预期输入:
getDatesFromRange( '2010-10-01', '2010-10-05' );
预期输出:
Array( '2010-10-01', '2010-10-02', '2010-10-03', '2010-10-04', '2010-10-05' )
答案 0 :(得分:365)
您还可以查看DatePeriod课程:
$period = new DatePeriod(
new DateTime('2010-10-01'),
new DateInterval('P1D'),
new DateTime('2010-10-05')
);
哪个应该为你提供一个包含DateTime对象的数组。
迭代
foreach ($period as $key => $value) {
//$value->format('Y-m-d')
}
答案 1 :(得分:148)
function createDateRangeArray($strDateFrom,$strDateTo)
{
// takes two dates formatted as YYYY-MM-DD and creates an
// inclusive array of the dates between the from and to dates.
// could test validity of dates here but I'm already doing
// that in the main script
$aryRange=array();
$iDateFrom=mktime(1,0,0,substr($strDateFrom,5,2), substr($strDateFrom,8,2),substr($strDateFrom,0,4));
$iDateTo=mktime(1,0,0,substr($strDateTo,5,2), substr($strDateTo,8,2),substr($strDateTo,0,4));
if ($iDateTo>=$iDateFrom)
{
array_push($aryRange,date('Y-m-d',$iDateFrom)); // first entry
while ($iDateFrom<$iDateTo)
{
$iDateFrom+=86400; // add 24 hours
array_push($aryRange,date('Y-m-d',$iDateFrom));
}
}
return $aryRange;
}
来源:http://boonedocks.net/mike/archives/137-Creating-a-Date-Range-Array-with-PHP.html
答案 2 :(得分:126)
这非常灵活。
/**
* Creating date collection between two dates
*
* <code>
* <?php
* # Example 1
* date_range("2014-01-01", "2014-01-20", "+1 day", "m/d/Y");
*
* # Example 2. you can use even time
* date_range("01:00:00", "23:00:00", "+1 hour", "H:i:s");
* </code>
*
* @author Ali OYGUR <alioygur@gmail.com>
* @param string since any date, time or datetime format
* @param string until any date, time or datetime format
* @param string step
* @param string date of output format
* @return array
*/
function date_range($first, $last, $step = '+1 day', $output_format = 'd/m/Y' ) {
$dates = array();
$current = strtotime($first);
$last = strtotime($last);
while( $current <= $last ) {
$dates[] = date($output_format, $current);
$current = strtotime($step, $current);
}
return $dates;
}
答案 3 :(得分:32)
请注意,ViNce提供的答案不包括该期间的结束日期。
如果您使用的是PHP 5.3+,最好的办法是使用这样的函数:
/**
* Generate an array of string dates between 2 dates
*
* @param string $start Start date
* @param string $end End date
* @param string $format Output format (Default: Y-m-d)
*
* @return array
*/
function getDatesFromRange($start, $end, $format = 'Y-m-d') {
$array = array();
$interval = new DateInterval('P1D');
$realEnd = new DateTime($end);
$realEnd->add($interval);
$period = new DatePeriod(new DateTime($start), $interval, $realEnd);
foreach($period as $date) {
$array[] = $date->format($format);
}
return $array;
}
然后,您可以按预期调用该函数:
getDatesFromRange('2010-10-01', '2010-10-05');
关于DatePeriod
类的注意事项:您可以使用DatePeriod的第4个参数来排除开始日期(DatePeriod::EXCLUDE_START_DATE
),但此时您不能包含结尾日期。
答案 4 :(得分:21)
简单但又喜欢魅力:
$period = new DatePeriod(new DateTime('2015-01-01'), new DateInterval('P1D'), new DateTime('2015-01-15 +1 day'));
foreach ($period as $date) {
$dates[] = $date->format("Y-m-d");
}
//ONLY SHOWING
echo '<pre>';
var_dump($dates);
echo '</pre>';
exit();
答案 5 :(得分:12)
这很简短,很好用,应该可以在PHP4 +中使用。
function getDatesFromRange($start, $end){
$dates = array($start);
while(end($dates) < $end){
$dates[] = date('Y-m-d', strtotime(end($dates).' +1 day'));
}
return $dates;
}
答案 6 :(得分:12)
查看this
function GetDays($sStartDate, $sEndDate){
// Firstly, format the provided dates.
// This function works best with YYYY-MM-DD
// but other date formats will work thanks
// to strtotime().
