I don't understand why strtotime is returning a date off by a few days.
$expiration = '2015-07-19'; // yyyy-mm-dd
$exp = strtotime($expiration);
$exp = date('F m, Y',$exp);
echo $exp; // returns "July 07, 2015" (NOT the 19th)
What am I missing?
UPDATE: Even if I do this:
echo date('F m, Y');
It says it's August 8 when today is August 21! Why!?!?!?
Here it is: Outputs July 19, 2015. See demo. You are using two characters to represent months in your date function. F and m are textual and numeric representations of a month. So change 'm' to 'd' to represent days. date("F d, Y") e.g. January 01, 2000
$expiration = '2015-07-19'; // yyyy-mm-dd
$exp = strtotime($expiration);
$exp = date('F d, Y',$exp);
echo $exp;
DEMO
Try this
$expiration = '2015-07-19'; // yyyy-mm-dd
$exp = strtotime($expiration);
//$exp = date('F m, Y',$exp);
$exp = date('jS F, Y',$exp);
Output
19th July, 2015
Related
I am currently displaying date in my php code like below
$date = $news_row['date'];
$newDate = date("l M d Y", strtotime($date));
its giving me output like
Monday Mar 08 2021
Instead I want output like
Monday Mar 08th 2021
I have found function
function ordinal($number) {
$ends = array('th','st','nd','rd','th','th','th','th','th','th');
if ((($number % 100) >= 11) && (($number%100) <= 13))
return $number. 'th';
else
return $number. $ends[$number % 10];
}
on this answer
But I am not getting idea how I can use it with my date format. Let me know if anyone here can help me for do it.
Thanks a lot!
$newDate = date('l M jS Y', strtotime($date));
Read about more datetime formats in PHP
https://www.php.net/manual/en/datetime.format.php
l => A full textual representation of the day of the week
M=> A short textual representation of a month, three letters,
j => Day of the month without leading zeros
S => English ordinal suffix for the day of the month, 2 characters
Y => A full numeric representation of a year, 4 digits
You can specify any character that you want:
$newDate = date("l M d\t\h Y", strtotime($date));
But if you want to get 1st, 2nd, 3rd, 4th ... you should use:
$newDate = date("l M jS Y", strtotime($date));
I want to implement a Javascript countdown timer that has the month value substracted by 1.
To get the date dynamically via PHP I use this code:
$date = "2014:3:19 00:00:00";
$newDate = date("Y, n, j, H, i", strtotime($date));
Which returns:
2014, 3, 9, 00
My question is how can I substract the value n by 1, so the final output will be always like this:
2014, (3-1), 9, 00
Here's the DateTime() way to do it (I used dashes instead of colons as that is the proper separator for date parts):
$date = "2014-3-19 00:00:00";
$date = (new DateTime($date))->modify('-1 month')->format("Y, n, j, H, i");
or
$date = "2014-3-19 00:00:00";
$date = (new DateTime($date))->diff(new DateInterval('P1M'))->format("Y, n, j, H, i");
If you mean minus one month, then you could do:
$date = "2014-3-19 00:00:00";
$newDate = date("Y, n, j, H, i", strtotime('-1 month', strtotime($date)));
And 2014, 1, 19, 00 will be 2013, 12, 19, 00 but not 2014, 0, 19, 00.
Update:
You want to pass a date to the jQuery plugin(jquery.magicbusmultimedia.net).
The plugin only ask you to pass a javascript Date object.
So you could do:
$('#myCounter').mbComingsoon(new Date(<?php echo strtotime($date); ?> * 1000));
$date = "2014-3-19 00:00:00";
$newDate = date("Y, n, j, H, i", strtotime('-1 month', strtotime($date)));
And 2014, 1, 19, 00 will be 2013, 12, 19, 00 but not 2014, 0, 19, 00.
This is weird, i'm trying to echo out a date like Monday, January 1, 2013, but its echoing out the wrong textual day. I don't have a clue why?
I have:
<?php echo date('l, F n, Y', strtotime($do['dueDate'])); ?>
And $do['dueDate'] is the date from the database of "2013-03-22". Its formatted as DATE in mysql.
When the above echos out it says: Friday, March 3, 2013
But march 3, 2013 is a sunday...
Try like this
echo date('l ,F j ,Y', strtotime($do['dueDate']));
Try this :
n --> Numeric representation of a month, without leading zeros --> 1 through 12
j --> Day of the month without leading zeros --> 1 to 31
<?php
$do['dueDate'] = "2013-03-22";
echo date('l, F j, Y', strtotime($do['dueDate']));
?>
Output :
Friday, March 22, 2013
use this
<?php
echo date('l, F j, Y', strtotime("2013-03-22")); // output Friday, March 22, 2013
working example http://codepad.viper-7.com/rF6w1U
http://php.net/manual/en/function.date.php
n is the numerical representation of the month.
try this:
$date = "2013-03-22";
echo date('l, F j, Y', strtotime($date));
outputs:
Friday, March 22, 2013
Is there anyway to convert the value of idate("z") to a date format that reads the Day, Month, and Year? My code looks like this:
$date_int = idate("z");
$date_text = strtotime($date_int);
$date = date("l, F j, Y", $date_text);
For some reason, it's still echoing Thursday, January 1, 1970.
Any ideas?
idate("z") is incorrect as that will return the day of the year. It seems like you want idate("U"), but in that case just use date() without the second parameter, it will assume time(). Example:
$date = date("l, F j, Y");
That should be all you need.
idate("z") will only return the day of the year. Today being 74. strtotime will not understand 74 as a parsing format. Therefore date() fails all together.
Assuming you want the date of the CURRENT YEAR:
$dayOfYear = 80;
print date("l, F j, Y", strtotime($dayOfYear - idate("z") . " day")); // Wednesday, March 21, 2012
$dayOfYear = 1;
print date("l, F j, Y", strtotime($dayOfYear - idate("z") . " day")) // Monday, January 2, 2012
How to find month no,name from week number using php
If you have the ISO week number, then to get the month (of the start of the week) you can use strtotime like:
// F = full name of month, n = month number without leading zero
echo date('F n', strtotime('2010-W50'));
Bear in mind that the ISO week might not be the same as your meaning of week, so read on.
If you want to count the whole weeks since January 1st of this year (regardless of what day of the week that is) then you could do as Adnan mentioned:
echo date('F n', strtotime('1 Jan + 50 weeks'));
echo date('F',strtotime('1 January 2010 +50 weeks'));
www.php.net/date
www.php.net/strtotime
Have a look at php date() - http://php.net/manual/en/function.date.php
Here are some good examples:
<?php
// set the default timezone to use. Available since PHP 5.1
date_default_timezone_set('UTC');
// Prints something like: Monday
echo date("l");
// Prints something like: Monday 8th of August 2005 03:12:46 PM
echo date('l jS \of F Y h:i:s A');
// Prints: July 1, 2000 is on a Saturday
echo "July 1, 2000 is on a " . date("l", mktime(0, 0, 0, 7, 1, 2000));
/* use the constants in the format parameter */
// prints something like: Mon, 15 Aug 2005 15:12:46 UTC
echo date(DATE_RFC822);
// prints something like: 2000-07-01T00:00:00+00:00
echo date(DATE_ATOM, mktime(0, 0, 0, 7, 1, 2000));
?>
$myDate = "2010-05-12";
$weekNumber = date("W", strtotime($myDate));
Just replace the "W" with the value you need. Full reference:
http://php.net/manual/en/function.date.php
If you have a week number, and want the date from it you can use:
date("d m Y", strtotime("1.1.2010 + 30 weeks"));