i write code in php and i get all days between 2date
function getDatesBetween2Dates($startTime, $endTime){
$day = 86400;
$format = 'Y-m-d';
$startTime = strtotime($startTime);
$endTime = strtotime($endTime);
$numDays = round(($endTime - $startTime) / $day) + 1;
$days = array();
for ($i = 0; $i < $numDays; $i++) {
$days[] = date($format, ($startTime + ($i * $day)));
}
return $days;
}
$days = getDatesBetween2Dates($_POST['periodfrom'], $_POST['periodto']);
foreach($days as $key => $value){
echo $daydatey = date('Y-m-d', strtotime($value));
//contain days between 2date from my form
}
$days2 = getDatesBetween2Dates($ppfr, $ppto);
foreach($days2 as $key2 => $value2){
echo $daydatey2 = date('Y-m-d', strtotime($value2));
}
Contain days between 2date from my DB
I just need to make comparison between $days and $days2 to check if any days in($days) exist or equal or like days in ($days2) just that. Please help.
I suggest you take a look at DateTime::diff / date_diff..
http://php.net/manual/en/datetime.diff.php (OOP)
http://php.net/manual/en/function.date-diff.php (Procedural)
This does however require php 5.3+ so ignore this answer if you are below that.
Example:
<?php
$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days');
?>
Related
I am getting the starting and ending dates from a form.
I need to put within an array all the date between the former two, including themselves.
I'm using a normal for loop and, at the same time, printing the result, to verify.
Everything appear to be alright.
But when I print_r the very array, I get only a series of equal dates. Which are all the same: last date + 1.
This is the code:
$date1 = date_create("2013-03-15");
$date2 = date_create("2013-03-22");
$diff = date_diff($date1, $date2);
echo $diff->format("%R%a days");
$diffDays = $diff->days;
echo $diffDays;
$dates = array();
$addDay = $date1;
for ($i = 0; $i < $diffDays; $i++) {
$dates[$i] = $addDay;
date_add($addDay, date_interval_create_from_date_string("1 day"));
echo "array: " . $i . " : " . date_format($dates[$i], 'Y-m-d');
}
print_r($dates);
PHP code demo
<?php
$dates = array();
$datetime1 = new DateTime("2013-03-15");
$datetime2 = new DateTime("2013-03-22");
$interval = $datetime1->diff($datetime2);
$days = (int) $interval->format('%R%a');
$currentTimestamp = $datetime1->getTimestamp();
$dates[] = date("Y-m-d H:i:s", $currentTimestamp);
for ($x = 0; $x < $days; $x++)
{
$currentTimestamp = strtotime("+1 day", $currentTimestamp);
$dates[] = date("Y-m-d H:i:s", $currentTimestamp);
}
print_r($dates);
I would do it that way
$startDate = new \DateTime("2017-03-15");
$endDate = new \DateTime("2017-03-22");
$dates = [];
$stop = false;
$date = $startDate;
while(!$stop){
$dates[] = $date->format('Y-m-d'); // or $dates[] = $date->format('Y-m-d H:i:s')
$date->modify('+1 day');
if($date->getTimestamp() > $endDate->getTimestamp()){
$stop = true;
}
}
print_r($dates);
I have a date like this
$start = strtotime('2010-01-01'); $end = strtotime('2010-01-25');
My question:
How can I calculate or count weekend from $start & $end date range..??
A more modern approach is using php's DateTime class. Below, you get an array with week numbers as keys. I added the counts of weeks and weekend days.
<?php
$begin = new DateTime('2010-01-01');
$end = new DateTime('2010-01-25');
$interval = new DateInterval('P1D');
$daterange = new DatePeriod($begin, $interval, $end);
$weekends = [];
foreach($daterange as $date) {
if (in_array($date->format('N'), [6,7])) {
$weekends[$date->format('W')][] = $date->format('Y-m-d');
}
}
print_r($weekends);
echo 'Number of weeks: ' . count($weekends);
echo 'Number of weekend days: ' . (count($weekends, COUNT_RECURSIVE) - count($weekends));
Note: if you're using PHP 5.3, use array() instead of block arrays [].
May be this code snippet will help:
<?php
//get current month for example
$beginday = date("Y-m-01");
$lastday = date("Y-m-t");
$nr_work_days = getWorkingDays($beginday, $lastday);
echo $nr_work_days;
function getWorkingDays($startDate, $endDate)
{
$begin = strtotime($startDate);
$end = strtotime($endDate);
if ($begin > $end) {
echo "startdate is in the future! <br />";
return 0;
} else {
$no_days = 0;
$weekends = 0;
while ($begin <= $end) {
$no_days++; // no of days in the given interval
$what_day = date("N", $begin);
if ($what_day > 5) { // 6 and 7 are weekend days
$weekends++;
};
$begin += 86400; // +1 day
};
$working_days = $no_days - $weekends;
return $working_days;
}
}
Another solution can be: (Get date range between two dates excluding weekends)
This might help maybe:
$start = strtotime('2010-01-01');
$end = strtotime('2010-01-25');
$differ = $end-$start;
$min = $differ/60;
$hrs = $min/60;
$days = $hrs/24;
$weeks = $days/7;
if(is_int($weeks))
$weeks++;
echo '<pre>';
print_r(ceil($weeks));
echo '</pre>';
Need some logic here:
Need to get day of month date("d")
What I know:
$year = 2013;
$month = 10;
$week_nr_of_month = 3; // from 1 to 6 weeks in month
$day_of_week = 0; // Sunday date("w")
Thanks for logic
Result must be: 13 October
This was fun to figure out.