$sStartDate = gmdate("Y-m-d", strtotime($sStartDate));
$sEndDate = gmdate("Y-m-d", strtotime($sEndDate));
// Start the variable off with the start date
$aDays[] = $sStartDate;
// Set a 'temp' variable, sCurrentDate, with
// the start date - before beginning the loop
$sCurrentDate = $sStartDate;
// While the current date is less than the end date
while($sCurrentDate < $sEndDate){
// Add a day to the current date
$sCurrentDate = gmdate("Y-m-d", strtotime("+1 day", strtotime($sCurrentDate)));
// Add this new day to the aDays array
$aDays[] = $sCurrentDate;
}
// Once the loop has finished, return the
// array of days.
return $aDays;
}
使用
GetDays('2007-01-01', '2007-01-31');
答案 7 :(得分:8)
完成此任务的许多方法,但最终都取决于您使用的PHP版本。以下是所有解决方案的摘要:
获取PHP版本:
echo phpinfo();
PHP 5.3 +
$period = new DatePeriod(
new DateTime('2010-10-01'),
new DateInterval('P1D'),
new DateTime('2010-10-05')
);
PHP 4 +
/**
* creating between two date
* @param string since
* @param string until
* @param string step
* @param string date format
* @return array
* @author Ali OYGUR <alioygur@gmail.com>
*/
function dateRange($first, $last, $step = '+1 day', $format = 'd/m/Y' ) {
$dates = array();
$current = strtotime($first);
$last = strtotime($last);
while( $current <= $last ) {
$dates[] = date($format, $current);
$current = strtotime($step, $current);
}
return $dates;
}
PHP&lt; 4 强>
你应该升级:)
答案 8 :(得分:7)
功能短。 PHP 5.3及以上。可以采取strtotime可以理解的任何日期格式的可选第三个参数。如果结束&lt;自动反转方向。启动。
function getDatesFromRange($start, $end, $format='Y-m-d') {
return array_map(function($timestamp) use($format) {
return date($format, $timestamp);
},
range(strtotime($start) + ($start < $end ? 4000 : 8000), strtotime($end) + ($start < $end ? 8000 : 4000), 86400));
}
测试:
date_default_timezone_set('Europe/Berlin');
print_r(getDatesFromRange( '2016-7-28','2016-8-2' ));
print_r(getDatesFromRange( '2016-8-2','2016-7-28' ));
print_r(getDatesFromRange( '2016-10-28','2016-11-2' ));
print_r(getDatesFromRange( '2016-11-2','2016-10-28' ));
print_r(getDatesFromRange( '2016-4-2','2016-3-25' ));
print_r(getDatesFromRange( '2016-3-25','2016-4-2' ));
print_r(getDatesFromRange( '2016-8-2','2016-7-25' ));
print_r(getDatesFromRange( '2016-7-25','2016-8-2' ));
输出:
Array ( [0] => 2016-07-28 [1] => 2016-07-29 [2] => 2016-07-30 [3] => 2016-07-31 [4] => 2016-08-01 [5] => 2016-08-02 )
Array ( [0] => 2016-08-02 [1] => 2016-08-01 [2] => 2016-07-31 [3] => 2016-07-30 [4] => 2016-07-29 [5] => 2016-07-28 )
Array ( [0] => 2016-10-28 [1] => 2016-10-29 [2] => 2016-10-30 [3] => 2016-10-31 [4] => 2016-11-01 [5] => 2016-11-02 )
Array ( [0] => 2016-11-02 [1] => 2016-11-01 [2] => 2016-10-31 [3] => 2016-10-30 [4] => 2016-10-29 [5] => 2016-10-28 )
Array ( [0] => 2016-04-02 [1] => 2016-04-01 [2] => 2016-03-31 [3] => 2016-03-30 [4] => 2016-03-29 [5] => 2016-03-28 [6] => 2016-03-27 [7] => 2016-03-26 [8] => 2016-03-25 )
Array ( [0] => 2016-03-25 [1] => 2016-03-26 [2] => 2016-03-27 [3] => 2016-03-28 [4] => 2016-03-29 [5] => 2016-03-30 [6] => 2016-03-31 [7] => 2016-04-01 [8] => 2016-04-02 )
Array ( [0] => 2016-08-02 [1] => 2016-08-01 [2] => 2016-07-31 [3] => 2016-07-30 [4] => 2016-07-29 [5] => 2016-07-28 [6] => 2016-07-27 [7] => 2016-07-26 [8] => 2016-07-25 )
