Calcul des jours de la semaine ayant un numéro de semaine

Question

Compte tenu d'un numéro de semaine, par ex. date -u +%W, comment calculez-vous les jours de cette semaine à partir de lundi?

Exemple rfc-3339 sortie pour la semaine 40:

2008-10-06 
2008-10-07 
2008-10-08 
2008-10-09 
2008-10-10 
2008-10-11 
2008-10-12 

Answer 1

PHP

$week_number = 40; 
$year = 2008; 
for($day=1; $day<=7; $day++) 
{ 
    echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n"; 
} 

---

Ci-dessous après parce que je suis un idiot qui n'a pas lu la question correctement, mais obtiendra les dates dans une semaine à partir de lundi, étant donné la date, pas le numéro de la semaine ..

En PHP, adapté de this post sur le PHP date manual page:

function week_from_monday($date) { 
    // Assuming $date is in format DD-MM-YYYY 
    list($day, $month, $year) = explode("-", $_REQUEST["date"]); 

    // Get the weekday of the given date 
    $wkday = date('l',mktime('0','0','0', $month, $day, $year)); 

    switch($wkday) { 
     case 'Monday': $numDaysToMon = 0; break; 
     case 'Tuesday': $numDaysToMon = 1; break; 
     case 'Wednesday': $numDaysToMon = 2; break; 
     case 'Thursday': $numDaysToMon = 3; break; 
     case 'Friday': $numDaysToMon = 4; break; 
     case 'Saturday': $numDaysToMon = 5; break; 
     case 'Sunday': $numDaysToMon = 6; break; 
    } 

    // Timestamp of the monday for that week 
    $monday = mktime('0','0','0', $month, $day-$numDaysToMon, $year); 

    $seconds_in_a_day = 86400; 

    // Get date for 7 days from Monday (inclusive) 
    for($i=0; $i<7; $i++) 
    { 
     $dates[$i] = date('Y-m-d',$monday+($seconds_in_a_day*$i)); 
    } 

    return $dates; 
} 

sortie de week_from_monday('07-10-2008') donne:

Array 
(
    [0] => 2008-10-06 
    [1] => 2008-10-07 
    [2] => 2008-10-08 
    [3] => 2008-10-09 
    [4] => 2008-10-10 
    [5] => 2008-10-11 
    [6] => 2008-10-12 
) 

Answer 2

Ce calcul varie largement selon l'endroit où vous vivez. Par exemple, en Europe, nous commençons la semaine avec un lundi, au dimanche des États-Unis est le premier jour de la semaine. Au Royaume-Uni la semaine 1 est le 1er janvier, les autres pays commencent la semaine 1 la semaine contenant le premier jeudi de l'année.

Vous trouverez des informations plus générales sur http://en.wikipedia.org/wiki/Week#Week_number

Answer 3

Si vous avez Zend Framework, vous pouvez utiliser la classe Zend_Date pour ce faire:

require_once 'Zend/Date.php'; 

$date = new Zend_Date(); 
$date->setYear(2008) 
    ->setWeek(40) 
    ->setWeekDay(1); 

$weekDates = array(); 

for ($day = 1; $day <= 7; $day++) { 
    if ($day == 1) { 
     // we're already at day 1 
    } 
    else { 
     // get the next day in the week 
     $date->addDay(1); 
    } 

    $weekDates[] = date('Y-m-d', $date->getTimestamp()); 
} 

echo '<pre>'; 
print_r($weekDates); 
echo '</pre>'; 

Answer 4

Cette fonction donnera les horodatages de jours de la semaine où $ date est trouvée. Si $ date n'est pas donné, il suppose "maintenant". Si vous préférez des dates lisibles aux horodatages, passez un format de date dans le deuxième paramètre. Si vous ne commencez pas votre semaine le lundi (chanceux), passez un jour différent pour le troisième paramètre.

function week_dates($date = null, $format = null, $start = 'monday') { 
    // is date given? if not, use current time... 
    if(is_null($date)) $date = 'now'; 

    // get the timestamp of the day that started $date's week... 
    $weekstart = strtotime('last '.$start, strtotime($date)); 

    // add 86400 to the timestamp for each day that follows it... 
    for($i = 0; $i < 7; $i++) { 
    $day = $weekstart + (86400 * $i); 
    if(is_null($format)) $dates[$i] = $day; 
    else $dates[$i] = date($format, $day); 
    } 

    return $dates; 
} 

Alors week_dates() devrait retourner quelque chose comme ...

