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I'm coding a script where I require to save the current date, and the date 1 month from that date. I am pretty sure that the time() variable works, but I am not sure how to +1 month onto that?
Any ideas, suggestions. Cheers!
Try this
$today = date("Y-m-d");
$date = date('Y-m-d', strtotime('+1 month', $today));
or use DateTime()
$dt1 = new DateTime();
$today = $dt1->format("Y-m-d");
$dt2 = new DateTime("+1 month");
$date = $dt2->format("Y-m-d");
$time = strtotime("2010-12-11");
$final = date("Y-m-d", strtotime("+1 month", $time));
(OR)
strtotime( "+1 month", strtotime( $time ) );
this returns a timestamp that can be used with the date function
Use this:
Current Date:
echo "Today is " . date("Y/m/d");
1 Month to the Current Date:
$time = strtotime(date("Y/m/d"));
$final = date("Y-m-d", strtotime("+1 month", $time));
<?php
$current_time = date("Y-M-d h:i:s",time()); // Getting Current Date & Time
print $current_time; // Current Date & Time Printing for display purpose
$future_timestamp = strtotime("+1 month"); // Getting timestamp of 1 month from now
$final_future = date("Y-M-d h:i:s",+$future_timestamp); // Getting Future Date & Time of 1 month from now
print $final_future; // Printing Future time for display purpose
?>
shorter : $today=date("Y-m-d"); $date=
This one liner worked for me:
$monthFromToday = date("Y-m-d", strtotime("+1 month", strtotime(date("Y/m/d"))));
The given answers may not give you the results you might expect or desire.
Consider:
$today = "29Jan2018";
$nextMonth = date('dMY', strtotime('+1 month', (strtotime($today))));
echo $today // yields 29Jan2018
echo $nextMonth // yields 01Mar2018
$today = date("Y-m-d");
$enddate = date('Y-m-01',strtotime($today. ' + 1 months'));
You could also consider using the Carbon package.
The solution would look like this:
use Carbon\Carbon
$now = Carbon::now;
$now->addMonth();
Here is the link for reference https://carbon.nesbot.com/docs/
I want to add number of days to current date:
I am using following code:
$i=30;
echo $date = strtotime(date("Y-m-d", strtotime($date)) . " +".$i."days");
But instead of getting proper date i am getting this:
2592000
Please suggest.
This should be
echo date('Y-m-d', strtotime("+30 days"));
strtotime
expects to be given a string containing a US English date format and will try to parse that format into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 UTC), relative to the timestamp given in now, or the current time if now is not supplied.
while date
Returns a string formatted according to the given format string using the given integer timestamp or the current time if no timestamp is given.
See the manual pages for
http://www.php.net/manual/en/function.strtotime.php
http://www.php.net/manual/en/function.date.php
and their function signatures.
This one might be good
function addDayswithdate($date,$days){
$date = strtotime("+".$days." days", strtotime($date));
return date("Y-m-d", $date);
}
$date = new DateTime();
$date->modify('+1 week');
print $date->format('Y-m-d H:i:s');
or print date('Y-m-d H:i:s', mktime(date("H"), date("i"), date("s"), date("m"), date("d") + 7, date("Y"));
$today=date('d-m-Y');
$next_date= date('d-m-Y', strtotime($today. ' + 90 days'));
echo $next_date;
You can add like this as well, if you want the date 5 days from a specific date :
You have a variable with a date like this (gotten from an input or DB or just hard coded):
$today = "2015-06-15"; // Or can put $today = date ("Y-m-d");
$fiveDays = date ("Y-m-d", strtotime ($today ."+5 days"));
echo $fiveDays; // Will output 2015-06-20
Keep in mind, the change of clock changes because of daylight saving time might give you some problems when only calculating the days.
