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Add X Days To Date, Excluding Weekends
(5 answers)
Closed 3 years ago.
I want to find an end date of my first date.
For example my first date is
2020-01-01 04:20:18
And I want to get 20 days after this first date but only the weekdays (exclude the saturday and sunday)
I created this
$expirationDate = date_format(date_add(date_create(date("Y-m-d H:i:s",strtotime($_POST['created_at']))),date_interval_create_from_date_string("20 day")),'Y-m-d H:i:s');
The result I got is
2020-01-21 04:20:18
The end date result should be
2020-01-28 04:20:18
Because date 4,5,11,12,18,19,25,26 is saturday and sunday.
Anyone can help me?
You can use a loop and check which day of the week it is to achieve this. I've left your original $expirationDate calculation in to show the difference in the output:
$dt = date("Y-m-d H:i:s", strtotime('2020-01-01 04:20:18'));
$expirationDate = date_format(date_add(date_create($dt),date_interval_create_from_date_string("20 day")),'Y-m-d H:i:s');
// Add 20 days
for ($i = 0; $i < 20; $i++) {
$dt = date("Y-m-d H:i:s", strtotime("+1 day", strtotime($dt)));
// Is it a Sunday or Saturday?
$day = date('w', strtotime($dt));
if ($day == 0 || $day == 6)
// Deduct one from loop counter
$i--;
}
// Output
echo $dt;
echo $expirationDate;
Output:
$dt = 2020-01-29 04:20:18 (correct)
$expirationDate = 2020-01-21 04:20:18 (incorrect)
How does it work?
It's quite straight-forward:
Store the start date in $dt
Loop from 0 to 19 (for 20 days)
Add one day to $dt
If the day is a Sunday (0) or a Saturday (6), subtract 1 from the loop counter ($i)
An Alternative Solution
As an alternative, this may be greatly simplified by simply adding four weeks to the start time:
$dt = date("Y-m-d H:i:s", strtotime('2020-01-01 04:20:18'));
$dt = date("Y-m-d H:i:s", strtotime("+4 week", strtotime($dt)));
echo $dt;
However, this could only work if that start date was guaranteed to be a weekday. Otherwise you'd end up with an end date on a weekend.
suppose I have an initial date whose year was prior to that of the current year and I want to repeat the event every 7 days but only in the current year.
How would I find the first occurrence in the current year?
I realize I can do it with a loop like this:
$reOccurringEvent =new DateTime('2013-12-01');
$interval = new DateInterval('P7D');
while($reOccurringEvent->format('Y') < date('Y') ){
$reOccurringEvent->add($interval);
}
echo $reOccurringEvent->format('d m Y'); //05 01 2014
But it strikes me there should be a more efficient way to achieve this rather than repeatedly adding an interval to the date (it would happen many times if the initial date was some years ago).
I was hoping to be able to calculate the number of times the interval should be added and just do it a single time.
I was thinking something like:
$date = new DateTime();
$diff = $date->diff($reOccurringEvent)->days%7;
But obviously that doesn't work and I can't quite figure out the logic of how to do it.
More generically, the algorithm would be to find the number of intervals between the given date and the last day of last year. Then multiplying the interval by the number of intervals + 1 to get the first interval of the current year.
$date1="12/9/2013";
$ts1 = strtotime($date1);
$ts2 = strtotime("12/31/" . Date("Y")-1);
//get the number of seconds between the date and first of the year
$seconds_diff = $ts2 - $ts1;
echo "$seconds_diff <br>";
//get the number of days
$dayDiff=$seconds_diff/86400;
//how many intervals?
$intervalDays = "10";
//get the number of intervals from start date to last day of last year
$numIntervals = floor($dayDiff/$intervalDays);
echo $numIntervals."<br>";
//now the total intervals to get into the current year is one more interval, turn this into days
$totIntervals= ($numIntervals* $intervalDays)+$intervalDays;
//Date Time date in question
$theDt = new DateTime($date1);
//Add the intervals we calculated to the date in question, and we have the first date of the interval for the current year...
$theDt->add(new DateInterval('P' . $totIntervals. 'D'));
echo "The first date of the intreval is: " . $theDt->format('Y-m-d');
I think, if you are doing 7 day intervals, you can find out the Day of week of your initial date, and then get the first date of the current year with that day of week...
Find out day of week: How to find the day of week from a date using PHP?
