php 给定周数计算一周中的天数
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Calculating days of week given a week number
提问by hendry
Given a week number, e.g. date -u +%W, how do you calculate the days in that week starting from Monday?
给定一个星期数，例如date -u +%W，您如何计算从星期一开始的那一周中的天数？
Example rfc-3339 output for week 40:
第 40 周的 rfc-3339 输出示例：
2008-10-06
2008-10-07
2008-10-08
2008-10-09
2008-10-10
2008-10-11
2008-10-12
回答by ConroyP
PHP
PHP
$week_number = 40;
$year = 2008;
for($day=1; $day<=7; $day++)
{
echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n";
}
下面的帖子是因为我是一个白痴，没有正确阅读问题，但是会在从星期一开始的一周内获取日期，给定日期，而不是周数。
In PHP, adapted from this poston the 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;
}
Output from week_from_monday('07-10-2008')gives:
从week_from_monday('07-10-2008')给出的输出：
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
)
回答by Shane
If you've got Zend Framework you can use the Zend_Date class to do this:
如果你有 Zend 框架，你可以使用 Zend_Date 类来做到这一点：
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>';
回答by vascowhite
Since this question and the accepted answer were posted the DateTimeclasses make this much simpler to do:-
由于此问题和已接受的答案已发布，因此DateTime类使此操作变得更加简单：-
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));
This has the added advantage of taking care of leap years etc..
这具有处理闰年等的额外优势。
See it working. Including the difficult weeks 1 and 53.
看到它工作。包括艰难的第 1 周和第 53 周。
回答by Martin Liesén
This calculation varies largely depending on where you live. For example, in Europe we start the week with a Monday, in US Sunday is the first day of the week. In UK week 1 is on Jan 1, others countries start week 1 on the week containing the first Thursday of the year.
这个计算在很大程度上取决于你住的地方。例如，在欧洲，我们从星期一开始一周，在美国，星期日是一周的第一天。在英国，第 1 周是 1 月 1 日，其他国家/地区从包含一年第一个星期四的那一周开始第 1 周。
You can find more general information at http://en.wikipedia.org/wiki/Week#Week_number
您可以在http://en.wikipedia.org/wiki/Week#Week_number 上找到更多一般信息
回答by Martin Liesén
This function will give the timestamps of days of the week in which $date is found. If $date isn't given, it assumes "now." If you prefer readable dates to timestamps, pass a date format into the second parameter. If you don't start your week on Monday (lucky), pass in a different day for the third parameter.
此函数将给出找到 $date 的星期几的时间戳。如果未给出 $date，则假定为“现在”。如果您更喜欢可读日期而不是时间戳，请将日期格式传递给第二个参数。如果您没有在星期一开始您的一周（幸运），请为第三个参数传入另一天。
function week_dates($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;
}
So week_dates()should return something like...
所以week_dates()应该返回类似...
Array (
[0] => 1234155600
[1] => 1234242000
[2] => 1234328400
[3] => 1234414800
[4] => 1234501200
[5] => 1234587600
[6] => 1234674000
)
回答by Nicolas
$week_number = 40;
$year = 2008;
for($day=1; $day<=7; $day++)
{
echo date('m/d/Y', strtotime($year."W".$week_number.$day))."\n";
}
This will fail if $week_numberis less than 10.
如果$week_number小于 10，这将失败。
//============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";
}
//==============================//
回答by Miton Leon
Another code hehe:
另一个代码嘿嘿：
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;
}
回答by Peter Breuls
$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_[$dayNumberOfWeek]->format('Y-m-d'));
echo '</pre>';
use as date() object from php date php
从 php date php 中用作 date() 对象
回答by joan16v
Another solution:
另一种解决方案：
//$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);
}
Example:
例子：
week_bounds("2014/02/10", $start, $end);
echo $start."<br>".$end;
Out:
出去：
2014-02-10
2014-02-16
回答by codepuppy
I had the same question only using strftime instead of date as my starting point i.e. having derived a week number from strftime using %W I wanted to know the date range for that week - Monday to Sunday (or indeed any starting day). A review of several similar posts and in particular trying out a couple of the above approaches didn't get me to the solution I wanted. Of course I may have misunderstood something but I couldn't get what I wanted.
我有同样的问题，只使用 strftime 而不是日期作为我的起点，即使用 %WI 从 strftime 导出周数想知道那一周的日期范围 - 周一到周日（或实际上任何开始日）。回顾几个类似的帖子，特别是尝试上述几种方法并没有让我找到我想要的解决方案。当然，我可能误解了一些东西，但我无法得到我想要的。
I would therefore like to share my solution.
因此，我想分享我的解决方案。
My first thought was that given the description of strftime %W is:
我的第一个想法是，鉴于 strftime %W 的描述是：
week number of the current year, starting with the first Monday as the first day of the first week
当前年份的周数，从第一个星期一开始作为第一周的第一天
if I established what the first Monday of each year is I could calculate an array of date ranges with an index equal to the value of %W. Thereafter I could call the function using strftime.
如果我确定每年的第一个星期一是什么，我可以计算一个日期范围数组，其索引等于 %W 的值。此后我可以使用 strftime 调用该函数。
So here goes:
所以这里是：
The Function:
功能：
<?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;
}
?>
Example:
例子：
// 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"];
Output:
输出：
> 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
I have tested this over several years including 2018 which is the next year when 1/1/2018 = Monday. Thus far seems to deliver the correct date range.
我已经对此进行了几年的测试，包括 2018 年，即 2018 年 1 月 1 日 = 星期一的下一年。到目前为止，似乎提供了正确的日期范围。
So I hope that this helps.
所以我希望这会有所帮助。
Regards
问候
