I want to always echo out the date 3 days from now. So right now I have:
$date = date("l F jS");
echo $date;
Which echos "Friday June 5th"
What exactly do I do so that it echos out "Monday June 8th" and tomorrow it echos out "Tuesday June 9th" (always 3 days ahead).
You can use strtotime() with a relative date format. When you pass a Unix timestamp as the second parameter to date() it will format that date.
$date = date("l F jS", strtotime('+3 days'));
echo $date;
Or if you prefer OOP use DateTime(). With DateTime() you can put the relative date format right into its constructor. It also handles things like daylight savings time which may come into play depending on what you're doing.
$date = new DateTime('+3 days');
echo $date->format("l F jS");
Related
I have a feed which gives feed in the following format: "Fri 14 Oct"
I want to see if today's date matches the date from the feed. My problem is the format of today's date/
$today = date("d m");
This outputs 17 10.
What is the best way to format $today so that it outputs Day (shorthand) space date (number) Month (shorthand) ?
how about:
$today = date("D j M");
As explained in date() reference manual.
Anyway you should be aware of timezone issues unless you are 100% sure that your server is in the same timezone of the feed you are comparing.
I would follow a different approach though, you can parse the feed's date using DateTime::createFromFormat() which also understand timezones, and then compare it with today's date.
$today = date("D d M");
PHP Date Documentation
<?php
// Prints the day
echo date("l") . "<br>";
// Prints the day, date, month, year, time, AM or PM
echo date("l jS \of F Y h:i:s A");
?>
For more details, please visit http://www.w3schools.com/php/func_date_date.asp
Can anybody tell me why strtotime() seems to be adding 1 day? This seems to only happen in the late afternoon (something like 7 or 8 PM), otherwise it says the correct day.
echo date('m/d/Y h:i:s a', time());
Output:
12/21/2015 08:34:43 pm
echo gmdate('l, F jS, Y', strtotime(date('m/d/Y h:i:s a', time())));
Output:
Tuesday, December 22nd, 2015
I would like the above output, however, I want today's date (the 21st not the 22nd).
Use date instead of gmdate.
You are using gmdate() which gets the date in UTC. The problem only happens late in the afternoon/evening because at those times it really is the next day in UTC time.
You're also doing too much work - you can simplify that line of code to this:
// echo gmdate('l, F jS, Y', strtotime(date('m/d/Y h:i:s a', time())));
echo date('l, F jS, Y');
Otherwise you've created a timestamp from a time string based on the current time stamp. You could just leave the second parameter to date empty and the current time "now" is assumed.
It is also very important to make sure you are calling date_default_timezone_set somewhere or that you have it configured in your php.ini.
This detail in your code...
echo gmdate('l, F jS, Y', strtotime(date('m/d/Y h:i:s a', time())));
(= the "gmdate") will always return Greenwich Mean Time (GMT), which is London/UK.
So change that to date(....
And add date_default_timezone_set('America/New_York'); anyway...
Decided to ultimately use:
$date = new DateTime(date('Y-m-d'), new DateTimeZone('America/New_York'));
$timestamp = $date->format('U');
$date = gmdate('l, F jS, Y', $timestamp);
based on Alexander's comment.
I just hope this question won't be marked as a duplicate because I've seen similar questions on stackoverflow but they all talk about adding days to the date, The problem here is that i want to add some particular months to a particular date which is gotten from my database I've tried adding it using strtotime() but the date just returns 1st January 1970, the code looks like this
<?php echo date('jS F Y', strtotime("$date +1 month")); ?>
//This is the value of date
$date = $student->date;
How to I add months to this particular date? Please note that the date is a timestamp in my database.Thanks
You have a Unix timestamp, not an actual date. Here I use the DateTime class to create a datetime object using that Unix timestamp. Then I can add a month to it and format the output.
$date = new DateTime('#'.$student->date);
$date->modify('+1 month');
echo $date-format('jS F Y');
If you want to stick to using date() and strtotime() you would use this:
echo date("jS F Y", strtotime("+1 month", $student->date));
strtotime() would take the starting date as the second parameter and then how you wish to modify it as your first parameter.
You should check out the documentation here,
But the just of it is the $date->add function. It allows you to add any amount of time to a timestamp using a DateInterval. Its a little tricky to get used to but here are a couple of examples:
<?php
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('PT10H30S'));
echo $date->format('Y-m-d H:i:s') . "\n";
$date = new DateTime('2000-01-01');
$date->add(new DateInterval('P7Y5M4DT4H3M2S'));
echo $date->format('Y-m-d H:i:s') . "\n";
?>
which outputs:
2000-01-01 10:00:30
2007-06-05 04:03:02
The date interval is formatted in years months days hours minuets seconds, simply put in the amount you want and it will add it, so in your case:
<?php
$date->add(new DateInterval('P1M'));
echo $date->format('Y-m-d H:i:s') . "\n";
?>
PHP's strtotime() function allows for a second parameter that allows you to set a relative date.
If you would like to add a month to tomorrow, here's how:
<?php
echo date("jS F Y", strtotime("+1 month", strtotime("2014-10-09")));
// returns: 9th November 2014
I have a string which is in the format: dd/mm/yy
e.g. 29/03/14
But when I print it using date() I get a completely different date!
What am I missing?
$endDate = "29/03/14";
echo date("jS F, Y", strtotime( $endDate ));
1st January, 1970
I even tried:
echo date("jS F, Y", strtotime( trim($endDate) ));
With no luck!
I am reading $endDate from a text file
What I am trying to do is find out if it is the last day of the month...
i.e. Current month is 03 - March has 31 days
The day of month in file is 29
This is not the last day of the month
Your date format is not valid.
Check the Date Formats
You might consider using date_create_from_format('d/m/y','29/03/14'); and work with the DateTime object.
Use DateTime::createFromFormat
$date = DateTime::createFromFormat('d/m/y', '29/03/14');
echo $date->format("jS F, Y");
For getting the last day
echo $date->format('t');
Your date format isn't recognized by strtotime(). Use DateTime class instead:
$dateObj = DateTime::createFromFormat('d/m/y', $str);
echo $dateObj->format('jS F, Y');
Output:
29th March, 2014
Demo
What I am trying to do is find out if it is the last day of the month...
For that, you just have to check if the given date is the same as the last day of the month (which can be obtained with t format character)
$str = '29/03/14';
$dateObj = DateTime::createFromFormat('d/m/y', $str);
if ($dateObj->format('d') == $dateObj->format('t')) {
echo 'Given date is the last day of the month';
}
How can I get what date it will be after 31 days starting with $startDate, where $startDate is a string of this format: YYYYMMDD.
Thank you.
strtotime will give you a Unix timestamp:
$date = '20101007';
$newDate = strtotime($date.' + 31 days');
you can then use date to format that into the same format, if that's what you need:
echo date('Ymd', $newDate);
If you're using PHP 5.3:
$date = new DateTime('20101007');
$date->add(new DateInterval('P31D'));
echo $date->format('Y-m-d');
The pre-5.3 date functions are lacking, to say the least. The DateTime stuff makes it much easier to deal with dates. http://us3.php.net/manual/en/book.datetime.php
Just a note that +1 month will also work if you want the same date on the next month and not 31 days exactly each time.
echo date('Y m d',strtotime('+31 Days'));