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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
How can I get the last day (Dec 31) of the current year as a date using PHP?
I tried the following but this doesn't work:
$year = date('Y');
$yearEnd = strtotime($year . '-12-31');
What I need is a date that looks like 2014-12-31 for the current year.
PHP strtotime() uses the current date/time as a basis (so it will use this current year), and you need date() to format:
$yearEnd = date('Y-m-d', strtotime('Dec 31'));
//or
$yearEnd = date('Y-m-d', strtotime('12/31'));
You can just concatenate actual year with required date
$year = date('Y') . '-12-31';
echo $year;
//output 2014-12-31
DateTime is perfectly capable of doing this:
$endOfYear = new \DateTime('last day of December this year');
You can even combine it with more modifiers to get the end of next year:
$endOfNextYear = new \DateTime('last day of December this year +1 years');
Another way to do this is use the Relative Formats of strtotime()
$yearEnd = date('Y-m-d', strtotime('last day of december this year'));
// or
$yearEnd = date('Y-m-d', strtotime('last day of december'));
How can I calculate the day of month in PHP with giving month, year, day of week and number of week.
Like, if I have September 2013 and day of week is Friday and number of week is 2, I should get 6. (9/6/2013 is Friday on the 2nd week.)
One way to achieve this is using relative formats for strtotime().
Unfortunately, it's not as straightforward as:
strtotime('Friday of second week of September 2013');
In order for the weeks to work as you mentioned, you need to call strtotime() again with a relative timestamp.
$first_of_month_timestamp = strtotime('first day of September 2013');
$second_week_friday = strtotime('+1 week, Friday', $first_of_month_timestamp);
echo date('Y-m-d', $second_week_friday); // 2013-09-13
Note: Since the first day of the month starts on week one, I've decremented the week accordingly.
I was going to suggest to just use strtotime() in this fashion:
$ts = strtotime('2nd friday of september 2013');
echo date('Y-m-d', $ts), PHP_EOL;
// outputs: 2013-09-13
It seems that this is not how you want the calendar to behave? But it is following a (proper) standard :)
This way its a little longer and obvious but it works.
/* INPUT */
$month = "September";
$year = "2013";
$dayWeek= "Friday";
$week = 2;
$start = strtotime("{$year}/{$month}/1"); //get first day of that month
$result = false;
while(true) { //loop all days of month to find expected day
if(date("w", $start) == $week && date("l", $start) == $dayWeek) {
$result = date("d", $start);
break;
}
$start += 60 * 60 * 24;
}
var_dump($result); // string(2) "06"
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
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