Find out date with that day of week for this year: Getting first weekday in a month with strtotime
Putting it together:
$date=Date("2/8/2012");
//Get the day of week for the date in question
$dayOfWeek = date('l', strtotime($date));
echo "The day of week for the given date is: $dayOfWeek <br>";
//Get the current year
$thisYear = date("Y");
echo "This year: $thisYear <br>";
//Create a date with the first occurence of the day of week of the given date for the current year
$firstOccurenceThisYear = date("m/d/y", strtotime("January " .$thisYear ." " . $dayOfWeek));
echo "The first interval of the year is: $firstOccurenceThisYear";
/*
Output:
This year: 2014
The day of week for the given date is: Wednesday
The first interval of the year is: 01/01/14
*/
Here is a slightly modified version of #Dan's second answer which worked well for me.
Benchmarks shown below.
$date="1985-02-18";
$intervalDays = "5";
//original version
$benchMark = microtime(true);
$dt1 = new DateTime($date);
$interval = new DateInterval("P{$intervalDays}D");
while ($dt1->format('Y') < date('Y')) {
$dt1->add($interval);
}
echo $dt1->format('d m Y') . '<br>';
echo microtime(true)-$benchMark.'<br>';
//new version
$benchMark = microtime(true);
$dt1 = new DateTime($date);
$dt2 = new DateTime("12/31/" . ((int) Date("Y") - 1));
$dayDiff = $dt1->diff($dt2)->days;
$numIntervals = floor($dayDiff / $intervalDays);
$totIntervals = ($numIntervals * $intervalDays) + $intervalDays;
$dt1->add(new DateInterval('P' . $totIntervals . 'D'));
echo $dt1->format('d m Y').'<br>';
echo microtime(true)-$benchMark.'<br>';
exit;
output
02 01 2014
0.0145111083984
02 01 2014
0.000123977661133
Hy, I have in database number of the week date ("W") and I want to display week interval like 28 Jan -> 3 Feb in this format, and I don't know if it's possible. Can you help?
Thanks!
Try this
$year = 2013;
$week_no = 6;
$week_start = new DateTime();
$week_start->setISODate($year,$week_no);
$week_end = clone $week_start;
$week_end = $week_end->add(new DateInterval("P1W"));
echo $week_start->format('d-M-Y') . " - ".$week_end->format('d-M-Y');
Transform your intervals into timestamps.
If its not the first day of the week get the first day of that week with strtotime "last sunday" (or monday) for the first date.
Do the same for the second date this time geting the last day of the week with "next saturday" (or sunday);
Get both dates W and make a mysql comparison between the weeks.
I've been working a lot with the DateTime class and recently ran into what I thought was a bug when adding months. After a bit of research, it appears that it wasn't a bug, but instead working as intended. According to the documentation found here:
Example #2 Beware when adding or
subtracting months
<?php
$date = new DateTime('2000-12-31');
$date->modify('+1 month');
echo $date->format('Y-m-d') . "\n";
$date->modify('+1 month');
echo $date->format('Y-m-d') . "\n";
?>
The above example will output:
2001-01-31
2001-03-03
Can anyone justify why this isn't considered a bug?
Furthermore does anyone have any elegant solutions to correct the issue and make it so +1 month will work as expected instead of as intended?
Why it's not a bug:
The current behavior is correct. The following happens internally:
+1 month increases the month number (originally 1) by one. This makes the date 2010-02-31.
The second month (February) only has 28 days in 2010, so PHP auto-corrects this by just continuing to count days from February 1st. You then end up at March 3rd.
How to get what you want:
To get what you want is by: manually checking the next month. Then add the number of days next month has.
I hope you can yourself code this. I am just giving what-to-do.
PHP 5.3 way:
To obtain the correct behavior, you can use one of the PHP 5.3's new functionality that introduces the relative time stanza first day of. This stanza can be used in combination with next month, fifth month or +8 months to go to the first day of the specified month. Instead of +1 month from what you're doing, you can use this code to get the first day of next month like this:
<?php
$d = new DateTime( '2010-01-31' );
$d->modify( 'first day of next month' );
echo $d->format( 'F' ), "\n";
?>
This script will correctly output February. The following things happen when PHP processes this first day of next month stanza:
next month increases the month number (originally 1) by one. This makes the date 2010-02-31.
first day of sets the day number to 1, resulting in the date 2010-02-01.
Here is another compact solution entirely using DateTime methods, modifying the object in-place without creating clones.