Here's a little php function which takes care of that:
function add_days($date, $days) {
$timeStamp = strtotime(date('Y-m-d',$date));
$timeStamp+= 24 * 60 * 60 * $days;
// ...clock change....
if (date("I",$timeStamp) != date("I",$date)) {
if (date("I",$date)=="1") {
// summer to winter, add an hour
$timeStamp+= 60 * 60;
} else {
// summer to winter, deduct an hour
$timeStamp-= 60 * 60;
} // if
} // if
$cur_dat = mktime(0, 0, 0,
date("n", $timeStamp),
date("j", $timeStamp),
date("Y", $timeStamp)
);
return $cur_dat;
}
You could also try:
$date->modify("+30 days");
You can do it by manipulating the timecode or by using strtotime(). Here's an example using strtotime.
$data['created'] = date('Y-m-d H:i:s', strtotime("+1 week"));
You can use strtotime()
$data['created'] = date('Y-m-d H:m:s', strtotime('+1 week'));
I know this is an old question, but for PHP <5.3 you could try this:
$date = '05/07/2013';
$add_days = 7;
$date = date('Y-m-d',strtotime($date) + (24*3600*$add_days)); //my preferred method
//or
$date = date('Y-m-d',strtotime($date.' +'.$add_days.' days');
You could use the DateTime class built in PHP. It has a method called "add", and how it is used is thoroughly demonstrated in the manual: http://www.php.net/manual/en/datetime.add.php
It however requires PHP 5.3.0.
$date = "04/28/2013 07:30:00";
$dates = explode(" ",$date);
$date = strtotime($dates[0]);
$date = strtotime("+6 days", $date);
echo date('m/d/Y', $date)." ".$dates[1];
You may try this.
$i=30;
echo date("Y-m-d",mktime(0,0,0,date('m'),date('d')+$i,date('Y')));
Simple and Best
echo date('Y-m-d H:i:s')."\n";
echo "<br>";
echo date('Y-m-d H:i:s', mktime(date('H'),date('i'),date('s'), date('m'),date('d')+30,date('Y')))."\n";
Try this
//add the two day
$date = "**2-4-2016**"; //stored into date to variable
echo date("d-m-Y",strtotime($date.**' +2 days'**));
//print output
**4-4-2016**
Use this addDate() function to add or subtract days, month or years (you will need the auxiliar function reformatDate() as well)
/**
* $date self explanatory
* $diff the difference to add or subtract: e.g. '2 days' or '-1 month'
* $format the format for $date
**/
function addDate($date = '', $diff = '', $format = "d/m/Y") {
if (empty($date) || empty($diff))
return false;
$formatedDate = reformatDate($date, $format, $to_format = 'Y-m-d H:i:s');
$newdate = strtotime($diff, strtotime($formatedDate));
return date($format, $newdate);
}
//Aux function
function reformatDate($date, $from_format = 'd/m/Y', $to_format = 'Y-m-d') {
$date_aux = date_create_from_format($from_format, $date);
return date_format($date_aux,$to_format);
}
Note: only for php >=5.3
Use the following code.
<?php echo date('Y-m-d', strtotime(' + 5 days')); ?>
Reference has found from here - How to Add Days to Current Date in PHP
Even though this is an old question, this way of doing it would take of many situations and seems to be robust. You need to have PHP 5.3.0 or above.
$EndDateTime = DateTime::createFromFormat('d/m/Y', "16/07/2017");
$EndDateTime->modify('+6 days');
echo $EndDateTime->format('d/m/Y');
You can have any type of format for the date string and this would work.
//Set time zone
date_default_timezone_set("asia/kolkata");
$pastdate='2016-07-20';
$addYear=1;
$addMonth=3;
$addWeek=2;
$addDays=5;
$newdate=date('Y-m-d', strtotime($pastdate.' +'.$addYear.' years +'.$addMonth. ' months +'.$addWeek.' weeks +'.$addDays.' days'));
echo $newdate;
Do not use php's date() function, it's not as accurate as the below solution and furthermore it is unreliable in the future.
Use the DateTime class
<?php
$date = new DateTime('2016-06-06'); // Y-m-d
$date->add(new DateInterval('P30D'));
echo $date->format('Y-m-d') . "\n";
?>
The reason you should avoid anything to do with UNIX timestamps (time(), date(), strtotime() etc) is that they will inevitably break in the year 2038 due to integer limitations.
