你好，如何找到两个日期之间的周数和每周一的日期。例如，从2009年10月7日到今天。
注意：考虑闰年和其他与日期相关的限制。
10个回答
84
这里有一个使用DateTime的替代方案:-
function datediffInWeeks($date1, $date2)
{
if($date1 > $date2) return datediffInWeeks($date2, $date1);
$first = DateTime::createFromFormat('m/d/Y', $date1);
$second = DateTime::createFromFormat('m/d/Y', $date2);
return floor($first->diff($second)->days/7);
}
var_dump(datediffInWeeks('1/2/2013', '6/4/2013'));// 21
- vascowhite
4
2我认为这个答案比被接受的答案更完整，特别是如果@Piskvor说datediff有漏洞的话。 - Hector Ordonez
1这里的 floor($first->diff($second)->days/7) 给我带来了问题...所以这可能取决于你的程序是否需要将2.8周视为3周或2周。 - Aukhan
@vascowhite ... 是的 :) 那天我修改了它以适应我的需求，我想通过 ceil() 来查看它。感谢你的片段！我认为它应该成为核心 PHP 的一部分...喜欢 moment.js 处理这些事情的方式 ;) - Aukhan
23
echo datediff('ww', '9 July 2003', '4 March 2004', false);
在以下网站中找到这个函数： http://www.addedbytes.com/code/php-datediff-function/
更新
链接已经失效（2017年9月），因此下面的函数来自于Web Archive：
<?php
/**
* @param $interval
* @param $datefrom
* @param $dateto
* @param bool $using_timestamps
* @return false|float|int|string
*/
function datediff($interval, $datefrom, $dateto, $using_timestamps = false)
{
/*
$interval can be:
yyyy - Number of full years
q - Number of full quarters
m - Number of full months
y - Difference between day numbers
(eg 1st Jan 2004 is "1", the first day. 2nd Feb 2003 is "33". The datediff is "-32".)
d - Number of full days
w - Number of full weekdays
ww - Number of full weeks
h - Number of full hours
n - Number of full minutes
s - Number of full seconds (default)
*/
if (!$using_timestamps) {
$datefrom = strtotime($datefrom, 0);
$dateto = strtotime($dateto, 0);
}
$difference = $dateto - $datefrom; // Difference in seconds
$months_difference = 0;
switch ($interval) {
case 'yyyy': // Number of full years
$years_difference = floor($difference / 31536000);
if (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom), date("j", $datefrom), date("Y", $datefrom)+$years_difference) > $dateto) {
$years_difference--;
}
if (mktime(date("H", $dateto), date("i", $dateto), date("s", $dateto), date("n", $dateto), date("j", $dateto), date("Y", $dateto)-($years_difference+1)) > $datefrom) {
$years_difference++;
}
$datediff = $years_difference;
break;
case "q": // Number of full quarters
$quarters_difference = floor($difference / 8035200);
while (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom)+($quarters_difference*3), date("j", $dateto), date("Y", $datefrom)) < $dateto) {
$months_difference++;
}
$quarters_difference--;
$datediff = $quarters_difference;
break;
case "m": // Number of full months
$months_difference = floor($difference / 2678400);
while (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom)+($months_difference), date("j", $dateto), date("Y", $datefrom)) < $dateto) {
$months_difference++;
}
$months_difference--;
$datediff = $months_difference;
break;
case 'y': // Difference between day numbers
$datediff = date("z", $dateto) - date("z", $datefrom);
break;
case "d": // Number of full days
$datediff = floor($difference / 86400);
break;
case "w": // Number of full weekdays
$days_difference = floor($difference / 86400);
$weeks_difference = floor($days_difference / 7); // Complete weeks
$first_day = date("w", $datefrom);
$days_remainder = floor($days_difference % 7);
$odd_days = $first_day + $days_remainder; // Do we have a Saturday or Sunday in the remainder?
