so currently I have 3 variables holding the day as an integer, month as a 3 letter representation of the month, and the year as 4 digits.
So for example
$day = '16';
$month = 'nov';
$year = '2013';
Now if I want to display it like
November 16th 2013
I was thinking I would have to generate a unix timestamp from my data, then use
date('F jS Y', $timestamp);
I'm not sure how to properly generate the unix timestamp though.
This will do what you want:
echo date('F jS Y', strtotime("$month $day $year"));
The key piece is the strtotime() function that will parse a date string into a timestamp for you.
Working example: http://3v4l.org/INV8d
Follow the OOP way , it's cool ;)
<?php
$day = '16';
$month = 'nov';
$year = '2013';
$newDate="$day-$month-$year";
$date_new = DateTime::createFromFormat('d-M-Y',$newDate);
echo $newformat=$date_new->format('F dS Y'); //November 16th 2013
?>
Related
I have a date $da = '18-Nov-2015'
I want month and year separate.I tried this.but it didn't work.
$month = date('F',strtotime($da));
$YEAR = date('Y',strtotime($da));
Give it try with below code:
$da = '18-Nov-2015';
$date = DateTime::createFromFormat('d-M-Y',$da);
echo $date->format("Y");
echo $date->format("F");
Note:DateTime We can create the object using arbitrary parameters like $date = DateTime::createFromFormat('d-M-Y', $weird_user_input); which can be formatted to unix timestamp or
whatever other date format We wish.
You can use the date_parse() function. It returns an array that contains the components of the date: day, month, year, hour, minute, and others.
Usage sample:
$da = '18-Nov-2015';
$dateComps = date_parse($da);
$year = $dateComps['year'];
$month = $dateComps['month'];
$day = $dateComps['day'];
//and so on, ...
use
$da=new DateTime($da);
$da->format('m-d');
hello try this hope you will get your answer what you want.
let me know if you got iy correct.
<?php
$dateValue = Date('Y-m-d');
$time=strtotime($dateValue);
$year=date("Y",$time);
$month=date("F",$time);
$date=date("d",$time);
?>
Almost you have done.This is how you can do it.
$year = date('Y', strtotime($da));
$month = date('m', strtotime($da));
F - A full textual representation of a month, such as January or March January through December
m - Numeric representation of a month, with leading zeros 01 through 12
M - A short textual representation of a month, three letters Jan through Dec
n - Numeric representation of a month, without leading zeros 1 through 12
Or else simply you can use explode method
$dateArray = explode('-', $da);
$dateArray[0] //date
$dateArray[1] //Month
$dateArray[2] //Year
Date reference : http://php.net/manual/en/function.date.php
i tried this in http://phpfiddle.org/
$da='18-Nov-2015';
$dateValue = strtotime($da);
$year = date('Y',$dateValue);
$monthName = date('F',$dateValue);
$day = date('d',$dateValue);
echo $monthName;
echo $year;
How would one go about converting a local name of the month, to the respective month number?
You could just make the translation manually, but I'm pretty sure PHP got some built in functions for this?
$monthname = "februar"; // norwegian name for february
$monthnumber = insertCorrectFunctionHere($monthname);
The goal is for $monthnumber to become '2' here (or '02'), since february is the second month.
Just use the date() function. Use the code below
<?php
$month_number = date('m', strtotime('1 February'));
echo $month_number; //Prints 02
Another way you can do it by making an array. Use the code below
<?php
$array = array("Januar"=>"01","Februar"=>"02","Mars"=>"03","April"=>"04","Mai"=>"05","Juni"=>"06","Juli"=>"07","August"=>"08","September"=>"09","Oktober"=>"10","November"=>"11","Desember");
$month_name = "February";
$month = ucwords(strtolower("Februar"));
$month_number = $array[$month];
echo $month_number; // Prints 02
Hope this helps you
First of all you need to convert the month to time format using strtotime function
The output of strtotime function can be passed to date function to generate the month in numerical format
you can use the following code
$date = 'january';
echo date('m', strtotime($date));
The output of the above code will be 01 refering to
try like this
$date = '1 February';
$month= date('m', strtotime($date));
echo $month;
I want to find the next occurrence of date from the current date.
For example, imagine I want to find 20th of the month from current date
If current date is 10th october then return the result 2014-10-20 (Y-m-d)
If current date is 22nd october then return the result 2014-11-20 (Y-m-d)
I created a solution just now using a while loop.
