Related
I can add any number of months to a date:
strtotime("+ 1 months", '2017-01-31') // result: 2017-03-03
But I want to do this without going to the next month.
in this case I want the result 2017-02-28, that is, the last day of the month before the target month.
There seems to be a lot of overcomplicating in these answers. You want the last day of the month before your target month, which is also always going to be 1 day before the first day of the target month. This can be expressed quite simply:
$months = 1;
$relativeto = '2017-01-31';
echo date(
'Y-m-d',
strtotime(
'-1 day',
strtotime(
date(
'Y-m-01',
strtotime("+ $months months", strtotime($relativeto))
)
)
)
);
Try using the DateTime object like so:
$dateTime = new \DateTime('2017-01-31');
$dateTime->add(new \DateInterval('P1M'));
$result = $dateTime->format('Y-m-d');
Use PHP's DateTime to accomplish this, specifically the modify() method.
$date = new DateTime('2006-12-12');
$date->modify('+1 month');
echo $date->format('Y-m-d');
If the idea here is not to allow the date to overflow into the next month, which PHP does, then you'll have to impose that constraint in your logic.
One approach is to check the modified month against the given month before returning the updated date.
function nextMonth(DateTimeImmutable $date) {
$nextMonth = $date->add(new DateInterval("P1M"));
$diff = $nextMonth->diff($date);
if ($diff->d > 0) {
$nextMonth = $nextMonth->sub(new DateInterval("P{$diff->d}D"));
}
return $nextMonth;
}
$date = new DateTimeImmutable("2017-01-31");
$nextMonth = nextMonth($date);
echo "{$date->format('Y-m-d')} - {$nextMonth->format('Y-m-d')}\n";
//2017-01-31 - 2017-02-28
$date = new DateTimeImmutable("2017-10-31");
$nextMonth = nextMonth($date);
echo "{$date->format('Y-m-d')} - {$nextMonth->format('Y-m-d')}\n";
//2017-10-31 - 2017-11-30
The nextMonth() function in the example above, imposes the constraint of not overflowing into the next month. Note that what PHP actually does is try to find the corresponding day, in the following consecutive number of months added, not just add a given number of months to the existing date. I simply undo this last part by subtracting any additional days beyond the 1 month interval in the function above.
So for example, strtotime("+1 month") for the date "2017-01-31", tells PHP find the 31st day that is +1 month from "2017-01-31". But of course, there are only 28 days in February, so PHP goes into March and counts up 3 more days to compensate.
In other words, it's not just add +1 month to this date, but add +1 month to this date and arrive at the same day of the month as is in the given date. Which is where the overflow happens.
Update
Since you've now made it clear that you actually want the last day of the month (not necessarily the same day of the next month without the overflow provision) you should instead just explicitly set the day of the month.
function nextMonth(DateTimeImmutable $date) {
$nextMonth = $date->setDate($date->format('Y'), $date->format('n') + 1, 1);
$nextMonth = $date->setDate($nextMonth->format('Y'), $nextMonth->format('n'), $nextMonth->format('t'));
return $nextMonth;
}
$date = new DateTimeImmutable("2017-02-28");
$nextMonth = nextMonth($date);
echo "{$date->format('Y-m-d')} - {$nextMonth->format('Y-m-d')}\n";
// 2017-02-28 - 2017-03-31
I am currently struggling with the following: I need to get the Date like "Sun, 29.05.2016 18:00" but all i have is the number of the week from the beginning of the year, the number of the day of week and the hour of the day.
Week: 21
Day: 7
Hour: 18
Now my question is how do i get the actual date and time with php's date functions or own code?
You are looking for DateTime::setISODate
$week_no = 21;
$day =7;
$hour = 18;
$d= new DateTime();
$d->setISODate(date('Y'),$week_no, $day);
$d->setTime ($hour,0);
echo $d->format('D, d.m.Y H:i');
Pretty straightforward with DateTime objects:
$week = 21;
$day = 7;
$hour = 18;
$dateFormat = sprintf('%d-W%02d-%d', (new DateTime())->format('Y'), $week, $day);
$x = new DateTime($dateFormat);
$x->setTime($hour, 0);
echo $x->format('Y-m-d H:i:s');
(assuming the current year)
Try with setISODate and DateTime. Online Check
Setting the 2016 from your desire output and use the hour directly to the format.