$dt = new DateTime('2012-01-31');
echo $dt->format('Y-m-d'), PHP_EOL;
$day = $dt->format('j');
$dt->modify('first day of +1 month');
$dt->modify('+' . (min($day, $dt->format('t')) - 1) . ' days');
echo $dt->format('Y-m-d'), PHP_EOL;
It outputs:
2012-01-31
2012-02-29
This may be useful:
echo Date("Y-m-d", strtotime("2013-01-01 +1 Month -1 Day"));
// 2013-01-31
echo Date("Y-m-d", strtotime("2013-02-01 +1 Month -1 Day"));
// 2013-02-28
echo Date("Y-m-d", strtotime("2013-03-01 +1 Month -1 Day"));
// 2013-03-31
echo Date("Y-m-d", strtotime("2013-04-01 +1 Month -1 Day"));
// 2013-04-30
echo Date("Y-m-d", strtotime("2013-05-01 +1 Month -1 Day"));
// 2013-05-31
echo Date("Y-m-d", strtotime("2013-06-01 +1 Month -1 Day"));
// 2013-06-30
echo Date("Y-m-d", strtotime("2013-07-01 +1 Month -1 Day"));
// 2013-07-31
echo Date("Y-m-d", strtotime("2013-08-01 +1 Month -1 Day"));
// 2013-08-31
echo Date("Y-m-d", strtotime("2013-09-01 +1 Month -1 Day"));
// 2013-09-30
echo Date("Y-m-d", strtotime("2013-10-01 +1 Month -1 Day"));
// 2013-10-31
echo Date("Y-m-d", strtotime("2013-11-01 +1 Month -1 Day"));
// 2013-11-30
echo Date("Y-m-d", strtotime("2013-12-01 +1 Month -1 Day"));
// 2013-12-31
My solution to the problem:
$startDate = new \DateTime( '2015-08-30' );
$endDate = clone $startDate;
$billing_count = '6';
$billing_unit = 'm';
$endDate->add( new \DateInterval( 'P' . $billing_count . strtoupper( $billing_unit ) ) );
if ( intval( $endDate->format( 'n' ) ) > ( intval( $startDate->format( 'n' ) ) + intval( $billing_count ) ) % 12 )
{
if ( intval( $startDate->format( 'n' ) ) + intval( $billing_count ) != 12 )
{
$endDate->modify( 'last day of -1 month' );
}
}
I agree with the sentiment of the OP that this is counter-intuitive and frustrating, but so is determining what +1 month means in the scenarios where this occurs. Consider these examples:
You start with 2015-01-31 and want to add a month 6 times to get a scheduling cycle for sending an email newsletter. With the OP's initial expectations in mind, this would return:
2015-01-31
2015-02-28
2015-03-31
2015-04-30
2015-05-31
2015-06-30
Right away, notice that we are expecting +1 month to mean last day of month or, alternatively, to add 1 month per iteration but always in reference to the start point. Instead of interpreting this as "last day of month" we could read it as "31st day of next month or last available within that month". This means that we jump from April 30th to May 31st instead of to May 30th. Note that this is not because it is "last day of month" but because we want "closest available to date of start month."
So suppose one of our users subscribes to another newsletter to start on 2015-01-30. What is the intuitive date for +1 month? One interpretation would be "30th day of next month or closest available" which would return:
2015-01-30
2015-02-28
2015-03-30
2015-04-30
2015-05-30
2015-06-30
This would be fine except when our user gets both newsletters on the same day. Let's assume that this is a supply-side issue instead of demand-side We're not worried that the user will be annoyed with getting 2 newsletters in the same day but instead that our mail servers can't afford the bandwidth for sending twice as many newsletters. With that in mind, we return to the other interpretation of "+1 month" as "send on the second to last day of each month" which would return:
2015-01-30
2015-02-27
2015-03-30
2015-04-29
2015-05-30
2015-06-29
Now we've avoided any overlap with the first set, but we also end up with April and June 29th, which certainly does match our original intuitions that +1 month simply should return m/$d/Y or the attractive and simple m/30/Y for all possible months. So now let's consider a third interpretation of +1 month using both dates:
Jan. 31st
2015-01-31
2015-03-03
2015-03-31
2015-05-01
2015-05-31
2015-07-01
Jan. 30th
2015-01-30
2015-03-02
2015-03-30
2015-04-30
2015-05-30
2015-06-30
The above has some issues. February is skipped, which could be a problem both supply-end (say if there is a monthly bandwidth allocation and Feb goes to waste and March gets doubled up on) and demand-end (users feel cheated out of Feb and perceive the extra March as attempt to correct mistake). On the other hand, notice that the two date sets:
never overlap
are always on the same date when that month has the date (so the Jan. 30 set looks pretty clean)
are all within 3 days (1 day in most cases) of what might be considered the "correct" date.
are all at least 28 days (a lunar month) from their successor and predecessor, so very evenly distributed.