Array ( 
    [0] => 1234155600 
    [1] => 1234242000 
    [2] => 1234328400 
    [3] => 1234414800 
    [4] => 1234501200 
    [5] => 1234587600 
    [6] => 1234674000 
) 

Answer 5

J'ai trouvé un problème avec cette solution. J'ai dû mettre à zéro le numéro de la semaine ou bien il était en train de se casser.

Ma solution ressemble à ceci maintenant:

$week_number = 40; 
$year = 2008; 
for($day=1; $day<=7; $day++) 
{ 
    echo date('m/d/Y', strtotime($year."W".str_pad($week_number,2,'0',STR_PAD_LEFT).$day))."\n"; 
} 

Answer 6

Pour ceux qui recherchent les jours de la semaine étant donné le numéro de la semaine (1-52) A partir d'un dimanche alors voici mon petit travail autour . Prend en compte la vérification de la semaine est dans la bonne gamme et remplit les valeurs 1-9 pour que tout fonctionne.

$week = 2; $year = 2009; 

$week = (($week >= 1) AND ($week <= 52))?($week-1):(1); 

$dayrange = array(7,1,2,3,4,5,6); 

for($count=0; $count<=6; $count++) { 
    $week = ($count == 1)?($week + 1): ($week); 
    $week = str_pad($week,2,'0',STR_PAD_LEFT); 
    echo date('d m Y', strtotime($year."W".$week.($dayrange[$count]))); } 

Answer 7

$week_number = 40; 
$year = 2008; 

for($day=1; $day<=7; $day++) 
{ 
    echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n"; 
} 

Cela échouera si $week_number est inférieur à 10.

//============Try this================// 

$week_number = 40; 
$year = 2008; 

if($week_number < 10){ 
    $week_number = "0".$week_number; 
} 

for($day=1; $day<=7; $day++) 
{ 
    echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n"; 
} 

//==============================// 

Answer 8

J'ai eu la même question que l'utilisation strftime au lieu de la date comme point de départ à savoir avoir dérivé plusieurs semaines de strftime en utilisant% WI voulait connaître la plage de dates pour cette semaine - du lundi au dimanche (voire tout jour de départ). Un examen de plusieurs postes similaires et en particulier l'essai de quelques-unes des approches ci-dessus ne m'a pas amené à la solution que je voulais. Bien sûr, j'ai peut-être mal compris quelque chose, mais je n'ai pas pu obtenir ce que je voulais.

Je voudrais donc partager ma solution.

Ma première pensée a été que compte tenu de la description de strftime% W est:

numéro de la semaine

de l'année en cours, en commençant par le premier lundi comme le premier jour de la première semaine

Si j'établis quel est le premier lundi de chaque année, je pourrais calculer un tableau de plages de dates avec un indice égal à la valeur de% W. Par la suite, je pourrais appeler la fonction en utilisant strftime.

Alors, voici:

La fonction:

<?php 

/* 
* function to establish scope of week given a week of the year value returned from strftime %W 
*/ 

// note strftime %W reports 1/1/YYYY as wk 00 unless 1/1/YYYY is a monday when it reports wk 01 
// note strtotime Monday [last, this, next] week - runs sun - sat 

function date_Range_For_Week($W,$Y){ 

// where $W = %W returned from strftime 
//  $Y = %Y returned from strftime 

    // establish 1st day of 1/1/YYYY 

    $first_Day_Of_Year = mktime(0,0,0,1,1,$Y); 

    // establish the first monday of year after 1/1/YYYY  

    $first_Monday_Of_Year = strtotime("Monday this week",(mktime(0,0,0,1,1,$Y))); 

    // Check for week 00 advance first monday if found 
    // We could use strtotime "Monday next week" or add 604800 seconds to find next monday 
    // I have decided to avoid any potential strtotime overhead and do the arthimetic 

    if (strftime("%W",$first_Monday_Of_Year) != "01"){ 
     $first_Monday_Of_Year += (60 * 60 * 24 * 7); 
    } 

    // create array to ranges for the year. Note 52 wks is the norm but it is possible to have 54 weeks 
    // in a given yr therefore allow for this in array index 