$gendate = new DateTime();
$gendate->setISODate(2016, 21, 7); //year , week num , day
echo $gendate->format('D, d.m.Y 18:00'); //Sun, 29.05.2016 18:00
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
PHP DateTime::modify adding and subtracting months
I have a starting date (i.e. 2011-01-30) and want to add 1 month.
The problem is in defining what a month is. So if I use the following code:
$d1 = DateTime::createFromFormat('Y-m-d H:i:s', '2011-01-30 15:57:57');
$d1->add(new DateInterval('P1M'));
echo $d1->format('Y-m-d H:i:s');
I get the following result: 2011-03-02 15:57:57
The problem is, that I need it to use the following rules:
If I add 1 month it will just add 1 on the month part and leave the day part
(2011-01-15 will become 2011-02-15)
If the day is not existing in the month we will end, we take the last existing day of it
(2011-01-30 will become 2011-02-28)
Is there a common function in php that can do this or do I have to code it by myself?
Maybe I'm just missing a parameter or something!?
It seems that there is no ready function for it, so I wrote it by myself.
This should solve my problem. Thanks anyway for your answers and comments.
If you will find some errors, please provide in the comments.
This function calculates the month and the year I will end in after adding some month. Then it checks if the date is right. If not, we have the case that the day is not in the target month, so we take the last day in the month instead.
Tested with PHP 5.3.10.
<?php
$monthToAdd = 1;
$d1 = DateTime::createFromFormat('Y-m-d H:i:s', '2011-01-30 15:57:57');
$year = $d1->format('Y');
$month = $d1->format('n');
$day = $d1->format('d');
$year += floor($monthToAdd/12);
$monthToAdd = $monthToAdd%12;
$month += $monthToAdd;
if($month > 12) {
$year ++;
$month = $month % 12;
if($month === 0)
$month = 12;
}
if(!checkdate($month, $day, $year)) {
$d2 = DateTime::createFromFormat('Y-n-j', $year.'-'.$month.'-1');
$d2->modify('last day of');
}else {
$d2 = DateTime::createFromFormat('Y-n-d', $year.'-'.$month.'-'.$day);
}
$d2->setTime($d1->format('H'), $d1->format('i'), $d1->format('s'));
echo $d2->format('Y-m-d H:i:s');
You have several alternatives besides DateInterval.
Here are examples that use strtotime():
http://www.brightcherry.co.uk/scribbles/php-adding-and-subtracting-dates/
// Subtracting days from a date
$date = "1998-08-14";
$newdate = strtotime ( '-3 day' , strtotime ( $date ) ) ;
$newdate = date ( 'Y-m-j' , $newdate );
echo $newdate;
...
// Subtracting months from a date
$date = "1998-08-14";
$newdate = strtotime ( '-3 month' , strtotime ( $date ) ) ;
$newdate = date ( 'Y-m-j' , $newdate );
echo $newdate;
Here's are some links for DateInterval:
http://php.net/manual/en/dateinterval.construct.php
What can go wrong when adding months with a DateInterval and DateTime::add?
Q: Exactly how do you define a "month"?
Def#1): a "month" is a "month" - regardless of #/days
Def#2): a "month" is 30 days (for example)
Def#3): a "month" is the #/days between the 1st Monday of
subsequent months
etc. etc
Q: What exactly is your "definition"?
OK - as far as I can tell, you still haven't clearly defined what you mean by "month".
But here's one more function that might help:
http://php.net/manual/en/function.getdate.php
/* Function to find the first and last day of the month from the given date.
*
* Author Binu v Pillai binupillai2003#yahoo.com
* #Param String yyyy-mm-dd
*
*/
function findFirstAndLastDay($anyDate)
{
//$anyDate = '2009-08-25'; // date format should be yyyy-mm-dd
list($yr,$mn,$dt) = split('-',$anyDate); // separate year, month and date
$timeStamp = mktime(0,0,0,$mn,1,$yr); //Create time stamp of the first day from the give date.
$firstDay = date('D',$timeStamp); //get first day of the given month
list($y,$m,$t) = split('-',date('Y-m-t',$timeStamp)); //Find the last date of the month and separating it
$lastDayTimeStamp = mktime(0,0,0,$m,$t,$y);//create time stamp of the last date of the give month
$lastDay = date('D',$lastDayTimeStamp);// Find last day of the month
$arrDay = array("$firstDay","$lastDay"); // return the result in an array format.
return $arrDay;
}
Let's say I have a date in the following format: 2010-12-11 (year-mon-day)
With PHP, I want to increment the date by one month, and I want the year to be automatically incremented, if necessary (i.e. incrementing from December 2012 to January 2013).