The maximum value of an integer is 2147483647 which converts to Tuesday, 19 January 2038 03:14:07 so come this time; this minute; this second; everything breaks
Source
Another example of why I stick to using DateTime is that it's actually able to calculate months correctly regardless of what the current date is:
$now = strtotime('31 December 2019');
for ($i = 1; $i <= 6; $i++) {
echo date('d M y', strtotime('-' . $i .' month', $now)) . PHP_EOL;
}
You'd get the following sequence of dates:
31 December
31 November
31 October
31 September
31 August
31 July
31 June
PHP conveniently recognises that three of these dates are illegal and converts them into its best guess, leaving you with:
01 Dec 19
31 Oct 19
01 Oct 19
31 Aug 19
31 Jul 19
01 Jul 19
How can I get the timestamp of 12 o'clock of today, yesterday and the day before yesterday by using strtotime() function in php?
12 o'clock is a variable and would be changed by user.
$hour = 12;
$today = strtotime($hour . ':00:00');
$yesterday = strtotime('-1 day', $today);
$dayBeforeYesterday = strtotime('-1 day', $yesterday);
strtotime supports a number of interesting modifiers that can be used:
$hour = 12;
$today = strtotime("today $hour:00");
$yesterday = strtotime("yesterday $hour:00");
$dayBeforeYesterday = strtotime("yesterday -1 day $hour:00");
echo date("Y-m-d H:i:s\n", $today);
echo date("Y-m-d H:i:s\n", $yesterday);
echo date("Y-m-d H:i:s\n", $dayBeforeYesterday);
It works as predicted:
2011-01-24 12:00:00
2011-01-23 12:00:00
2011-01-22 12:00:00
OO Equivalent
$iHour = 12;
$oToday = new DateTime();
$oToday->setTime($iHour, 0);
$oYesterday = clone $oToday;
$oYesterday->modify('-1 day');
$oDayBefore = clone $oYesterday;
$oDayBefore->modify('-1 day');
$iToday = $oToday->getTimestamp();
$iYesterday = $oYesterday->getTimestamp();
$iDayBefore = $oDayBefore->getTimestamp();
echo "Today: $iToday\n";
echo "Yesterday: $iYesterday\n";
echo "Day Before: $iDayBefore\n";
You can easily find out any date using DateTime object, It is so flexible
$yesterday = new DateTime('yesterday');
echo $yesterday->format('Y-m-d');
$firstModayOfApril = new DateTime('first monday of april');
echo $firstModayOfApril->format('Y-m-d');
$nextMonday = new DateTime('next monday');
echo $nextMonday->format('Y-m-d');
to get start of day yesterday
$oDate = new DateTime();
$oDate->modify('-1 day');
echo $oDate->format('Y-m-d 00:00:00');
result
2014-11-05 00:00:00
All the answers here are too long and bloated, everyone loves one-lines ;)
$yesterday = Date('Y-m-d', strtotime('-1 day'));
(Or if you are American you can randomize the date unit order to m/d/y (or whatever you use) and use Cups, galloons, feet and horses as units...)
As of PHP 7 you can write something like this:
$today = new \DateTime();
$yesterday = (clone $today)->modify('-1 day');
$dayBefore = (clone $yesterday)->modify('-1 day');
// Then call ->format('Y-m-d 00:00:00'); on each objects
you can also use new DateTime("now") for today new DateTime("1 day ago") for yesterday or all can be parse by strtotime php function.
Then format as you want.
$timeStamp = time();
// $timeStamp = time() - 86400;
if (date('d.m.Y', $timeStamp) == date('d.m.Y')) {
echo 'Today';
} elseif (date('d.m.Y', $time) == date('d.m.Y', strtotime('-1 day'))) {
echo 'Yesterday';
}
I want to add number of days to current date:
I am using following code:
$i=30;
echo $date = strtotime(date("Y-m-d", strtotime($date)) . " +".$i."days");
But instead of getting proper date i am getting this:
2592000
Please suggest.
This should be
echo date('Y-m-d', strtotime("+30 days"));
strtotime
expects to be given a string containing a US English date format and will try to parse that format into a Unix timestamp (the number of seconds since January 1 1970 00:00:00 UTC), relative to the timestamp given in now, or the current time if now is not supplied.
while date
Returns a string formatted according to the given format string using the given integer timestamp or the current time if no timestamp is given.