if ($odd_days > 7) { // Sunday
$days_remainder--;
}
if ($odd_days > 6) { // Saturday
$days_remainder--;
}
$datediff = ($weeks_difference * 5) + $days_remainder;
break;
case "ww": // Number of full weeks
$datediff = floor($difference / 604800);
break;
case "h": // Number of full hours
$datediff = floor($difference / 3600);
break;
case "n": // Number of full minutes
$datediff = floor($difference / 60);
break;
default: // Number of full seconds (default)
$datediff = $difference;
break;
}
return $datediff;
}
- Tom Gullen
5
3许多时间范围都存在一个不太明显的 bug，包括 2012 年、2016 年、2020 年和 2024 年，这个 bug 就是二月份的 29 号。 - Piskvor left the building
4如果链接失效的话，包含
datediff函数的代码会很有用。 - AbcAeffchen13我到底是怎样在谷歌上搜索两个日期之间相差几周的函数，然后却在StackOverflow上看到一条评论，里面链接着我自己的网站和我已经写好解决这个问题的代码？我觉得我快疯了…… - Dave Child
@Piskvor 像这样的复杂函数，你应该拥有某种可重现代码来证明它的正确性。它运行得很好，我没有发现任何错误。 - raveren
@DaveChild发现了一个可以计算两个日期之间周数的函数，你看过吗？ - Harry Bosh
23
$diff = strtotime($dateto, 0) - strtotime($datefrom, 0);
echo floor($diff / 604800);
- silent
3
1存在夏令时问题。可能的解决方案：使用协调世界时（UTC）工作，使用 round 而非 floor，或者使用 DateTime::diff 而非减法运算。 - Brilliand
DST @Brilliand 是什么？ - Radmation
1@Radmation 夏令时 - Brilliand
3
这个做得很好
。function weeks_between($datefrom, $dateto)
{
$datefrom = DateTime::createFromFormat('d/m/Y H:i:s',$datefrom);
$dateto = DateTime::createFromFormat('d/m/Y H:i:s',$dateto);
$interval = $datefrom->diff($dateto);
$week_total = $interval->format('%a')/7;
return floor($week_total);
}
weeks_between("25/02/2000 11:30:00","05/06/2015 11:29:59")
- Cyril Giacopino
2
以下函数计算两个时间戳之间的“口语周”（例如，如果您在星期六，则下周一是下一周）。
function days_between($datefrom,$dateto){
$fromday_start = mktime(0,0,0,date("m",$datefrom),date("d",$datefrom),date("Y",$datefrom));
$diff = $dateto - $datefrom;
$days = intval( $diff / 86400 ); // 86400 / day
if( ($datefrom - $fromday_start) + ($diff % 86400) > 86400 )
$days++;
return $days;
}
function weeks_between($datefrom, $dateto)
{
$day_of_week = date("w", $datefrom);
$fromweek_start = $datefrom - ($day_of_week * 86400) - ($datefrom % 86400);
$diff_days = days_between($datefrom, $dateto);
$diff_weeks = intval($diff_days / 7);
$seconds_left = ($diff_days % 7) * 86400;
if( ($datefrom - $fromweek_start) + $seconds_left > 604800 )
$diff_weeks ++;
return $diff_weeks;
}
- Mat
1
我更新了上面Tom的代码，使函数可以自动检测字符串或整数日期/时间。
<?php
/**
* @param $interval
* @param $datefrom
* @param $dateto
* @param bool $using_timestamps // Removed 08-29-2019 Jay Simons - Now auto-detects
* @return false|float|int|string
*/
function datediff($interval, $datefrom, $dateto)
{
/*
$interval can be:
yyyy - Number of full years
q - Number of full quarters
m - Number of full months
y - Difference between day numbers
(eg 1st Jan 2004 is "1", the first day. 2nd Feb 2003 is "33". The datediff is "-32".)