Regards.
$time = strtotime("2010.12.11");
$final = date("Y-m-d", strtotime("+1 month", $time));
// Finally you will have the date you're looking for.
I needed similar functionality, except for a monthly cycle (plus months, minus 1 day). After searching S.O. for a while, I was able to craft this plug-n-play solution:
function add_months($months, DateTime $dateObject)
{
$next = new DateTime($dateObject->format('Y-m-d'));
$next->modify('last day of +'.$months.' month');
if($dateObject->format('d') > $next->format('d')) {
return $dateObject->diff($next);
} else {
return new DateInterval('P'.$months.'M');
}
}
function endCycle($d1, $months)
{
$date = new DateTime($d1);
// call second function to add the months
$newDate = $date->add(add_months($months, $date));
// goes back 1 day from date, remove if you want same day of month
$newDate->sub(new DateInterval('P1D'));
//formats final date to Y-m-d form
$dateReturned = $newDate->format('Y-m-d');
return $dateReturned;
}
Example:
$startDate = '2014-06-03'; // select date in Y-m-d format
$nMonths = 1; // choose how many months you want to move ahead
$final = endCycle($startDate, $nMonths); // output: 2014-07-02
Use DateTime::add.
$start = new DateTime("2010-12-11", new DateTimeZone("UTC"));
$month_later = clone $start;
$month_later->add(new DateInterval("P1M"));
I used clone because add modifies the original object, which might not be desired.
strtotime( "+1 month", strtotime( $time ) );
this returns a timestamp that can be used with the date function
You can use DateTime::modify like this :
$date = new DateTime('2010-12-11');
$date->modify('+1 month');
See documentations :
https://php.net/manual/en/datetime.modify.php
https://php.net/manual/en/class.datetime.php
UPDATE january 2021 : correct mistakes raised by comments
This solution has some problems for months with 31 days like May etc.
Exemple : this jumps from 31st May to 1st July which is incorrect.
To correct that, you can create this custom function
function addMonths($date,$months){
$init=clone $date;
$modifier=$months.' months';
$back_modifier =-$months.' months';
$date->modify($modifier);
$back_to_init= clone $date;
$back_to_init->modify($back_modifier);
while($init->format('m')!=$back_to_init->format('m')){
$date->modify('-1 day') ;
$back_to_init= clone $date;
$back_to_init->modify($back_modifier);
}
}
Then you can use it like that :
$date = new DateTime('2010-05-31');
addMonths($date, 1);
print_r($date);
//DateTime Object ( [date] => 2010-06-30 00:00:00.000000 [timezone_type] => 3 [timezone] => Europe/Berlin )
This solution was found in PHP.net posted by jenspj : https://www.php.net/manual/fr/datetime.modify.php#107592
(date('d') > 28) ? date("mdY", strtotime("last day of next month")) : date("mdY", strtotime("+1 month"));
This will compensate for February and the other 31 day months. You could of course do a lot more checking to to get more exact for 'this day next month' relative date formats (which does not work sadly, see below), and you could just as well use DateTime.
Both DateInterval('P1M') and strtotime("+1 month") are essentially blindly adding 31 days regardless of the number of days in the following month.
2010-01-31 => March 3rd
2012-01-31 => March 2nd (leap year)
Please first you set your date format as like 12-12-2012
After use this function it's work properly;
$date = date('d-m-Y',strtotime("12-12-2012 +2 Months");
Here 12-12-2012 is your date and +2 Months is increment of the month;
You also increment of Year, Date
strtotime("12-12-2012 +1 Year");
Ans is 12-12-2013
I use this way:-
$occDate='2014-01-28';
$forOdNextMonth= date('m', strtotime("+1 month", strtotime($occDate)));
//Output:- $forOdNextMonth=02
/*****************more example****************/
$occDate='2014-12-28';
$forOdNextMonth= date('m', strtotime("+1 month", strtotime($occDate)));
//Output:- $forOdNextMonth=01
//***********************wrong way**********************************//
$forOdNextMonth= date('m', strtotime("+1 month", $occDate));
//Output:- $forOdNextMonth=02; //instead of $forOdNextMonth=01;
//******************************************************************//
Just updating the answer with simple method for find the date after no of months. As the best answer marked doesn't give the correct solution.