$oldd= "2014-06-20";
$newdate = date("Y-m-d",strtotime($oldd."+1months"));
while(strtotime($newdate) <= strtotime(date("Y-m-d")))
{
$newdate = date("Y-m-d",strtotime($newdate."+1months"));
}
echo $newdate;
manually and pass that to strtotime(). The time information you need to extract from the reference time string. Like this:
$refdate = '2014-02-25 10:30:00';
$timestamp = strtotime($refdate);
echo date('Y-m-d H:i:s',
strtotime("next Thursday " . date('H:i:s', $timestamp), $timestamp)
);
same results could be achieved using string concatenation:
echo date('Y-m-d', strtotime("next Thursday", $timestamp)
. ' ' . date('H:i:s', $timestamp);
another way, you can use methods of DateTime object, PHP has really rich API in dealing with date time.
$current_date = new DateTime('2014-06-20');
if ($current_date->format('d') >= 20) {
// $current_date->modify('last day of this month')->modify("+20 days");
$current_date->modify('first day of next month')->modify("+19 days");
}else{
$current_date->modify('first day of this month')->modify("+19 days");
}
echo $current_date->format("Y-m-d");
http://php.net/manual/en/datetime.modify.php
http://php.net/manual/en/datetime.formats.relative.php
So, in PHP i'm trying to return a date value for last year based on the same day of week.
EX: (Monday) 2011-12-19 inputted should return (Monday) 2010-12-20.
I was just doing it simply by -364 but then that was failing on leap years. I came across another function :
$newDate = $_POST['date'];
$newDate = strtotime($newDate);
$oldDate = strtotime('-1 year',$newDate);
$newDayOfWeek = date('w',$oldDate);
$oldDayOfWeek = date('w',$newDate);
$dayDiff = $oldDayOfWeek-$newDayOfWeek;
$oldDate = strtotime("$dayDiff days",$oldDate);
echo 'LAST YEAR DAY OF WEEK DATE = ' . date('Ymd', $oldDate);
however, that is failing when you try to input a Sunday date, as it does a 0 (sunday) minus 6 (saturday of last year date), and returns with a value T-6. IE inputting 2011-12-25 gets you 2010-12-19 instead of 2011-12-26.
I'm kind of stumped to find a good solution in php that will work for leap years and obviously all days of the week.
Any suggestions?
Thanks!
How about this, using PHP's DateTime functionality:
$date = new DateTime('2011-12-25'); // make a new DateTime instance with the starting date
$day = $date->format('l'); // get the name of the day we want
$date->sub(new DateInterval('P1Y')); // go back a year
$date->modify('next ' . $day); // from this point, go to the next $day
echo $date->format('Ymd'), "\n"; // ouput the date
$newDate = '2011-12-19';
date_default_timezone_set('UTC');
$newDate = strtotime($newDate);
$oldDate = strtotime('last year', $newDate);
$oldDate = strtotime(date('l', $newDate), $oldDate);
$dateFormat = 'Y-m-d l w W';
echo "This date: ", date($dateFormat, $newDate), "\n";
echo "Old date : ", date($dateFormat, $oldDate);
That gives:
This date: 2011-12-19 Monday 1 51
Old date : 2010-12-20 Monday 1 51
Use strtotime() to get a date, for the same week last year.
Use the format {$year}-W{$week}-{$weekday}, like this:
echo date("Y-m-d", strtotime("2010-W12-1"));
And you can do that for as long back you wan't:
<?php
for($i = 2011; $i > 2000; $i--)
echo date("Y-m-d", strtotime($i."-W12-1"));
?>
Make it easier :)
echo date('Y-m-d (l, W)').<br/>;
echo date('Y-m-d (l, W)', strtotime("-52 week"));
Edit: I forgot to write output: :)
2015-05-06 (Wednesday, 19)
2014-05-07 (Wednesday, 19)
<?php
$date = "2020-01-11";
$newdate = date("Y-m-d",strtotime ( '-1 year' , strtotime ( $date ) )) ;
echo $newdate;
?>
ref https://www.nicesnippets.com/blog/how-to-get-previous-year-from-date-in-php
I want to get last month's date. I wrote this out:
$prevmonth = date('M Y');
Which gives me the current month/year. I can't tell if I should use strtotime, mktime. Something to the timestamp? Do I need to add something afterwards to reset so that the date isn't set to last month throughout everything for all timestamps on my site? I'm trying to RTM but it's hard for me to figure this out.
It's simple to get last month date
echo date("Y-n-j", strtotime("first day of previous month"));
echo date("Y-n-j", strtotime("last day of previous month"));
at November 3 returns
2014-10-1
2014-10-31
echo strtotime("-1 month");
That will output the timestamp for last month exactly. You don't need to reset anything afterwards. If you want it in an English format after that, you can use date() to format the timestamp, ie:
echo date("Y-m-d H:i:s",strtotime("-1 month"));
$prevmonth = date('M Y', strtotime("last month"));
Incorrect answers are:
$lastMonth = date('M Y', strtotime("-1 month"));
$lastDate = date('Y-m', strtotime('last month'));
The reason is if current month is 30+ days but previous month is 29 and less $lastMonth will be the same as current month.
e.g.