Array ( [0] => 2016-07-25 [1] => 2016-07-26 [2] => 2016-07-27 [3] => 2016-07-28 [4] => 2016-07-29 [5] => 2016-07-30 [6] => 2016-07-31 [7] => 2016-08-01 [8] => 2016-08-02 )
答案 9 :(得分:6)
您必须添加 $ end->修改(“ +1天”) 以包括间隔的最后一天,例如一月如果不使用Modify()方法,则将有31天而不是30天。 此版本的代码将包含间隔的最后一天:
$begin = new DateTime( '2018-08-01' );
$end = new DateTime( '2018-08-31' );
$end = $end->modify( '+1 day' );
$interval = new DateInterval('P1D');
$daterange = new DatePeriod($begin, $interval ,$end);
foreach($daterange as $date){
echo $date->format("Ymd") . "<br>";
}
答案 10 :(得分:5)
这是一个函数,它会返回两个方向的日期范围,并且它适用于PHP >=5.2.2:
function createRange($start, $end, $format = 'Y-m-d') {
$start = new DateTime($start);
$end = new DateTime($end);
$invert = $start > $end;
$dates = array();
$dates[] = $start->format($format);
while ($start != $end) {
$start->modify(($invert ? '-' : '+') . '1 day');
$dates[] = $start->format($format);
}
return $dates;
}
使用示例:
print_r(createRange('2010-10-01', '2010-10-05'));
/*Array
(
[0] => 2010-10-01
[1] => 2010-10-02
[2] => 2010-10-03
[3] => 2010-10-04
[4] => 2010-10-05
)*/
print_r(createRange('2010-10-05', '2010-10-01', 'j M Y'));
/*Array
(
[0] => 5 Oct 2010
[1] => 4 Oct 2010
[2] => 3 Oct 2010
[3] => 2 Oct 2010
[4] => 1 Oct 2010
)*/
答案 11 :(得分:4)
使用DateTime对象的PHP 5.2解决方案。但是startDate 必须在endDate之前。
function createRange($startDate, $endDate) {
$tmpDate = new DateTime($startDate);
$tmpEndDate = new DateTime($endDate);
$outArray = array();
do {
$outArray[] = $tmpDate->format('Y-m-d');
} while ($tmpDate->modify('+1 day') <= $tmpEndDate);
return $outArray;
}
使用:
$dates = createRange('2010-10-01', '2010-10-05');
$ date包含:
Array( '2010-10-01', '2010-10-02', '2010-10-03', '2010-10-04', '2010-10-05' )
答案 12 :(得分:4)
// Specify the start date. This date can be any English textual format
$date_from = "2018-02-03";
$date_from = strtotime($date_from); // Convert date to a UNIX timestamp
// Specify the end date. This date can be any English textual format
$date_to = "2018-09-10";
$date_to = strtotime($date_to); // Convert date to a UNIX timestamp
// Loop from the start date to end date and output all dates inbetween
for ($i=$date_from; $i<=$date_to; $i+=86400) {
echo date("Y-m-d", $i).'<br />';
}
答案 13 :(得分:3)
以下是使用Carbon https://github.com/briannesbitt/Carbon执行此操作的方法:
public function buildDateRangeArray($first, $last)
{
while ($first <= $last) {
$dates[] = $first->toDateString();
$first->addDay();
}
return $dates;
}
当然,这可以调整为不使用碳。传递给函数的$ first和$ last参数是Carbon实例。
答案 14 :(得分:3)
<?
print_r(getDatesFromRange( '2010-10-01', '2010-10-05' ));
function getDatesFromRange($startDate, $endDate)
{
$return = array($startDate);
$start = $startDate;
$i=1;
if (strtotime($startDate) < strtotime($endDate))
{
while (strtotime($start) < strtotime($endDate))
{
$start = date('Y-m-d', strtotime($startDate.'+'.$i.' days'));
$return[] = $start;
$i++;
}
}
return $return;
}
答案 15 :(得分:3)
function createDateRangeArray($start, $end) {
// Modified by JJ Geewax
$range = array();
if (is_string($start) === true) $start = strtotime($start);
if (is_string($end) === true ) $end = strtotime($end);
if ($start > $end) return createDateRangeArray($end, $start);
do {
$range[] = date('Y-m-d', $start);
$start = strtotime("+ 1 day", $start);