See the manual pages for
http://www.php.net/manual/en/function.strtotime.php
http://www.php.net/manual/en/function.date.php
and their function signatures.
This one might be good
function addDayswithdate($date,$days){
$date = strtotime("+".$days." days", strtotime($date));
return date("Y-m-d", $date);
}
$date = new DateTime();
$date->modify('+1 week');
print $date->format('Y-m-d H:i:s');
or print date('Y-m-d H:i:s', mktime(date("H"), date("i"), date("s"), date("m"), date("d") + 7, date("Y"));
$today=date('d-m-Y');
$next_date= date('d-m-Y', strtotime($today. ' + 90 days'));
echo $next_date;
You can add like this as well, if you want the date 5 days from a specific date :
You have a variable with a date like this (gotten from an input or DB or just hard coded):
$today = "2015-06-15"; // Or can put $today = date ("Y-m-d");
$fiveDays = date ("Y-m-d", strtotime ($today ."+5 days"));
echo $fiveDays; // Will output 2015-06-20
Keep in mind, the change of clock changes because of daylight saving time might give you some problems when only calculating the days.
Here's a little php function which takes care of that:
function add_days($date, $days) {
$timeStamp = strtotime(date('Y-m-d',$date));
$timeStamp+= 24 * 60 * 60 * $days;
// ...clock change....
if (date("I",$timeStamp) != date("I",$date)) {
if (date("I",$date)=="1") {
// summer to winter, add an hour
$timeStamp+= 60 * 60;
} else {
// summer to winter, deduct an hour
$timeStamp-= 60 * 60;
} // if
} // if
$cur_dat = mktime(0, 0, 0,
date("n", $timeStamp),
date("j", $timeStamp),
date("Y", $timeStamp)
);
return $cur_dat;
}
You could also try:
$date->modify("+30 days");
You can do it by manipulating the timecode or by using strtotime(). Here's an example using strtotime.
$data['created'] = date('Y-m-d H:i:s', strtotime("+1 week"));
You can use strtotime()
$data['created'] = date('Y-m-d H:m:s', strtotime('+1 week'));
I know this is an old question, but for PHP <5.3 you could try this:
$date = '05/07/2013';
$add_days = 7;
$date = date('Y-m-d',strtotime($date) + (24*3600*$add_days)); //my preferred method
//or
$date = date('Y-m-d',strtotime($date.' +'.$add_days.' days');
You could use the DateTime class built in PHP. It has a method called "add", and how it is used is thoroughly demonstrated in the manual: http://www.php.net/manual/en/datetime.add.php
It however requires PHP 5.3.0.
$date = "04/28/2013 07:30:00";
$dates = explode(" ",$date);
$date = strtotime($dates[0]);
$date = strtotime("+6 days", $date);
echo date('m/d/Y', $date)." ".$dates[1];
You may try this.
$i=30;
echo date("Y-m-d",mktime(0,0,0,date('m'),date('d')+$i,date('Y')));
Simple and Best
echo date('Y-m-d H:i:s')."\n";
echo "<br>";
echo date('Y-m-d H:i:s', mktime(date('H'),date('i'),date('s'), date('m'),date('d')+30,date('Y')))."\n";
Try this
//add the two day
$date = "**2-4-2016**"; //stored into date to variable
echo date("d-m-Y",strtotime($date.**' +2 days'**));
//print output
**4-4-2016**
Use this addDate() function to add or subtract days, month or years (you will need the auxiliar function reformatDate() as well)
/**
* $date self explanatory
* $diff the difference to add or subtract: e.g. '2 days' or '-1 month'
* $format the format for $date
**/
function addDate($date = '', $diff = '', $format = "d/m/Y") {
if (empty($date) || empty($diff))
return false;
$formatedDate = reformatDate($date, $format, $to_format = 'Y-m-d H:i:s');
$newdate = strtotime($diff, strtotime($formatedDate));
return date($format, $newdate);
}
//Aux function
function reformatDate($date, $from_format = 'd/m/Y', $to_format = 'Y-m-d') {
$date_aux = date_create_from_format($from_format, $date);
return date_format($date_aux,$to_format);
}
Note: only for php >=5.3
Use the following code.