Given the last two sets, it would not be difficult to simply roll back one of the dates if it falls outside of the actual following month (so roll back to Feb 28th and April 30th in the first set) and not lose any sleep over the occasional overlap and divergence from the "last day of month" vs "second to last day of month" pattern. But expecting the library to choose between "most pretty/natural", "mathematical interpretation of 02/31 and other month overflows", and "relative to first of month or last month" is always going to end with someone's expectations not being met and some schedule needing to adjust the "wrong" date to avoid the real-world problem that the "wrong" interpretation introduces.
So again, while I also would expect +1 month to return a date that actually is in the following month, it is not as simple as intuition and given the choices, going with math over the expectations of web developers is probably the safe choice.
Here's an alternative solution that is still as clunky as any but I think has nice results:
foreach(range(0,5) as $count) {
$new_date = clone $date;
$new_date->modify("+$count month");
$expected_month = $count + 1;
$actual_month = $new_date->format("m");
if($expected_month != $actual_month) {
$new_date = clone $date;
$new_date->modify("+". ($count - 1) . " month");
$new_date->modify("+4 weeks");
}
echo "* " . nl2br($new_date->format("Y-m-d") . PHP_EOL);
}
It's not optimal but the underlying logic is : If adding 1 month results in a date other than the expected next month, scrap that date and add 4 weeks instead. Here are the results with the two test dates:
Jan. 31st
2015-01-31
2015-02-28
2015-03-31
2015-04-28
2015-05-31
2015-06-28
Jan. 30th
2015-01-30
2015-02-27
2015-03-30
2015-04-30
2015-05-30
2015-06-30
(My code is a mess and wouldn't work in a multi-year scenario. I welcome anyone to rewrite the solution with more elegant code so long as the underlying premise is kept intact, i.e. if +1 month returns a funky date, use +4 weeks instead.)
In conjunction with shamittomar's answer, it could then be this for adding months "safely":
/**
* Adds months without jumping over last days of months
*
* #param \DateTime $date
* #param int $monthsToAdd
* #return \DateTime
*/
public function addMonths($date, $monthsToAdd) {
$tmpDate = clone $date;
$tmpDate->modify('first day of +'.(int) $monthsToAdd.' month');
if($date->format('j') > $tmpDate->format('t')) {
$daysToAdd = $tmpDate->format('t') - 1;
}else{
$daysToAdd = $date->format('j') - 1;
}
$tmpDate->modify('+ '. $daysToAdd .' days');
return $tmpDate;
}
I made a function that returns a DateInterval to make sure that adding a month shows the next month, and removes the days into the after that.
$time = new DateTime('2014-01-31');
echo $time->format('d-m-Y H:i') . '<br/>';
$time->add( add_months(1, $time));
echo $time->format('d-m-Y H:i') . '<br/>';
function add_months( $months, \DateTime $object ) {
$next = new DateTime($object->format('d-m-Y H:i:s'));
$next->modify('last day of +'.$months.' month');
if( $object->format('d') > $next->format('d') ) {
return $object->diff($next);
} else {
return new DateInterval('P'.$months.'M');
}
}
This is an improved version of Kasihasi's answer in a related question. This will correctly add or subtract an arbitrary number of months to a date.