    $week_Start = array(); 
    $week_End = array();   

    for($i=0;$i<=53;$i++){ 

     if ($i == 0){ 
      if ($first_Day_Of_Year != $first_Monday_Of_Year){ 
       $week_Start[$i] = $first_Day_Of_Year; 
       $week_End[$i] = $first_Monday_Of_Year - (60 * 60 * 24 * 1); 
      } else { 
       // %W returns no week 00 
       $week_Start[$i] = 0; 
       $week_End[$i] = 0;        
      } 
      $current_Monday = $first_Monday_Of_Year; 
     } else { 
      $week_Start[$i] = $current_Monday; 
      $week_End[$i] = $current_Monday + (60 * 60 * 24 * 6); 
      // find next monday 
      $current_Monday += (60 * 60 * 24 * 7); 
      // test for end of year 
      if (strftime("%W",$current_Monday) == "01"){ $i = 999; }; 
     } 
    }; 

    $result = array("start" => strftime("%a on %d, %b, %Y", $week_Start[$W]), "end" => strftime("%a on %d, %b, %Y", $week_End[$W])); 

    return $result; 

    } 

?> 

Exemple:

// usage example 

//assume we wish to find the date range of a week for a given date July 12th 2011 

$Y = strftime("%Y",mktime(0,0,0,7,12,2011)); 
$W = strftime("%W",mktime(0,0,0,7,12,2011)); 

// use dynamic array variable to check if we have range if so get result if not run function 

$date_Range = date_Range . "$Y"; 

isset(${$date_Range}) ? null : ${$date_Range} = date_Range_For_Week($W, $Y); 

echo "Date sought: " . strftime(" was %a on %b %d, %Y, %X time zone: %Z",mktime(0,0,0,7,12,2011)) . "<br/>"; 
echo "start of week " . $W . " is " . ${$date_Range}["start"] . "<br/>"; 
echo "end of week " . $W . " is " . ${$date_Range}["end"]; 

Sortie:

> Date sought: was Tue on Jul 12, 2011, 00:00:00 time zone: GMT Daylight 
> Time start of week 28 is Mon on 11, Jul, 2011 end of week 28 is Sun on 
> 17, Jul, 2011 

J'ai testé sur plusieurs années, dont 2018 est l'année suivante quand 1/1/2018 = lundi. Jusqu'à présent, semble fournir la plage de dates correcte.

Donc j'espère que cela aidera.

Cordialement

Answer 9

Depuis cette question et la réponse acceptée ont été affichés les DateTime cours font beaucoup plus simple à faire: -

function daysInWeek($weekNum) 
{ 
    $result = array(); 
    $datetime = new DateTime('00:00:00'); 
    $datetime->setISODate((int)$datetime->format('o'), $weekNum, 1); 
    $interval = new DateInterval('P1D'); 
    $week = new DatePeriod($datetime, $interval, 6); 

    foreach($week as $day){ 
     $result[] = $day->format('D d m Y H:i:s'); 
    } 
    return $result; 
} 

var_dump(daysInWeek(24)); 

Ceci a l'avantage de prendre soin des années bissextiles etc ..

See it working. Y compris les semaines difficiles 1 et 53.

Answer 10

Une autre solution:

//$date Date in week 
//$start Week start (out) 
//$end Week end (out) 

function week_bounds($date, &$start, &$end) { 
    $date = strtotime($date); 
    $start = $date; 
    while(date('w', $start)>1) { 
     $start -= 86400; 
    } 
    $end = date('Y-m-d', $start + (6*86400)); 
    $start = date('Y-m-d', $start); 
} 

Exemple:

week_bounds("2014/02/10", $start, $end); 
echo $start."<br>".$end; 

Out:

2014-02-10 
2014-02-16 

Answer 11

<?php 
    $iWeeksAgo = 5;// need weeks ago 
    $sWeekDayStartOn = 0;// 0 - Sunday, 1 - Monday, 2 - Tuesday 
    $aWeeksDetails = getWeekDetails($iWeeksAgo, $sWeekDayStartOn); 

    print_r($aWeeksDetails); 
    die('end of line of getWeekDetails '); 

    function getWeekDetails($iWeeksAgo, $sWeekDayStartOn){ 
     $date = new DateTime(); 
     $sCurrentDate = $date->format('W, Y-m-d, w'); 
     #echo 'Current Date (Week of the year, YYYY-MM-DD, day of week): ' . $sCurrentDate . "\n"; 

     $iWeekOfTheYear = $date->format('W');// Week of the Year i.e. 19-Feb-2014 = 08 
     $iDayOfWeek = $date->format('w');// day of week for the current month i.e. 19-Feb-2014 = 4 
     $iDayOfMonth = $date->format('d'); // date of the month i.e. 19-Feb-2014 = 19 

     $iNoDaysAdd = 6;// number of days adding to get last date of the week i.e. 19-Feb-2014 + 6 days = 25-Feb-2014 