<?php echo date('Y-m-d', strtotime(' + 5 days')); ?>
Reference has found from here - How to Add Days to Current Date in PHP
Even though this is an old question, this way of doing it would take of many situations and seems to be robust. You need to have PHP 5.3.0 or above.
$EndDateTime = DateTime::createFromFormat('d/m/Y', "16/07/2017");
$EndDateTime->modify('+6 days');
echo $EndDateTime->format('d/m/Y');
You can have any type of format for the date string and this would work.
//Set time zone
date_default_timezone_set("asia/kolkata");
$pastdate='2016-07-20';
$addYear=1;
$addMonth=3;
$addWeek=2;
$addDays=5;
$newdate=date('Y-m-d', strtotime($pastdate.' +'.$addYear.' years +'.$addMonth. ' months +'.$addWeek.' weeks +'.$addDays.' days'));
echo $newdate;
Do not use php's date() function, it's not as accurate as the below solution and furthermore it is unreliable in the future.
Use the DateTime class
<?php
$date = new DateTime('2016-06-06'); // Y-m-d
$date->add(new DateInterval('P30D'));
echo $date->format('Y-m-d') . "\n";
?>
The reason you should avoid anything to do with UNIX timestamps (time(), date(), strtotime() etc) is that they will inevitably break in the year 2038 due to integer limitations.
The maximum value of an integer is 2147483647 which converts to Tuesday, 19 January 2038 03:14:07 so come this time; this minute; this second; everything breaks
Source
Another example of why I stick to using DateTime is that it's actually able to calculate months correctly regardless of what the current date is:
$now = strtotime('31 December 2019');
for ($i = 1; $i <= 6; $i++) {
echo date('d M y', strtotime('-' . $i .' month', $now)) . PHP_EOL;
}
You'd get the following sequence of dates:
31 December
31 November
31 October
31 September
31 August
31 July
31 June
PHP conveniently recognises that three of these dates are illegal and converts them into its best guess, leaving you with:
01 Dec 19
31 Oct 19
01 Oct 19
31 Aug 19
31 Jul 19
01 Jul 19
Given a date MM-dd-yyyy format, can someone help me get the first day of the week?
Here is what I am using...
$day = date('w');
$week_start = date('m-d-Y', strtotime('-'.$day.' days'));
$week_end = date('m-d-Y', strtotime('+'.(6-$day).' days'));
$day contains a number from 0 to 6 representing the day of the week (Sunday = 0, Monday = 1, etc.).
$week_start contains the date for Sunday of the current week as mm-dd-yyyy.
$week_end contains the date for the Saturday of the current week as mm-dd-yyyy.
Very simple to use strtotime function:
echo date("Y-m-d", strtotime('monday this week')), "\n";
echo date("Y-m-d", strtotime('sunday this week')), "\n";
It differs a bit across PHP versions:
Output for 5.3.0 - 5.6.6, php7#20140507 - 20150301, hhvm-3.3.1 - 3.5.1
2015-03-16
2015-03-22
Output for 4.3.5 - 5.2.17
2015-03-23
2015-03-22
Output for 4.3.0 - 4.3.4
2015-03-30
2015-03-29
Comparing at Edge-Cases
Relative descriptions like this week have their own context. The following shows the output for this week monday and sunday when it's a monday or a sunday:
$date = '2015-03-16'; // monday
echo date("Y-m-d", strtotime('monday this week', strtotime($date))), "\n";
echo date("Y-m-d", strtotime('sunday this week', strtotime($date))), "\n";
$date = '2015-03-22'; // sunday
echo date("Y-m-d", strtotime('monday this week', strtotime($date))), "\n";
echo date("Y-m-d", strtotime('sunday this week', strtotime($date))), "\n";
Againt it differs a bit across PHP versions:
Output for 5.3.0 - 5.6.6, php7#20140507 - 20150301, hhvm-3.3.1 - 3.5.1
2015-03-16
2015-03-22
2015-03-23
2015-03-29
Output for 4.3.5 - 5.0.5, 5.2.0 - 5.2.17
2015-03-16
2015-03-22
2015-03-23
2015-03-22
Output for 5.1.0 - 5.1.6
2015-03-23
2015-03-22
2015-03-23
2015-03-29
Output for 4.3.0 - 4.3.4
2015-03-23
2015-03-29
2015-03-30
2015-03-29
strtotime('this week', time());
Replace time(). Next sunday/last monday methods won't work when the current day is sunday/monday.