d - Number of full days
w - Number of full weekdays
ww - Number of full weeks
h - Number of full hours
n - Number of full minutes
s - Number of full seconds (default)
*/
/*Remove
if (!$using_timestamps) {
$datefrom = strtotime($datefrom, 0);
$dateto = strtotime($dateto, 0);
}
*/
// Auto-detect string date or int date:
$dateto_str = strtotime($dateto);
$datefrom_str = strtotime($datefrom);
if ($dateto_str) $dateto = $dateto_str;
if ($datefrom_str) $datefrom = $datefrom_str;
$difference = $dateto - $datefrom; // Difference in seconds
$months_difference = 0;
switch ($interval) {
case 'yyyy': // Number of full years
$years_difference = floor($difference / 31536000);
if (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom), date("j", $datefrom), date("Y", $datefrom)+$years_difference) > $dateto) {
$years_difference--;
}
if (mktime(date("H", $dateto), date("i", $dateto), date("s", $dateto), date("n", $dateto), date("j", $dateto), date("Y", $dateto)-($years_difference+1)) > $datefrom) {
$years_difference++;
}
$datediff = $years_difference;
break;
case "q": // Number of full quarters
$quarters_difference = floor($difference / 8035200);
while (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom)+($quarters_difference*3), date("j", $dateto), date("Y", $datefrom)) < $dateto) {
$months_difference++;
}
$quarters_difference--;
$datediff = $quarters_difference;
break;
case "m": // Number of full months
$months_difference = floor($difference / 2678400);
while (mktime(date("H", $datefrom), date("i", $datefrom), date("s", $datefrom), date("n", $datefrom)+($months_difference), date("j", $dateto), date("Y", $datefrom)) < $dateto) {
$months_difference++;
}
$months_difference--;
$datediff = $months_difference;
break;
case 'y': // Difference between day numbers
$datediff = date("z", $dateto) - date("z", $datefrom);
break;
case "d": // Number of full days
$datediff = floor($difference / 86400);
break;
case "w": // Number of full weekdays
$days_difference = floor($difference / 86400);
$weeks_difference = floor($days_difference / 7); // Complete weeks
$first_day = date("w", $datefrom);
$days_remainder = floor($days_difference % 7);
$odd_days = $first_day + $days_remainder; // Do we have a Saturday or Sunday in the remainder?
if ($odd_days > 7) { // Sunday
$days_remainder--;
}
if ($odd_days > 6) { // Saturday
$days_remainder--;
}
$datediff = ($weeks_difference * 5) + $days_remainder;
break;
case "ww": // Number of full weeks
$datediff = floor($difference / 604800);
break;
case "h": // Number of full hours
$datediff = floor($difference / 3600);
break;
case "n": // Number of full minutes
$datediff = floor($difference / 60);
break;
default: // Number of full seconds (default)
$datediff = $difference;
break;
}
return $datediff;
}
?>
- Jay
0
$formt_start_date = new DateTime($start_date);
$formt_end_date = new DateTime($end_date);
for ($i = $formt_start_date; $i < $formt_end_date; $i->modify('+7 day')) {
$week_frequency[] = new DateTime($i->format("Y-m-d"));
}
- Rajendra Rajput
1
1虽然这段代码可能解决了提问者的问题，但最好添加上下文或解释为什么这可以解决问题，以便未来的访问者可以从您的帖子中学习，并将此知识应用于自己的问题。高质量的答案更有可能被点赞，并为SO作为平台的价值和质量做出贡献。如果有帮助，您还可以添加文档链接。 - SherylHohman
0
This will count a single day as one week and if you select sunday of one week and monday of another. though only 2 days are there but of different weeks therefore 2 weeks will be counted.