<?php
$year = 2013;
$month = 10;
$week_nr_of_month = 3; // from 1 to 6 weeks in month
$day_of_week = 0; // Sunday date("w")
$start = new DateTime();
$start->setDate($year, $month, '1');
$end = clone $start;
$end->modify('last day of this month');
$interval = new DateInterval('P1D');
$period = new DatePeriod($start, $interval, $end);
foreach ($period as $date) {
if ($date->format('w') == $day_of_week) {
$wom = week_of_month($date->format('Y-m-d'));
if ($week_nr_of_month == $wom) {
echo $date->format('Y-m-d');
}
}
}
function week_of_month($date) {
$dt = new DateTime($date);
$bg = clone $dt;
$bg->modify('first day of this month');
$day_of_first = $bg->format('N');
$day_of_month = $dt->format('j');
return floor(($day_of_first + $day_of_month - 1) / 7) + 1;
}
See it in action
Used this answer for inspiration to get the week number for the date.
strtotime may be of help. Can't test right now, but...
$ordinal = array("first","second","third","fourth","fifth");
$weekdays = array("monday","tuesday","wednesday","thursday","friday","saturday","sunday");
$timestamp = strtotime($year."-".$month." ".$ordinal[$week_nr_of_month]." ".$weekdays[$day_of_week]);
This question already has answers here:
PHP: Return all dates between two dates in an array [duplicate]
(26 answers)
Closed 4 years ago.
Is there an easy way to get a list of days between two dates in PHP?
I would like to have something like this in the end:
(pseudocode)
date1 = 29/08/2013
date2 = 03/09/2013
resultArray = functionReturnDates(date1, date2);
and the resulting array would contain:
resultArray[0] = 29/08/2013
resultArray[1] = 30/08/2013
resultArray[2] = 31/08/2013
resultArray[3] = 01/09/2013
resultArray[4] = 02/09/2013
resultArray[5] = 03/09/2013
for example.
$date1 = '29/08/2013';
$date2 = '03/09/2013';
function returnDates($fromdate, $todate) {
$fromdate = \DateTime::createFromFormat('d/m/Y', $fromdate);
$todate = \DateTime::createFromFormat('d/m/Y', $todate);
return new \DatePeriod(
$fromdate,
new \DateInterval('P1D'),
$todate->modify('+1 day')
);
}
$datePeriod = returnDates($date1, $date2);
foreach($datePeriod as $date) {
echo $date->format('d/m/Y'), PHP_EOL;
}
function DatePeriod_start_end($begin,$end){
$begin = new DateTime($begin);
$end = new DateTime($end.' +1 day');
$daterange = new DatePeriod($begin, new DateInterval('P1D'), $end);
foreach($daterange as $date){
$dates[] = $date->format("Y-m-d");
}
return $dates;
}
dunno if this is at all practical, but it works pretty straight-forward
$end = '2013-08-29';
$start = '2013-08-25';
$datediff = strtotime($end) - strtotime($start);
$datediff = floor($datediff/(60*60*24));
for($i = 0; $i < $datediff + 1; $i++){
echo date("Y-m-d", strtotime($start . ' + ' . $i . 'day')) . "<br>";
}
Try this:
function daysBetween($start, $end)
$dates = array();
while($start <= $end)
{
array_push(
$dates,
date(
'dS M Y',
$start
)
);
$start += 86400;
}
return $dates;
}
$start = strtotime('2009-10-20');
$end = strtotime('2009-10-25');
var_dump(daysBetween($start,$end));
$datearray = array();
$date = $date1;
$days = ceil(abs($date2 - $date1) / 86400) + 1;//no of days
for($i = 1;$i <= $days; $i++){
array_push($datearray,$date);
$date = $date+86400;
}
foreach($datearray as $days){
echo date('Y-m-d, $days);
}
if
$_POST['SelectedDate1'] = 2013/08/05
and
$_POST['SelectedDate2'] = 2013/08/07
How can I set a variable which gives me back the number of days (2 in this case) to then echo it as result
I'm looking for a solution that can cover any calendar combination.
Is there any global function in php.
I think, in the following Documentation on PHP.net is exactly what you're trying to do.
http://nl3.php.net/manual/en/datetime.diff.php
<?php
$datetime1 = new DateTime('2009-10-11');
$datetime2 = new DateTime('2009-10-13');
$interval = $datetime1->diff($datetime2);
echo $interval->format('%R%a days');
?>
In your case:
<?php
$first = new DateTime($_POST['SelectedDate1']);
$second = new DateTime($_POST['SelectedDate2']);
$passed = $first->diff($second);
var_dump($passed->format('%R%a days'));
For more formats, next to %R%a, see: http://nl3.php.net/manual/en/function.date.php
<?php
$days = (strtotime($_POST['SelectedDate2']) - strtotime($_POST['SelectedDate1'])) / 86400;
example:
<?php
$_POST['SelectedDate1'] = '2013/08/05' ;
$_POST['SelectedDate2'] = '2013/08/07' ;
$days = (strtotime($_POST['SelectedDate2']) - strtotime($_POST['SelectedDate1'])) / 86400;
var_export($days);
// output: 2
I use this function that I've found on this forum but I don't remember where :
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;
}
it returns a range of date as an array, then you just have to get the count
DateTime class is created for this:
$date1 = new DateTime('2013/08/05');
$date2 = new DateTime('2013/08/07');
$diff = $date1->diff($date2);
echo $diff->days;