<?php
$date = date('2020-05-31');
$current = date("m",strtotime($date));
$next = date("m",strtotime($date."+1 month"));
if($current==$next-1){
$needed = date('Y-m-d',strtotime($date." +1 month"));
}else{
$needed = date('Y-m-d', strtotime("last day of next month",strtotime($date)));
}
echo "Date after 1 month from 2020-05-31 would be : $needed";
?>
If you want to get the date of one month from now you can do it like this
echo date('Y-m-d', strtotime('1 month'));
If you want to get the date of two months from now, you can achieve that by doing this
echo date('Y-m-d', strtotime('2 month'));
And so on, that's all.
Thanks Jason, your post was very helpful. I reformatted it and added more comments to help me understand it all. In case that helps anyone, I have posted it here:
function cycle_end_date($cycle_start_date, $months) {
$cycle_start_date_object = new DateTime($cycle_start_date);
//Find the date interval that we will need to add to the start date
$date_interval = find_date_interval($months, $cycle_start_date_object);
//Add this date interval to the current date (the DateTime class handles remaining complexity like year-ends)
$cycle_end_date_object = $cycle_start_date_object->add($date_interval);
//Subtract (sub) 1 day from date
$cycle_end_date_object->sub(new DateInterval('P1D'));
//Format final date to Y-m-d
$cycle_end_date = $cycle_end_date_object->format('Y-m-d');
return $cycle_end_date;
}
//Find the date interval we need to add to start date to get end date
function find_date_interval($n_months, DateTime $cycle_start_date_object) {
//Create new datetime object identical to inputted one
$date_of_last_day_next_month = new DateTime($cycle_start_date_object->format('Y-m-d'));
//And modify it so it is the date of the last day of the next month
$date_of_last_day_next_month->modify('last day of +'.$n_months.' month');
//If the day of inputted date (e.g. 31) is greater than last day of next month (e.g. 28)
if($cycle_start_date_object->format('d') > $date_of_last_day_next_month->format('d')) {
//Return a DateInterval object equal to the number of days difference
return $cycle_start_date_object->diff($date_of_last_day_next_month);
//Otherwise the date is easy and we can just add a month to it
} else {
//Return a DateInterval object equal to a period (P) of 1 month (M)
return new DateInterval('P'.$n_months.'M');
}
}
$cycle_start_date = '2014-01-31'; // select date in Y-m-d format
$n_months = 1; // choose how many months you want to move ahead
$cycle_end_date = cycle_end_date($cycle_start_date, $n_months); // output: 2014-07-02
function dayOfWeek($date){
return DateTime::createFromFormat('Y-m-d', $date)->format('N');
}
Usage examples:
echo dayOfWeek(2016-12-22);
// "4"
echo dayOfWeek(date('Y-m-d'));
// "4"
$date = strtotime("2017-12-11");
$newDate = date("Y-m-d", strtotime("+1 month", $date));
If you want to increment by days you can also do it
$date = strtotime("2017-12-11");
$newDate = date("Y-m-d", strtotime("+5 day", $date));
For anyone looking for an answer to any date format.
echo date_create_from_format('d/m/Y', '15/04/2017')->add(new DateInterval('P1M'))->format('d/m/Y');
Just change the date format.
//ECHO MONTHS BETWEEN TWO TIMESTAMPS
$my_earliest_timestamp = 1532095200;
$my_latest_timestamp = 1554991200;
echo '<pre>';
echo "Earliest timestamp: ". date('c',$my_earliest_timestamp) ."\r\n";
echo "Latest timestamp: " .date('c',$my_latest_timestamp) ."\r\n\r\n";
echo "Month start of earliest timestamp: ". date('c',strtotime('first day of '. date('F Y',$my_earliest_timestamp))) ."\r\n";
echo "Month start of latest timestamp: " .date('c',strtotime('first day of '. date('F Y',$my_latest_timestamp))) ."\r\n\r\n";
echo "Month end of earliest timestamp: ". date('c',strtotime('last day of '. date('F Y',$my_earliest_timestamp)) + 86399) ."\r\n";
echo "Month end of latest timestamp: " .date('c',strtotime('last day of '. date('F Y',$my_latest_timestamp)) + 86399) ."\r\n\r\n";
$sMonth = strtotime('first day of '. date('F Y',$my_earliest_timestamp));
$eMonth = strtotime('last day of '. date('F Y',$my_earliest_timestamp)) + 86399;
$xMonth = strtotime('+1 month', strtotime('first day of '. date('F Y',$my_latest_timestamp)));
while ($eMonth < $xMonth) {
echo "Things from ". date('Y-m-d',$sMonth) ." to ". date('Y-m-d',$eMonth) ."\r\n\r\n";
$sMonth = $eMonth + 1; //add 1 second to bring forward last date into first second of next month.