Keep it simple :
<?php
$dateTime = new \DateTime('2020-04-01');
$monday = clone $dateTime->modify(('Sunday' == $dateTime->format('l')) ? 'Monday last week' : 'Monday this week');
$sunday = clone $dateTime->modify('Sunday this week');
?>
Source : PHP manual
NB: as some user commented the $dateTime value will be modified.
$givenday = date("w", mktime(0, 0, 0, MM, dd, yyyy));
This gives you the day of the week of the given date itself where 0 = Sunday and 6 = Saturday. From there you can simply calculate backwards to the day you want.
This question needs a good DateTime answer:-
function firstDayOfWeek($date)
{
$day = DateTime::createFromFormat('m-d-Y', $date);
$day->setISODate((int)$day->format('o'), (int)$day->format('W'), 1);
return $day->format('m-d-Y');
}
var_dump(firstDayOfWeek('06-13-2013'));
Output:-
string '06-10-2013' (length=10)
This will deal with year boundaries and leap years.
EDIT: the below link is no longer running on the version of PHP stated. It is running on PHP 5.6 which improves the reliability of strtotime, but isn't perfect! The results in the table are live results from PHP 5.6.
For what it's worth, here is a breakdown of the wonky behavior of strtotime when determining a consistent frame of reference:
http://gamereplays.org/reference/strtotime.php
Basically only these strings will reliably give you the same date, no matter what day of the week you're currently on when you call them:
strtotime("next monday");
strtotime("this sunday");
strtotime("last sunday");
Assuming Monday as the first day of the week, this works:
echo date("M-d-y", strtotime('last monday', strtotime('next week', time())));
The following code should work with any custom date, just uses the desired date format.
$custom_date = strtotime( date('d-m-Y', strtotime('31-07-2012')) );
$week_start = date('d-m-Y', strtotime('this week last monday', $custom_date));
$week_end = date('d-m-Y', strtotime('this week next sunday', $custom_date));
echo '<br>Start: '. $week_start;
echo '<br>End: '. $week_end;
I tested the code with PHP 5.2.17 Results:
Start: 30-07-2012
End: 05-08-2012
How about this?
$first_day_of_week = date('m-d-Y', strtotime('Last Monday', time()));
$last_day_of_week = date('m-d-Y', strtotime('Next Sunday', time()));
This is what I am using to get the first and last day of the week from any date.
In this case, monday is the first day of the week...
$date = date('Y-m-d') // you can put any date you want
$nbDay = date('N', strtotime($date));
$monday = new DateTime($date);
$sunday = new DateTime($date);
$monday->modify('-'.($nbDay-1).' days');
$sunday->modify('+'.(7-$nbDay).' days');
Here I am considering Sunday as first & Saturday as last day of the week.
$m = strtotime('06-08-2012');
$today = date('l', $m);
$custom_date = strtotime( date('d-m-Y', $m) );
if ($today == 'Sunday') {
$week_start = date("d-m-Y", $m);
} else {
$week_start = date('d-m-Y', strtotime('this week last sunday', $custom_date));
}
if ($today == 'Saturday') {
$week_end = date("d-m-Y", $m);
} else {
$week_end = date('d-m-Y', strtotime('this week next saturday', $custom_date));
}
echo '<br>Start: '. $week_start;
echo '<br>End: '. $week_end;
Output :
Start: 05-08-2012
End: 11-08-2012
How about this?
