2020年)时，上面给出的大多数例子都会产生问题。因此，每隔四年，你就会经历一个星期的差异。此代码不会：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|633002|$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|code-block|syntax|javascript|633003|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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:

<pre><code>$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();
$tempDatum-&gt;setISODate($thisYear, $weeknr);
$tempDatum_start = $tempDatum-&gt;format('Y-m-d');
$tempDatum-&gt;setISODate($thisYear, $weeknr, 7);
$tempDatum_end = $tempDatum-&gt;format('Y-m-d');

echo $tempDatum_start //will output the date of monday
echo $tempDatum_end // will output the date of sunday
</code></pre>

blocks|key|633063|text|当你需要一年又一周的时候，就会变得更加困难。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|633064|试着找出哪一周是01.01.2017。|633065|(这是2016年52周，从2016年12月26日-+Sun+01.01.2017开始)。|633066|经过更长时间的搜索，我发现|633067|strftime('%25G-%25V',strtotime("2017-01-01"))|code-block|syntax|javascript|633068|结果:+2016-52|633069|633070|https://www.php.net/manual/de/function.strftime.php|offset|length|633071|ISO-8601:1988给定年份的周编号，从一年的第一周开始，至少有4个工作日，星期一是一周的开始。(01到53)|633072|633073|mysql中的等价物是DATE_FORMAT(date，'%25x-%25v')+https://www.w3schools.com/sql/func_mysql_date_format.asp|633074|周，其中星期一是一周的第一天(01到53)。|633075|633076|找不到具有date()或DateTime的相应解决方案。|633077|至少没有像“%2B1天，上周一”这样的解决方案。|633078|entityMap|0|LINK|mutability|MUTABLE|url|1|https://www.w3schools.com/sql/func_mysql_date_format.asp^0|0|0|0|0|0|0|0|0|1F|0|0|0|0|11|1K|1|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1F|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1G|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|1H|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|1I|8|@]|9|@]|A|$K|L]]|$1|M|3|N|5|6|7|1J|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]|$1|P|3|Q|5|6|7|1L|8|@]|9|@$R|1M|S|1N|1|1O]]|A|$]]|$1|T|3|U|5|6|7|1P|8|@]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|W|3|X|5|6|7|1R|8|@]|9|@$R|1S|S|1T|1|1U]]|A|$]]|$1|Y|3|Z|5|6|7|1V|8|@]|9|@]|A|$]]|$1|10|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]|$1|11|3|12|5|6|7|1X|8|@]|9|@]|A|$]]|$1|13|3|14|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|15|3|-4|5|6|7|1Z|8|@]|9|@]|A|$]]]|16|$17|$5|18|19|1A|A|$1B|Q]]|1C|$5|18|19|1A|A|$1B|1D]]]]

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
<pre><code>strftime('%G-%V',strtotime(&quot;2017-01-01&quot;))
</code></pre>
Result: 2016-52
<hr />
<a href="https://www.php.net/manual/de/function.strftime.php" rel="nofollow noreferrer">https://www.php.net/manual/de/function.strftime.php</a> 
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)
<hr />
The equivalent in mysql is DATE_FORMAT(date, '%x-%v')
<a href="https://www.w3schools.com/sql/func_mysql_date_format.asp" rel="nofollow noreferrer">https://www.w3schools.com/sql/func_mysql_date_format.asp</a> 
Week where Monday is the first day of the week (01 to 53).
<hr />
Could not find a corresponding solution with date() or DateTime. 
At least not without solutions like &quot;+1day, last monday&quot;.

blocks|key|417391|text|规则是一年的第一周是包含一年的第一个星期四的那一周。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417392|我个人使用Zend_Date进行这种计算，要得到今天的星期就是这样简单。如果您使用日期，它们还有很多其他有用的函数。|417393|$now+=+Zend_Date::now();
$week+=+$now->get(Zend_Date::WEEK);
//+10|code-block|syntax|javascript|417394|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

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.

<pre><code>$now = Zend_Date::now();
$week = $now-&gt;get(Zend_Date::WEEK);
// 10
</code></pre>

blocks|key|417460|text|你的代码可以工作，但是你需要翻转第四个和第五个参数。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417461|我会这样做的|417462|$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}.";|code-block|syntax|javascript|417463|此外，在不查看代码的一周后，您的变量名称将会让您感到困惑，您应该考虑阅读http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/|offset|length|417464|entityMap|0|LINK|mutability|MUTABLE|url|http://net.tutsplus.com/tutorials/php/why-youre-a-bad-php-programmer/^0|0|0|0|10|1X|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@$K|Y|L|Z|1|10]]|A|$]]|$1|M|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

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

I would do it this way

<pre><code>$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}.";
</code></pre>

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

blocks|key|417787|text|要获得2018-12-31日期的正确周计数，请使用以下代码|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417788|$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'))+%2B+1;
}else{
++++$yr_count+=+date('y',strtotime('2018-12-31'));
}|code-block|syntax|javascript|417789|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

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

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


if($week_count=='01' &amp;&amp; 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'));
}
</code></pre>

blocks|key|417953|text|非常简单，只有一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417954|<?php+$date=date("W");+echo+"Week+"+.+$date;+?>"|code-block|syntax|javascript|417955|例如，您还可以像我需要的图形那样，减去前一周的值，如下所示：|417956|<?php+$date=date("W");+echo+"Week+"+.+($date+-+1);+?>|417957|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Very simple
Just one line:
<pre><code>&lt;?php $date=date(&quot;W&quot;); echo &quot;Week &quot; . $date; ?&gt;&quot;
</code></pre>
You can also, for example like I needed for a graph, subtract to get the previous week like:
<pre><code>&lt;?php $date=date(&quot;W&quot;); echo &quot;Week &quot; . ($date - 1); ?&gt;
</code></pre>

blocks|key|417619|text|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("%25a");
$diff+=+$diff/7;
$week+=+intval($diff)+%2B+1;
echo+$week;
//Returns+2.|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|417620|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>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 = $date2-&gt;diff($date1)-&gt;format("%a");
$diff = $diff/7;
$week = intval($diff) + 1;
echo $week;
//Returns 2.
</code></pre>

blocks|key|417622|text|尝试此解决方案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|417623|date(+'W',+strtotime(+"2017-01-01+%2B+1+day"+)+);|code-block|syntax|javascript|417624|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

try this solution

<pre><code>date( 'W', strtotime( "2017-01-01 + 1 day" ) );
</code></pre>

I want to take a date and work out its week number.

So far, I have the following. It is returning 24 when it should be 42.

<pre><code>&lt;?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;
?&gt;
</code></pre>

Is it wrong and a coincidence that the digits are reversed? Or am I nearly there?

Get week number (in the year) from a date PHP

我想找个日期，算出它的星期数。到目前为止，我有以下几点。它返回24，而它应该是42。<?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

问从日期PHP中获取周数(在一年中)
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回答 15

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问从日期PHP中获取周数(在一年中)
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