我正在尝试查找给定月份和年份的每周时段。日期应从星期一开始,到星期日结束。如果该月的第一天是星期日(2011年5月),它应该是第一个元素。
2011年5月
2012年9月
我正在使用此函数计算两个日期的周数 - 即。这个月的第一天和一个月的最后一天。
public function getWeekNumbers($startDate, $endDate)
{
$p = new DatePeriod(
new DateTime($startDate),
new DateInterval('P1W'),
new DateTime($endDate)
);
$weekNumberList = array();
foreach ($p as $w)
{
$weekNumber = $w->format('W');
$weekNumberList[] = ltrim($weekNumber, '0');
}
return $weekNumberList;
}
奇怪的是,对于1月份,当我期待[1,2,3,4,5]时,它会返回[52,1,2,3,4]的周数。
一旦我有周数,我就像这样使用它们:
//The following loop will populate the dataset with the entire month's durations - regardless if hours were worked or not.
$firstDayOfMonth = date('Y-m-d', strtotime("first day of {$this->year}-{$monthName}"));
$lastDayOfMonth = date('Y-m-d', strtotime("last day of {$this->year}-{$monthName}"));
foreach ($this->getWeekNumbers($firstDayOfMonth, $lastDayOfMonth) as $key => $weekId)
{
// find first mоnday of the year
$firstMon = strtotime("mon jan {$this->year}");
// calculate how many weeks to add
$weeksOffset = $weekId - date('W', $firstMon);
$beginDays = $weeksOffset * 7;
$endDays = ($weeksOffset * 7) + 6;
$searchedMon = strtotime(date('Y-m-d', $firstMon) . " +{$beginDays} days");
$searchedSun = strtotime(date('Y-m-d', $firstMon) . " +{$endDays} days");
echo date("M d", $searchedMon) . " - " . date("M d", $searchedSun);
}
因为getWeekNumbers函数没有返回我期望的周数,所以上述函数的输出是
并不奇怪请注意,第1行(12月24日 - 12月30日)是当年(2012年)的结束,而不是去年年底(2011年)。
理想情况下,我希望它看起来像
有什么想法吗?谢谢!
答案 0 :(得分:3)
如果您需要选定月份的所有周数以及所选周的所有日期,那么这就是您所需要的:
function getWeekDays($month, $year)
{
$p = new DatePeriod(
DateTime::createFromFormat('!Y-n-d', "$year-$month-01"),
new DateInterval('P1D'),
DateTime::createFromFormat('!Y-n-d', "$year-$month-01")->add(new DateInterval('P1M'))
);
$datesByWeek = array();
foreach ($p as $d) {
$dateByWeek[ $d->format('W') ][] = $d;
}
return $dateByWeek;
}
getWeekDays()函数返回多维数组。第一个关键是周数。 2级是数组,其日期保存为DateTime对象。
获取示例:
print_r( getWeekDays(5, 2011) ); # May 2011
print_r( getWeekDays(9, 2012) ); # Sep 2012
我有一点额外的时间,所以我写了一个例子; - )
$datesByWeek = getWeekDays(8, 2012);
$o = '<table border="1">';
$o.= '<tr><th>Week</th><th>Monday</th><th>Tuesday</th><th>Wednesday</th><th>Thursday</th><th>Friday</th><th>Saturday</th><th>Sunday</th></tr>';
foreach ($datesByWeek as $week => $dates) {
$firstD = $dates[0];
$lastD = $dates[count($dates)-1];
$o.= "<tr>";
$o.= "<td>" . $firstD->format('M d') . ' - ' . $lastD->format('M d') . "</td>";
$N = $firstD->format('N');
for ($i = 1; $i < $N; $i++) {
$o.= "<td>-</td>";
}
foreach ($dates as $d) {
$o.= "<td>" . $d->format('d.') . " / 0.00</td>";
# for selected date do you magic
}
$N = $lastD->format('N');
for ($i = $N; $i < 7; $i++) {
$o.= "<td>-</td>";
}
$o.= "</tr>";
}
$o.= '</table>';
echo $o;
输出如下:
答案 1 :(得分:2)
这完美! phpfiddle here
<?php
// start and end must be timestamps !!!!
$start = 1346976000; // Thu 2012-09-06
$end = 1348704000; // Tue 2012-09-26
// generate the weeks
$weeks = generateweeks($start, $end);
// diaplay the weeks
echo 'From: '.fDate($start).'<br>';
foreach ($weeks as $week){
echo fDate($week['start']).' '.fDate($week['end']).'<br>';
}
echo 'To: '.fDate($end).'<br>';
/* outputs this:
From: Thu 2012-09-06
Thu 2012-09-06 Sun 2012-09-09
Mon 2012-09-10 Sun 2012-09-16
Mon 2012-09-17 Sun 2012-09-23
Mon 2012-09-24 Wed 2012-09-26
To: Wed 2012-09-26
*/
// $start and $end must be unix timestamps (any range)
// returns an array of arrays with 'start' and 'end' elements set
// for each week (or part of week) for the given interval
// return values are also in timestamps
function generateweeks($start,$end){
$ret = array();
$start = E2D($start);
$end = E2D($end);
$ns = nextSunday($start);
while(true){
if($ns>=$end) {insert($ret,$start,$end);return $ret;}
insert($ret,$start,$ns);
$start = $ns +1;
$ns+=7;
}
}
// helper function to append the array and convert back to unix timestamp
function insert(&$arr, $start, $end){$arr[] = array('start'=>D2E($start), 'end'=>D2E($end));}
// recives any date on CD format returns next Sunday on CD format
function nextSunday($Cdate){return $Cdate + 6 - $Cdate % 7;}
// recives any date on CD format returns previous Monday on CD format // finaly not used here
function prevMonday($Cdate){return $Cdate - $Cdate % 7;}
// recives timestamp returns CD
function E2D($what){return floor($what/86400)+2;} // floor may be optional in some circunstances
// recives CD returns timestamp
function D2E($what){return ($what-2)*86400;} // 24*60*60
// just format the timestamp for output, you can adapt it to your needs
function fDate($what){return date('D Y-m-d',$what);}
答案 2 :(得分:1)
以下假设用户可以选择他们想要运行报告的月份和年份(月份为1-12,年份为YYYY)。可能有一种更优雅的方式,但这似乎对我有用。此外,在帖子的顶部,您说您希望周数为周一至周日。但是,您在底部的示例/屏幕截图显示周是星期六到星期六。以下是周一至周日的原定目标。
$month = $_POST["month"];
$year = $_POST["year"];
$endDate = date("t", strtotime($year."-".$month."-01"));
$dayOfWeekOfFirstOfMonth = date("w", strtotime($year."-".$month."-01"));
$lastDayOfFirstWeek = 8 - $dayOfWeekOfFirstOfMonth;
$weeksArray = array(array("firstDay"=>1, "lastDay"=>$lastDayOfFirstWeek));
$loopDate = $lastDayOfFirstWeek + 1;
while($loopDate < $endDate)
{
$weeksArray[] = array("firstDay"=>$loopDate, "lastDay"=>($loopDate+6 > $endDate ? $endDate : $loopDate+6));
$loopDate+=7;
}
foreach($weeksArray as $week)
{
echo date("M d", strtotime($year."-".$month."-".$week["firstDay"])) . " - " . date("M d", strtotime($year."-".$month."-".$week["lastDay"])) . "\n";
}