$day_of_week = date('N', strtotime($string_date));
$week_first_day = date('Y-m-d', strtotime($string_date . " - " . ($day_of_week - 1) . " days"));
$week_last_day = date('Y-m-d', strtotime($string_date . " + " . (7 - $day_of_week) . " days"));
Just use date($format, strtotime($date,' LAST SUNDAY + 1 DAY'));
Try this:
function week_start_date($wk_num, $yr, $first = 1, $format = 'F d, Y')
{
$wk_ts = strtotime('+' . $wk_num . ' weeks', strtotime($yr . '0101'));
$mon_ts = strtotime('-' . date('w', $wk_ts) + $first . ' days', $wk_ts);
return date($format, $mon_ts);
}
$sStartDate = week_start_date($week_number, $year);
$sEndDate = date('F d, Y', strtotime('+6 days', strtotime($sStartDate)));
(from this forum thread)
This is the shortest and most readable solution I found:
<?php
$weekstart = strtotime('monday this week');
$weekstop = strtotime('sunday this week 23:59:59');
//echo date('d.m.Y H:i:s', $weekstart) .' - '. date('d.m.Y H:i:s', $weekstop);
?>
strtotime is faster than new DateTime()->getTimestamp().
$monday = date('d-m-Y',strtotime('last monday',strtotime('next monday',strtotime($date))));
You have to get next monday first then get the 'last monday' of next monday. So if the given date is monday it will return the same date not last week monday.
$string_date = '2019-07-31';
echo $day_of_week = date('N', strtotime($string_date));
echo $week_first_day = date('Y-m-d', strtotime($string_date . " - " . ($day_of_week - 1) . " days"));
echo $week_last_day = date('Y-m-d', strtotime($string_date . " + " . (7 - $day_of_week) . " days"));
Given PHP version pre 5.3 following function gives you a first day of the week of given date (in this case - Sunday, 2013-02-03):
<?php
function startOfWeek($aDate){
$d=strtotime($aDate);
return strtotime(date('Y-m-d',$d).' - '.date("w",$d).' days');
}
echo(date('Y-m-d',startOfWeek("2013-02-07")).'
');
?>
$today_day = date('D'); //Or add your own date
$start_of_week = date('Ymd');
$end_of_week = date('Ymd');
if($today_day != "Mon")
$start_of_week = date('Ymd', strtotime("last monday"));
if($today_day != "Sun")
$end_of_week = date('Ymd', strtotime("next sunday"));
If you want Monday as the start of your week, do this:
$date = '2015-10-12';
$day = date('N', strtotime($date));
$week_start = date('Y-m-d', strtotime('-'.($day-1).' days', strtotime($date)));
$week_end = date('Y-m-d', strtotime('+'.(7-$day).' days', strtotime($date)));
A smart way of doing this is to let PHP handle timezone differences and Daylight Savings Time (DST). Let me show you how to do this.
This function will generate all days from Monday until Friday, inclusive (handy for generating work week days):
class DateTimeUtilities {
public static function getPeriodFromMondayUntilFriday($offset = 'now') {
$now = new \DateTimeImmutable($offset, new \DateTimeZone('UTC'));
$today = $now->setTime(0, 0, 1);
$daysFromMonday = $today->format('N') - 1;
$monday = $today->sub(new \DateInterval(sprintf('P%dD', $daysFromMonday)));
$saturday = $monday->add(new \DateInterval('P5D'));
return new \DatePeriod($monday, new \DateInterval('P1D'), $saturday);
}
}
foreach (DateTimeUtilities::getPeriodFromMondayUntilFriday() as $day) {
print $day->format('c');
print PHP_EOL;
}
This will return datetimes Monday-Friday for current week. To do the same for an arbitrary date, pass a date as a parameter to DateTimeUtilities ::getPeriodFromMondayUntilFriday, thus:
foreach (DateTimeUtilities::getPeriodFromMondayUntilFriday('2017-01-02T15:05:21+00:00') as $day) {
print $day->format('c');
print PHP_EOL;
}
//prints
//2017-01-02T00:00:01+00:00
//2017-01-03T00:00:01+00:00
//2017-01-04T00:00:01+00:00
//2017-01-05T00:00:01+00:00
//2017-01-06T00:00:01+00:00
Only interested in Monday, as the OP asked?