$eMonth = strtotime('last day of '. date('F Y',$sMonth)) + 86399;
}
I find the mtkime() function works really well for this:
$start_date="2021-10-01";
$start_date_plus_a_month=date("Y-m-d", mktime(0, 0, 0, date("m",strtotime($start_date))+1, date("d",strtotime($start_date)), date("Y",strtotime($start_date))));
result: 2021-11-01
I like to subtract 1 from the 'day' to produce '2021-10-31' which can be useful if you want to display a range across 12 months, e.g. Oct 1, 2021 to Sep 30 2022
$start_date_plus_a_year=date("Y-m-d", mktime(0, 0, 0, date("m",strtotime($start_date))+12, date("d",strtotime($start_date))-1, date("Y",strtotime($start_date))));
result: 2022-09-30
The correct answer to the exact question asked is Giuseppe Canale's answer from earlier. I'm going to answer a slightly more generic question of how to increment the date by an arbitrary number of months, however.
<?php
/**
* Will return a timestamp corresponding to first day of the month that is N months into the future.
* #param int $months_later number of months into the future: 0 for current one
* #param string $today if supplied will be used as the "now" time
* #return int
*/
function rel_month_to_time($months_later, $today=null) {
if ($months_later===0) {
return is_null($today) ? time() : strtotime($today);
}
return strtotime('first day of next month', rel_month_to_time($months_later-1, $today));
}
As is many times the case, you can use recursion for these "human problems" like calendars. The above can be used to return a timestamp corresponding to "next month" -- the way we humans think of it.
<?php echo date('Y-m-d', rel_month_to_time(1, '2023-01-30'));
// 2023-02-01
As pointed by #NetVicious i corrected the code, it should work with all dates, some example:
2013-01-30 will be 2013-02-28
2013-05-15 will be 2013-05-15
2013-05-31 will be 2013-06-30
This code uses the DateTime class to create a new date object, then it adds 1 month to the date using the modify method. Next, it gets the day of the next month using the format method. If the next month's day doesn't match the original day, it modifies the date to the last day of the previous month using the modify method.
$original_date = "2013-01-30";
$original_day = date("d", strtotime($original_date));
$date = new DateTime($original_date);
$date->modify('+1 month');
$next_month_day = $date->format('d');
if ($next_month_day != $original_day) {
$date->modify('last day of previous month');
}
$new_date = $date->format('Y-m-d');
echo $new_date;
All presented solutions are not working properly.
strtotime() and DateTime::add or DateTime::modify give sometime invalid results.