public static function addMonths($monthToAdd, $date) {
$d1 = new DateTime($date);
$year = $d1->format('Y');
$month = $d1->format('n');
$day = $d1->format('d');
if ($monthToAdd > 0) {
$year += floor($monthToAdd/12);
} else {
$year += ceil($monthToAdd/12);
}
$monthToAdd = $monthToAdd%12;
$month += $monthToAdd;
if($month > 12) {
$year ++;
$month -= 12;
} elseif ($month < 1 ) {
$year --;
$month += 12;
}
if(!checkdate($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);
}
return $d2->format('Y-m-d');
}
For example:
addMonths(-25, '2017-03-31')
will output:
'2015-02-28'
I found a shorter way around it using the following code:
$datetime = new DateTime("2014-01-31");
$month = $datetime->format('n'); //without zeroes
$day = $datetime->format('j'); //without zeroes
if($day == 31){
$datetime->modify('last day of next month');
}else if($day == 29 || $day == 30){
if($month == 1){
$datetime->modify('last day of next month');
}else{
$datetime->modify('+1 month');
}
}else{
$datetime->modify('+1 month');
}
echo $datetime->format('Y-m-d H:i:s');
Here is an implementation of an improved version of Juhana's answer in a related question:
<?php
function sameDateNextMonth(DateTime $createdDate, DateTime $currentDate) {
$addMon = clone $currentDate;
$addMon->add(new DateInterval("P1M"));
$nextMon = clone $currentDate;
$nextMon->modify("last day of next month");
if ($addMon->format("n") == $nextMon->format("n")) {
$recurDay = $createdDate->format("j");
$daysInMon = $addMon->format("t");
$currentDay = $currentDate->format("j");
if ($recurDay > $currentDay && $recurDay <= $daysInMon) {
$addMon->setDate($addMon->format("Y"), $addMon->format("n"), $recurDay);
}
return $addMon;
} else {
return $nextMon;
}
}
This version takes $createdDate under the presumption that you are dealing with a recurring monthly period, such as a subscription, that started on a specific date, such as the 31st. It always takes $createdDate so late "recurs on" dates won't shift to lower values as they are pushed forward thru lesser-valued months (e.g., so all 29th, 30th or 31st recur dates won't eventually get stuck on the 28th after passing thru a non-leap-year February).
Here is some driver code to test the algorithm:
$createdDate = new DateTime("2015-03-31");
echo "created date = " . $createdDate->format("Y-m-d") . PHP_EOL;
$next = sameDateNextMonth($createdDate, $createdDate);
echo " next date = " . $next->format("Y-m-d") . PHP_EOL;
foreach(range(1, 12) as $i) {
$next = sameDateNextMonth($createdDate, $next);
echo " next date = " . $next->format("Y-m-d") . PHP_EOL;
}
Which outputs:
created date = 2015-03-31
next date = 2015-04-30
next date = 2015-05-31
next date = 2015-06-30
next date = 2015-07-31
next date = 2015-08-31
next date = 2015-09-30
next date = 2015-10-31
next date = 2015-11-30
next date = 2015-12-31
next date = 2016-01-31
next date = 2016-02-29
next date = 2016-03-31
next date = 2016-04-30
$ds = new DateTime();
$ds->modify('+1 month');
$ds->modify('first day of this month');
If you just want to avoid skipping a month you can perform something like this to get the date out and run a loop on the next month reducing the date by one and rechecking until a valid date where $starting_calculated is a valid string for strtotime (i.e. mysql datetime or "now"). This finds the very end of the month at 1 minute to midnight instead of skipping the month.
$start_dt = $starting_calculated;
$next_month = date("m",strtotime("+1 month",strtotime($start_dt)));
$next_month_year = date("Y",strtotime("+1 month",strtotime($start_dt)));
$date_of_month = date("d",$starting_calculated);
if($date_of_month>28){
$check_date = false;
while(!$check_date){
$check_date = checkdate($next_month,$date_of_month,$next_month_year);
$date_of_month--;
}
$date_of_month++;
$next_d = $date_of_month;
}else{
$next_d = "d";
}
$end_dt = date("Y-m-$next_d 23:59:59",strtotime("+1 month"));
Extension for DateTime class which solves problem of adding or subtracting months
https://gist.github.com/66Ton99/60571ee49bf1906aaa1c
If using strtotime() just use $date = strtotime('first day of +1 month');
I needed to get a date for 'this month last year' and it becomes unpleasant quite quickly when this month is February in a leap year. However, I believe this works... :-/ The trick seems to be to base your change on the 1st day of the month.
$this_month_last_year_end = new \DateTime();
$this_month_last_year_end->modify('first day of this month');
$this_month_last_year_end->modify('-1 year');
$this_month_last_year_end->modify('last day of this month');
$this_month_last_year_end->setTime(23, 59, 59);
$month = 1; $year = 2017;
echo date('n', mktime(0, 0, 0, $month + 2, -1, $year));
will output 2 (february). will work for other months too.
$current_date = new DateTime('now');
$after_3_months = $current_date->add(\DateInterval::createFromDateString('+3 months'));
For days:
$after_3_days = $current_date->add(\DateInterval::createFromDateString('+3 days'));
Important:
The method add() of DateTime class modify the object value so after calling add() on a DateTime Object it returns the new date object and also it modify the object it self.
you can actually do it with just date() and strtotime() as well. For example to add 1 month to todays date:
date("Y-m-d",strtotime("+1 month",time()));
if you are wanting to use the datetime class thats fine too but this is just as easy. more details here
$date = date('Y-m-d', strtotime("+1 month"));
echo $date;
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));