$monday = DateTimeUtilities::getPeriodFromMondayUntilFriday('2017-01-02T15:05:21+00:00')->getStartDate()->format('c');
print $monday;
// prints
//2017-01-02T00:00:01+00:00
You parse the date using strptime() and use date() on the result:
date('N', strptime('%m-%d-%g', $dateString));
<?php
/* PHP 5.3.0 */
date_default_timezone_set('America/Denver'); //Set apprpriate timezone
$start_date = strtotime('2009-12-15'); //Set start date
//Today's date if $start_date is a Sunday, otherwise date of previous Sunday
$today_or_previous_sunday = mktime(0, 0, 0, date('m', $start_date), date('d', $start_date), date('Y', $start_date)) - ((date("w", $start_date) ==0) ? 0 : (86400 * date("w", $start_date)));
//prints 12-13-2009 (month-day-year)
echo date('m-d-Y', $today_or_previous_sunday);
?>
(Note: MM, dd and yyyy in the Question are not standard php date format syntax - I can't be sure what is meant, so I set the $start_date with ISO year-month-day)
I've come against this question a few times and always surprised the date functions don't make this easier or clearer. Here's my solution for PHP5 that uses the DateTime class:
/**
* #param DateTime $date A given date
* #param int $firstDay 0-6, Sun-Sat respectively
* #return DateTime
*/
function getFirstDayOfWeek(DateTime $date, $firstDay = 0) {
$offset = 7 - $firstDay;
$ret = clone $date;
$ret->modify(-(($date->format('w') + $offset) % 7) . 'days');
return $ret;
}
Necessary to clone to avoid altering the original date.
Another way to do it....
$year = '2014';
$month = '02';
$day = '26';
$date = DateTime::createFromFormat('Y-m-d H:i:s', $year . '-' . $month . '-' . $day . '00:00:00');
$day = date('w', $date->getTimestamp());
// 0=Sunday 6=Saturday
if($day!=0){
$newdate = $date->getTimestamp() - $day * 86400; //86400 seconds in a day
// Look for DST change
if($old = date('I', $date->getTimestamp()) != $new = date('I', $newdate)){
if($old == 0){
$newdate -= 3600; //3600 seconds in an hour
} else {
$newdate += 3600;
}
}
$date->setTimestamp($newdate);
}
echo $date->format('D Y-m-d H:i:s');
The easiest way to get first day(Monday) of current week is:
strtotime("next Monday") - 604800;
where 604800 - is count of seconds in 1 week(60*60*24*7).
This code get next Monday and decrease it for 1 week. This code will work well in any day of week. Even if today is Monday.
I found this quite frustrating given that my timezone is Australian and that strtotime() hates UK dates.
If the current day is a Sunday, then strtotime("monday this week") will return the day after.
To overcome this:
Caution: This is only valid for Australian/UK dates
$startOfWeek = (date('l') == 'Monday') ? date('d/m/Y 00:00') : date('d/m/Y', strtotime("last monday 00:00"));
$endOfWeek = (date('l') == 'Sunday') ? date('d/m/Y 23:59:59') : date('d/m/Y', strtotime("sunday 23:59:59"));
Here's a one liner for the first day of last week, and the last day of last week as a DateTime object.
$firstDay = (new \DateTime())->modify(sprintf('-%d day', date('w') + 7))
->setTime(0, 0, 0);
$lastDay = (new \DateTime())->modify(sprintf('-%d day', date('w') + 1))
->setTime(23, 59, 59);
Just to note that timestamp math can also be a solution. If you have in mind that 01.jan 1970 was a Thursday, then start of a week for any given date can be calculated with:
function weekStart($dts)
{ $res = $dts - ($dts+date('Z',$dts)+259200)%604800;
return $res + 3600*(date('I',$dts)-date('I',$res));
}
It is predictable for any timestamp and php version, using date-func ('Z', 'I') only for timezone and daylight-saving offsets. And it produces same results as:
strtotime(date('Y-m-d', $dts).' - '.(date('N', $dts)-1.' days');
and with (the best and the most elegant) mentioned:
strtotime('monday this week', $dts);