}
while($start < $end);
return $range;
}
来源:http://boonedocks.net/mike/archives/137-Creating-a-Date-Range-Array-with-PHP.html
答案 16 :(得分:2)
// will return dates array
function returnBetweenDates( $startDate, $endDate ){
$startStamp = strtotime( $startDate );
$endStamp = strtotime( $endDate );
if( $endStamp > $startStamp ){
while( $endStamp >= $startStamp ){
$dateArr[] = date( 'Y-m-d', $startStamp );
$startStamp = strtotime( ' +1 day ', $startStamp );
}
return $dateArr;
}else{
return $startDate;
}
}
returnBetweenDates( '2014-09-16', '2014-09-26' );
// print_r( returnBetweenDates( '2014-09-16', '2014-09-26' ) );
它会返回如下所示的数组:
Array
(
[0] => 2014-09-16
[1] => 2014-09-17
[2] => 2014-09-18
[3] => 2014-09-19
[4] => 2014-09-20
[5] => 2014-09-21
[6] => 2014-09-22
[7] => 2014-09-23
[8] => 2014-09-24
[9] => 2014-09-25
[10] => 2014-09-26
)
答案 17 :(得分:2)
$report_starting_date=date('2014-09-16');
$report_ending_date=date('2014-09-26');
$report_starting_date1=date('Y-m-d',strtotime($report_starting_date.'-1 day'));
while (strtotime($report_starting_date1)<strtotime($report_ending_date))
{
$report_starting_date1=date('Y-m-d',strtotime($report_starting_date1.'+1 day'));
$dates[]=$report_starting_date1;
}
print_r($dates);
// dates ('2014-09-16', '2014-09-26')
//print result Array
(
[0] => 2014-09-16
[1] => 2014-09-17
[2] => 2014-09-18
[3] => 2014-09-19
[4] => 2014-09-20
[5] => 2014-09-21
[6] => 2014-09-22
[7] => 2014-09-23
[8] => 2014-09-24
[9] => 2014-09-25
[10] => 2014-09-26
)
答案 18 :(得分:2)
我认为这是最简短的答案
根据需要编辑代码
0%{?suse_version} < 01315
答案 19 :(得分:1)
//To find dates between two dates as an array
$dates = new DatePeriod(
new DateTime('2018-12-01'), new DateInterval('P1D'), new DateTime('2018-12-25')
);
//To display all dates
foreach ($dates as $key => $value) {
echo $value->format('Y-m-d')
}
答案 20 :(得分:1)
这是另一种解决方案。请检查一下。
$first = '10/30/2017'; //starting date
$last= '10/11/2017'; //ending date
$first_time_arr=explode('/',$first);
$last_time_arr=explode('/',$last);
//create timestamp of starting date
$start_timestamp=mktime(0,0,0, $first_time_arr[0], $first_time_arr[1],$first_time_arr[2]);
//create timestamp of ending date
$end_timestamp=mktime(0,0,0, $last_time_arr[0], $last_time_arr[1],$last_time_arr[2]);
$date_arr=array();
for($i=$start_timestamp;$i<=$end_timestamp;$i=$i+86400){
$date_arr[]=date("Y-m-d",$i); //this will save all dates in array
}
答案 21 :(得分:0)
我当天的PHP发现是array_push()
返回数组中新的元素数。
我设法检查结束日期匹配,增加$ x,并在空循环的两部分条件语句中推送新元素。
function getDatesFromRange($a,$b,$x=0,$dates=[]){
while(end($dates)!=$b && $x=array_push($dates,date("Y-m-d",strtotime("$a +$x day"))));
return $dates;
}
var_export(getDatesFromRange('2010-10-01','2010-10-05'));
在这个页面上我最相似的功能是drolex's(在我写下我之后,如果你相信我,我实际上并没有找到它)。我在大小日期范围内进行了一些速度测试,它们似乎经常相互击败 - 所以我称它们是相同的表演者。以下是其他一些比较:
date()
,strtotime()
和两个数组函数。$x
。$date
数组中,所以我可以在函数参数中声明它并备用该行(同样使用$x
)。**只是几个重要的注意事项:
1-日期字符串必须在被送入函数之前进行验证。
2-上述功能只能处理前进移动日期范围。
如果您想要向后移动日期范围,只需在函数调用中反转日期顺序,然后在$x=
之后添加一个减号。(非常漂亮,是吗?)
function getDatesFromRange($a,$b,$x=0,$dates=[]){
while(end($dates)!=$b && $x=-array_push($dates,date("Y-m-d",strtotime("$a +$x day"))));
return $dates;
}
var_export(getDatesFromRange('2010-10-05','2010-10-01'));
还有一个扩展/考虑......