Examples:
- 31.08.2019 + 1 month gives 01.10.2019 instead 30.09.2019
- 29.02.2020 + 1 year gives 01.03.2021 instead 28.02.2021
(tested on PHP 5.5, PHP 7.3)
Below is my function based on idea posted by Angelo that solves the problem:
// $time - unix time or date in any format accepted by strtotime() e.g. 2020-02-29
// $days, $months, $years - values to add
// returns new date in format 2021-02-28
function addTime($time, $days, $months, $years)
{
// Convert unix time to date format
if (is_numeric($time))
$time = date('Y-m-d', $time);
try
{
$date_time = new DateTime($time);
}
catch (Exception $e)
{
echo $e->getMessage();
exit;
}
if ($days)
$date_time->add(new DateInterval('P'.$days.'D'));
// Preserve day number
if ($months or $years)
$old_day = $date_time->format('d');
if ($months)
$date_time->add(new DateInterval('P'.$months.'M'));
if ($years)
$date_time->add(new DateInterval('P'.$years.'Y'));
// Patch for adding months or years
if ($months or $years)
{
$new_day = $date_time->format("d");
// The day is changed - set the last day of the previous month
if ($old_day != $new_day)
$date_time->sub(new DateInterval('P'.$new_day.'D'));
}
// You can chage returned format here
return $date_time->format('Y-m-d');
}
Usage examples:
echo addTime('2020-02-29', 0, 0, 1); // add 1 year (result: 2021-02-28)
echo addTime('2019-08-31', 0, 1, 0); // add 1 month (result: 2019-09-30)
echo addTime('2019-03-15', 12, 2, 1); // add 12 days, 2 months, 1 year (result: 2019-09-30)
put a date in input box then click the button get day from date in jquery
$(document).ready( function() {
$("button").click(function(){
var day = ["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"];
var a = new Date();
$(".result").text(day[a.getDay()]);
});
});
<?php
$selectdata ="select fromd,tod from register where username='$username'";
$q=mysqli_query($conm,$selectdata);
$row=mysqli_fetch_array($q);
$startdate=$row['fromd'];
$stdate=date('Y', strtotime($startdate));
$endate=$row['tod'];
$enddate=date('Y', strtotime($endate));
$years = range ($stdate,$enddate);
echo '<select name="years" class="form-control">';
echo '<option>SELECT</option>';
foreach($years as $year)
{ echo '<option value="'.$year.'"> '.$year.' </option>'; }
echo '</select>'; ?>
I want to get the date for current day in php. what i tried is here...
echo $x."<br>";
echo date("D",$x)."<br>";
But the output was
21-02-10
Thu
It is giving correct date but not the correct day value.Why..?
What I want day is the date for monday for the current week which can be generated on any day of the week. so what I did was, I'm taking the today's day and comparing with (Mon,Tue.... Sun) and respectively creating a timestamp using
case "Mon":
$startdate1=date("d-m-y");
$parts = explode('-',$startdate1);
$startdate2 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+1),$parts[2]));
$startdate3 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+2),$parts[2]));
$startdate4 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+3),$parts[2]));
$startdate5 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+4),$parts[2]));
$startdate6 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+5),$parts[2]));
$startdate7 = date('d-m-Y',mktime(0,0,0,$parts[1],($parts[0]+6),$parts[2]));
$dates=array(1 => $startdate1,$startdate2,$startdate3,$startdate4,$startdate5,$startdate6,$startdate7);
$i=1;
while( $i <= 7 )
{
echo $dates[$i];
$i++;
}
break;
$date is the final array respective to today that has to be returned. Is there any other better method to do this operation.
I tried this to get current day.
echo date('l'); // output: current day.
How about this:
//today is monday
if (1 == date('N')){
$monday = time();
}else{
$monday = strtotime('last Monday');
}
for ($i = 0; $i < 7; $i++){
echo date('d-m-Y', $monday) . '<br>';
$monday = strtotime('tomorrow', $monday);
}
First find Monday, if it is not today, then print 7 dates
What I want day is the date for monday
for the current week which can be
generated on any day of the week.
That's what you want. $mday is the month day of this week's Monday. Nevermind if it's not positive, mktime will handle that right. $monday has the timestamp of the Monday's midnight.
$now = getdate();
$mday = $now['mday'] - ($now['wday'] + 6) % 7;
$monday = mktime(0, 0, 0, $now['mon'], $mday, $now['year']);
echo(date('d-m-y', $monday));
What i did to resolve it is used the date format ('d-m-Y') instead of ('d-m-y') in date function, which was causing the problem. Hence strtotime accepted the format and gave the correct result for
$t=date('d-m-Y');
echo date("D",strtotime($t));
I use the function date and path to it the "D" that refere to the current day , and it works with me
$today = date("D");
and to get the full info about the current date
$today = date("D M j G:i:s T Y"); // Sat Mar 10 17:16:18 MST 2001
what i tried is here...
echo date("D",$x)."<br>";
date expects a timestamp (int) value as the second parameter. Your $x is a string containing an ambiguous date format. Convert that date into a timestamp first, using strptime or strtotime and use the date function correctly to get the correct day value.
Regarding your second part, you don't need to (and shouldn't) check the day name to calculate the correct Monday, Tuesday etc. A more efficient approach is for example using strtotime to get last Monday etc.
You are likely passing a string as timestamp
echo $x."<br>";
echo date("D",$x)."<br>";
Remove $x and it will output the correct day or change it to
$x = '21-02-2010';
echo date('D', strtotime($x));