     $date->sub(new DateInterval("P{$iDayOfWeek}D"));// getting start date of the week 
     $sStartDateOfWeek = $date->format('Y-m-d');// getting start date of the week 

     $date->add(new DateInterval("P{$iNoDaysAdd}D"));// getting end date of the week 
     $sEndDateOfWeek = $date->format('Y-m-d');// getting end date of the week 

     $iWeekOfTheYearWeek = (string) $date->format('YW');//week of the year 
     $iWeekOfTheYearWeekWithPeriod = (string) $date->format('Y-W');//week of the year with year 

     //To check uncomment 
     #echo "Start Date/End Date of Current week($iWeekOfTheYearWeek), week with - ($iWeekOfTheYearWeekWithPeriod) : " . $sStartDateOfWeek . ',' . $sEndDateOfWeek . "\n"; 

     $iDaysAgo = ($iWeeksAgo*7) + $iNoDaysAdd + $sWeekDayStartOn;// getting 4 weeks ago i.e. no. of days to substract 

     $date->sub(new DateInterval("P{$iDaysAgo}D"));// getting 4 weeks ago i.e. no. of days to substract 
     $sStartDateOfWeekAgo = $date->format('Y-m-d');// getting 4 weeks ago start date i.e. 19-Jan-2014 

     $date->add(new DateInterval("P{$iNoDaysAdd}D")); // getting 4 weeks ago end date i.e. 25-Jan-2014 
     $sEndDateOfWeekAgo = $date->format('Y-m-d');// getting 4 weeks ago start date i.e. 25-Jan-2014 

     $iProccessedWeekAgoOfTheYear = (string) $date->format('YW');//ago week of the year 
     $iProccessedWeekOfTheYearWeekAgo = (string) $date->format('YW');//ago week of the year with year 
     $iProccessedWeekOfTheYearWeekWithPeriodAgo = (string) $date->format('Y-W');//ago week of the year with year 

     //To check uncomment 
     #echo "Start Date/End Date of week($iProccessedWeekOfTheYearWeekAgo), week with - ($iProccessedWeekOfTheYearWeekWithPeriodAgo) ago: " . $sStartDateOfWeekAgo . ',' . $sEndDateOfWeekAgo . "\n"; 

     $aWeeksDetails = array ('weeksago' => $iWeeksAgo, 'currentweek' => $iWeekOfTheYear, 'currentdate' => $sCurrentDate, 'startdateofcurrentweek' => $sStartDateOfWeek, 'enddateofcurrentweek' => $sEndDateOfWeek, 
           'weekagoyearweek' => $iProccessedWeekAgoOfTheYear, 'startdateofagoweek' => $sStartDateOfWeekAgo, 'enddateofagoweek' => $sEndDateOfWeekAgo); 

     return $aWeeksDetails; 
    } 
?> 

Answer 12

Un autre code hehe:

public function getAllowedDays($year, $week) { 
    $weekDaysArray = array(); 
    $dto = new \DateTime(); 
    $dto->setISODate($year, $week); 

    for($i = 0; $i < 7; $i++) { 
     array_push($weekDaysArray, $dto->format('Y-m-d')); 
     $dto->modify("+1 days"); 
    } 

    return $weekDaysArray; 
} 

Answer 13

$year  = 2016; //enter the year 
$wk_number = 46; //enter the weak nr 

$start = new DateTime($year.'-01-01 00:00:00'); 
$end = new DateTime($year.'-12-31 00:00:00'); 

$start_date = $start->format('Y-m-d H:i:s'); 

$output[0]= $start;  
$end = $end->format('U');  
$x = 1; 

//create array full of data objects 
for($i=0;;$i++){ 
    if($i == intval(date('z',$end)) || $i === 365){ 
     break; 
    } 
    $a = new DateTime($start_date); 
    $b = $a->modify('+1 day'); 
    $output[$x]= $a;   
    $start_date = $b->format('Y-m-d H:i:s'); 
    $x++; 
}  

//create a object to use 
for($i=0;$i<count($output);$i++){ 
    if(intval ($output[$i]->format('W')) === $wk_number){ 
     $output_[$output[$i]->format('N')]  = $output[$i]; 
    } 
} 

$dayNumberOfWeek = 1; //enter the desired day in 1 = Mon -> 7 = Sun 

echo '<pre>'; 
print_r($output_[$dayNumberOfWeek]->format('Y-m-d')); 
echo '</pre>'; 

utilisation comme date() objet de php date php