If $currentMonth = '30/03/2016';
echo $lastMonth = date('m-Y', strtotime("-1 month")); => 03-2016
echo $lastDate = date('Y-m', strtotime('last month')); => 2016-03
The correct answer will be:
echo date("m-Y", strtotime("first day of previous month")); => 02-2016
echo sprintf("%02d",date("m")-1) . date("-Y"); => 02-2016
echo date("m-Y",mktime(0,0,0,date("m")-1,1,date("Y"))); => 02-2016
echo date('Y',strtotime("-1 year")); //last year<br>
echo date('d',strtotime("-1 day")); //last day<br>
echo date('m',strtotime("-1 month")); //last month<br>
Found this one wrong when the previous months is shorter than current.
echo date("Y-m-d H:i:s",strtotime("-1 month"));
Try on March the 30th and you will get 2012-03-01 instead of 2012-02...
Working out on better solution...
if you want to get just previous month, then you can use as like following
$prevmonth = date('M Y', strtotime('-1 months'));
if you want to get same days of previous month, Then you can use as like following ..
$prevmonth = date('M Y d', strtotime('-1 months'));
if you want to get last date of previous month , Then you can use as like following ...
$prevmonth = date('M Y t', strtotime('-1 months'));
if you want to get first date of previous month , Then you can use as like following ...
$prevmonth = date('M Y 1', strtotime('-1 months'));
public function getLastMonth() {
$now = new DateTime();
$lastMonth = $now->sub(new DateInterval('P1M'));
return $lastMonth->format('Ym');
}
Use this short code to get previous month for any given date:
$tgl = '25 january 2012';
$prevmonth = date("M Y",mktime(0,0,0,date("m", strtotime($tgl))-1,1,date("Y", strtotime($tgl))));
echo $prevmonth;
The result is December 2011.
Works on a month with shorter day with previous month.
$lastMonth = date('M Y', strtotime("-1 month"));
var_dump($lastMonth);
$lastMonth = date('M Y', mktime(0, 0, 0, date('m') - 1, 1, date('Y')));
var_dump($lastMonth);
Simply get last month.
Example:
Today is: 2020-09-02
Code:
echo date('Y-m-d', strtotime(date('Y-m-d')." -1 month"));
Result:
2020-08-02
You can use strtotime, which is great in this kind of situations :
$timestamp = strtotime('-1 month');
var_dump(date('Y-m', $timestamp));
Will get you :
string '2009-11' (length=7)
$time = mktime(0, 0, 0, date("m"),date("d")-date("t"), date("Y"));
$lastMonth = date("d-m-Y", $time);
OR
$lastMonth = date("m-Y", mktime() - 31*3600*24);
works on 30.03.2012
Oh I figured this out, please ignore unless you have the same problem i did in which case:
$prevmonth = date("M Y",mktime(0,0,0,date("m")-1,1,date("Y")));
This question is quite old but here goes anyway. If you're trying to get just the previous month and the day does not matter you can use this:
$date = '2014-01-03';
$dateTime = new DateTime($date);
$lastMonth = $dateTime->modify('-' . $dateTime->format('d') . ' days')->format('F Y');
echo $lastMonth; // 'December 2013'
The best solution I have found is this:
function subtracMonth($currentMonth, $monthsToSubtract){
$finalMonth = $currentMonth;
for($i=0;$i<$monthsToSubtract;$i++) {
$finalMonth--;
if ($finalMonth=='0'){
$finalMonth = '12';
}
}
return $finalMonth;
}
So if we are in 3(March) and we want to subtract 5 months that would be
subtractMonth(3,5);
which would give 10(October). If the year is also desired, one could do this:
function subtracMonth($currentMonth, $monthsToSubtract){
$finalMonth = $currentMonth;
$totalYearsToSubtract = 0;
for($i=0;$i<$monthsToSubtract;$i++) {
$finalMonth--;
if ($finalMonth=='0'){
$finalMonth = '12';
$totalYearsToSubtract++;
}
}
//Get $currentYear
//Calculate $finalYear = $currentYear - $totalYearsToSubtract
//Put resulting $finalMonth and $finalYear into an object as attributes
//Return the object
}
It works for me:
Today is: 31/03/2012
echo date("Y-m-d", strtotime(date('m', mktime() - 31*3600*24).'/01/'.date('Y').' 00:00:00')); // 2012-02-01
echo date("Y-m-d", mktime() - 31*3600*24); // 2012-02-29
If you want to get first date of previous month , Then you can use as like following ... $prevmonth = date('M Y 1', strtotime('-1 months')); what? first date will always be 1 :D