想象一下,你有一个多文化(或草率)的用户群,你的功能必须能够以不同的有效格式接收开始和结束日期,你需要能够以任何有效的格式输出数组吗?通过微调,我为此提供了解决方案。
通过&#34;有效&#34;我的意思是YYYY-MM-DD
,MM/DD/YYY
和DD-MM-YYYY
,这些是全球范围内广受欢迎的标准,如果需要其他格式,则可用性将归结为strtotime
&#39;理解它。
这是Demo。
代码:
function getDatesFromRange($a,$b,$format='Y-m-d',$dates=[],$x=0){
while(date($format,strtotime(end($dates)))!=date($format,strtotime($b)) && $x=array_push($dates,date($format,strtotime("$a +$x day"))));
return $dates;
}
$formats=array("Computer"=>'Y-m-d',"American"=>'m/d/Y','Non-American'=>'d-m-Y');
$start='15-02-2017'; // Non-American formatted start date
$end='2017-02-27'; // Computer formatted start date
foreach($formats as $label=>$format){
echo "<br>$label<br>";
var_export(getDatesFromRange($start,$end,$format));
echo "<br>";
}
输出
Computer
array ( 0 => '2017-02-15', 1 => '2017-02-16', 2 => '2017-02-17', 3 => '2017-02-18',
4 => '2017-02-19', 5 => '2017-02-20', 6 => '2017-02-21', 7 => '2017-02-22',
8 => '2017-02-23', 9 => '2017-02-24', 10 => '2017-02-25', 11 => '2017-02-26',
12 => '2017-02-27', )
American
array ( 0 => '02/15/2017', 1 => '02/16/2017', 2 => '02/17/2017', 3 => '02/18/2017',
4 => '02/19/2017', 5 => '02/20/2017', 6 => '02/21/2017', 7 => '02/22/2017',
8 => '02/23/2017', 9 => '02/24/2017', 10 => '02/25/2017', 11 => '02/26/2017',
12 => '02/27/2017', )
Non-American
array ( 0 => '15-02-2017', 1 => '16-02-2017', 2 => '17-02-2017', 3 => '18-02-2017',
4 => '19-02-2017', 5 => '20-02-2017', 6 => '21-02-2017', 7 => '22-02-2017',
8 => '23-02-2017', 9 => '24-02-2017', 10 => '25-02-2017', 11 => '26-02-2017',
12 => '27-02-2017', )
现在有些人因为一些错误的行为而不能100%信任strtotime()。我想我已经读到,当试图从闰日跳一个月时它会肮脏。但是,除非有人能够重现它以证明我是错的,否则当你只增加一天时,strtotime()永远不会让你失望。
答案 22 :(得分:0)
function datesbetween ($date1,$date2)
{
$dates= array();
for ($i = $date1
; $i<= $date1
; $i=date_add($i, date_interval_create_from_date_string('1 days')) )
{
$dates[] = clone $i;
}
return $dates;
}
答案 23 :(得分:0)
public static function countDays($date1,$date2)
{
$date1 = strtotime($date1); // or your date as well
$date2 = strtotime($date2);
$datediff = $date1 - $date2;
return floor($datediff/(60*60*24));
}
public static function dateRange($date1,$date2)
{
$count = static::countDays($date1,$date2) + 1;
$dates = array();
for($i=0;$i<$count;$i++)
{
$dates[] = date("Y-m-d",strtotime($date2.'+'.$i.' days'));
}
return $dates;
}
答案 24 :(得分:-2)
为了使Mostafa的答案完整,这绝对是最简单,最有效的方法:
function getDatesFromRange($start_date, $end_date, $date_format = 'Y-m-d')
{
$dates_array = array();
for ($x = strtotime($start_date); $x <= strtotime($end_date); $x += 86400) {
array_push($dates_array, date($date_format, $x));
}
return $dates_array;
}
// see the dates in the array
print_r( getDatesFromRange('2017-02-09', '2017-02-19') );
如果在调用函数时添加第三个参数,甚至可以更改默认输出日期格式,否则它将使用默认格式设置为“Y-m-d”。
我希望它有所帮助:)
答案 25 :(得分:-3)
function getWeekdayDatesFrom($format, $start_date_epoch, $end_date_epoch, $range) {
$dates_arr = array();
if( ! $range) {
$range = round(abs($start_date_epoch-$end_date_epoch)/86400) + 1;
} else {
$range = $range + 1; //end date inclusive
}
$current_date_epoch = $start_date_epoch;
for($i = 1; $i <= $range; $i+1) {
$d = date('N', $current_date_epoch);
if($d <= 5) { // not sat or sun
$dates_arr[] = "'".date($format, $current_date_epoch)."'";
}
$next_day_epoch = strtotime('+'.$i.'day', $start_date_epoch);
$i++;
$current_date_epoch = $next_day_epoch;
}
return $dates_arr;
}
答案 26 :(得分:-4)
$arr = range(strtotime("2013-12-01"),strtotime("2013-12-31"), "86400");
array_walk_recursive($arr, function(&$element) { $element = date("Y-m-d", $element); });
print_r ($arr);