Also works for different years
function return_week($datee){
$duedt = explode("-", $datee);
$date = mktime(0, 0, 0, $duedt[1], $duedt[2], $duedt[0]);
$week = (int)date('W', $date);
return $week ;
}
function weeks_between($strtDate, $endDate){
// input dates of format yyyy-mm-dd between which you want to get number of weeks
$start_date = DateTime::createFromFormat("Y-m-d", $strtDate);
$end_date = DateTime::createFromFormat("Y-m-d", $endDate);
$start_year = $start_date->format("Y");
$end_year = $end_date->format("Y");
$start_month = $start_date->format("m");
$end_month = $end_date->format("m");
$week1 = return_week($strtDate);
$week2 = return_week($endDate);
$diff = $end_year - $start_year;;
if($end_month<$start_month && $start_year!=$end_year ){
$week_diff = (52*$diff-$week1)+$week2+1;
return ($start_year==$end_year)?$week_diff:abs($week_diff);
}else{
$week_diff = $week2-$week1+1;
return ($start_year==$end_year)?$week_diff:abs($week_diff)+52*$diff;
}
}
//call to function
weeks_between(""2018-10-21,"2019-10-21");
- Aditi Joshi
-1
PHP两个日期之间周数问题
以下是解决方案（根据https://stackoverflow.com/questions/27560850/php-number-of-week-between-2-dates-issue中提出的一个问题进行了修正），用于查找日期范围之间的周数。
由于下面的解决方案使用日期差异，为了包含起始日期所在的第一周，必须使用floor函数；为了包括结束日期所在的周数，必须使用ceil函数，才能返回正确的周数。
$strtDate = '2014-01-01';
$endDate = '2015-03-17';
$startDateWeekCnt = round(floor( date('d',strtotime($strtDate)) / 7)) ;
// echo $startDateWeekCnt ."\n";
$endDateWeekCnt = round(ceil( date('d',strtotime($endDate)) / 7)) ;
//echo $endDateWeekCnt. "\n";
$datediff = strtotime(date('Y-m',strtotime($endDate))."-01") - strtotime(date('Y-m',strtotime($strtDate))."-01");
$totalnoOfWeek = round(floor($datediff/(60*60*24)) / 7) + $endDateWeekCnt - $startDateWeekCnt ;
echo $totalnoOfWeek ."\n";
- Rakesh Lamp Stack
-1
在PHP中获取两个日期之间的所有周数
getNoOfWeek()将返回带有年份的周编号数组。您可以通过week_text_alter()更改此数组的格式。
function getNoOfWeek($startDate, $endDate){
// convert date in valid format
$startDate = date("Y-m-d", strtotime($startDate));
$endDate = date("Y-m-d", strtotime($endDate));
$yearEndDay = 31;
$weekArr = array();
$startYear = date("Y", strtotime($startDate));
$endYear = date("Y", strtotime($endDate));
if($startYear != $endYear) {
$newStartDate = $startDate;
for($i = $startYear; $i <= $endYear; $i++) {
if($endYear == $i) {
$newEndDate = $endDate;
} else {
$newEndDate = $i."-12-".$yearEndDay;
}
$startWeek = date("W", strtotime($newStartDate));
$endWeek = date("W", strtotime($newEndDate));
if($endWeek == 1){
$endWeek = date("W", strtotime($i."-12-".($yearEndDay-7)));
}
$tempWeekArr = range($startWeek, $endWeek);
array_walk($tempWeekArr, "week_text_alter",
array('pre' => 'Week ', 'post' => " '". substr($i, 2, 2) ));
$weekArr = array_merge($weekArr, $tempWeekArr);
$newStartDate = date("Y-m-d", strtotime($newEndDate . "+1 days"));
}
} else {
$startWeek = date("W", strtotime($startDate));
$endWeek = date("W", strtotime($endDate));
$endWeekMonth = date("m", strtotime($endDate));
if($endWeek == 1 && $endWeekMonth == 12){
$endWeek = date("W", strtotime($endYear."-12-".($yearEndDay-7)));
}
$weekArr = range($startWeek, $endWeek);
array_walk($weekArr, "week_text_alter",
array('pre' => 'Week ', 'post' => " '". substr($startYear, 2, 2)));
}
$weekArr = array_fill_keys($weekArr, 0);
return $weekArr;
}
function week_text_alter(&$item1, $key, $prefix)
{
$item1 = $prefix['pre']. $item1 . $prefix['post'];
}
Output 1 -
$weekArr = getNoOfWeek('2014-01-01', '2014-02-10');
Array
(
[Week 1 '14] => 0
[Week 2 '14] => 0
[Week 3 '14] => 0
[Week 4 '14] => 0
[Week 5 '14] => 0
[Week 6 '14] => 0
[Week 7 '14] => 0
)
Output 2 -
$weekArr = getNoOfWeek('2013-12-01', '2014-02-10');
Array
(
[Week 48 '13] => 0
[Week 49 '13] => 0
[Week 50 '13] => 0
[Week 51 '13] => 0
[Week 52 '13] => 0
[Week 1 '14] => 0
[Week 2 '14] => 0
[Week 3 '14] => 0
[Week 4 '14] => 0
[Week 5 '14] => 0
[Week 6 '14] => 0
[Week 7 '14] => 0
)
希望这能对你有所帮助。
- Gaurav Porwal
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