I have a variable
$date
Which contains a date in the following format:
'Monday, 24 August 2015'
Or as a strtotime:
'1440716400'
Id like to change it to the following format:
'24/08/2015'
How can I do this please?
Simply date()
$timestamp = strtotime('Monday, 24 August 2015');
echo date("d/m/Y", $timestamp); // Here you put the format you want
Try
date("d/m/Y", strtotime('Monday, 24 August 2015'));
Related
I have a query that inserts date in this format
$time = date("m-d-y");
But when I Fetch and output date it shows wrong date. real: October 2020 but outputs: January 1970
$time = $row1['time'];
$newDate = date('F Y', strtotime($time));
echo $newDate;
How do I output date in this format: 20 OCT, 2020
Try not to store date/time using varchar datatype. If you can change it, please do. However, if you can't change database structure, you can use date_create_from_format() to create a DateTime object from a custom format:
echo date_create_from_format('m-d-y', "10-29-20")->format('F Y');
Output:
October 2020
Edit: Change the format part to ->format('d M, Y') to match your desired format
Output:
29 Oct, 2020
I have a variable holding time in the format Apr 25, 2017 12:00:00 AM I wish to convert it to 2017-04-25 .My result should be without time
You can convert your string formatted date, to timestamp with strtotime() function. Then you can parameter date() function with converted timestamp, and format it with first parameter.
More about formatting:
http://php.net/manual/en/function.date.php
print date('Y-m-d',strtotime('Apr 25, 2017 12:00:00'));
You need to check the date method of PHP
$date = 'May 25, 2017 12:00:00 AM';
$date = date('Y-m-d', strtotime($date));
echo $date;
Try DateTime class:
$datestr = 'Apr 25, 2017 12:00:00 AM';
$date = DateTime::createFromFormat('M d, Y h:i:s A', $datestr);
echo $date->format('Y-m-d');
output
2017-04-25
Use the date() function:
$old_date="Apr 25, 2017 12:00:00 AM";
$new_date=date('Y-m-d', strtotime($old_date));
echo $new_date;
I'd like to use the datetime->modify function on a date string that's formatted like "21 Jan 2016". When I use the datetime->modify and add 1 day, it gives me a result of 30 Apr 2017. I know that if I don't use the short month name and use a number instead (i.e. 01), it will work fine but I would like to get it work this way with short month name. Is this possible?
Please see code below:
<?php
$date = "21 Jan 2016"; // this is my date string
$newdate = new DateTime($date );
$date2 = $newdate->modify('+1 day'); // add 1 day to date string
echo $date2->format("d-M-Y");
?>
RESULT is:
30-Apr-2017
RESULT WANTED
22-Jan-2016
The problem is that you are trying to create a DateTime object from a non-ISO format. That's that part that is not working.
Take a look at: http://php.net/manual/ro/datetime.createfromformat.php
You will need to have something like
DateTime::createFromFormat('d M Y', '21 Jan 2016');
Full example:
$tomorrow = DateTime::createFromFormat('d M Y', '21 Jan 2016')->modify('+1 day')->format("d-M-Y");
echo($tomorrow);
The format of the $date variable is incorrect. Off the top of my head, there are two easy ways to fix this:
Set $date = "Jan 21, 2016"
Set $date = "21-Jan 2016"
More options: https://secure.php.net/manual/en/datetime.formats.date.php
Your date format was wrong. That's all.
I can get the for example 19 March of specific date with this code:
$date = strtotime(" 19 March", $current_time);
For example if I gave the unix timestamp of 1st of January of 2010 as an input, It gave me 19 March of 2010. But also if I gave the unix timestamp of 20 March of 2010,I still get 19 March 2010. What I want is to get the next 19 March which in this case, It would be 19 March of 2011.
How can I do that?
Using PHP DateTime this can be achieved as follows:
// New DateTime object
$date = new DateTime('2010-03-19');
// Add a year
$date->add(new DateInterval('P1Y'));
// Output timestamp
echo $date->getTimestamp();
You can do something like as
$get = "19 March";
$given_date = "01 January 2010";
$date_month = date('d F',strtotime($given_date));
$year = date('Y',strtotime($given_date));
if(strtotime($given_date) - strtotime($date_month) < 0){
echo date('l,d F Y',strtotime("$get $year"));
}else{
echo date('l,d F Y',strtotime("$get ".($year+1)));
}
You should first get year from specified date. Then after you can create 19 march date with year and use strtotime() to get timestamp.
//add format according to your current_time variable format
$date = DateTime::createFromFormat("Y-m-d", $current_time);
echo $date->format("Y");
$fixed_date = strtotime($date->format("Y")."-03-19");
You can specify how many days or week you want to add or subtract from a day, as well as set the time with these functions
$nextUpdate = new DateTime("+5 day 1:00 pm");
echo $nextUpdate->getTimestamp();
$nextWeek = new DateTime("+1 week 9:00 am");
echo $nextWeek->getTimestamp();
I need to find a date after a certain number of weeks. For example, if the start date is Saturday, 31 October 2009, and if I choose 16 weeks then I need to find the date after sixteen Saturdays.
Thanks in advance.
You can use strtotime:
// Oct 31 2009 plus 16 weeks
echo date('Y-m-d', strtotime('2009-10-31 +16 week')); // outputs 2010-02-20
// next week
echo date('Y-m-d', strtotime('+1 week')); // outputs 2009-11-07
strtotime() is what you want: it takes an expression and turns it into a date object. Also see date():
$date = '31 october 2009';
$modification = '+ 16 weeks';
echo date('Y-m-d (l)', strtotime("$date")); //2009-10-31 (Saturday)
echo date('Y-m-d (l)', strtotime("$date $modification")); // 2010-02-20 (Saturday)
If you're using PHP 5.2.0 or later, you can use the DateTime class
<?php
$date = new DateTime('31 October 2009'); // This accepts any format that strtotime() does.
// Now add 16 weeks
$date->modify('+16 weeks');
// Now you can output it however you wish
echo $date->format('Y-m-d');
?>
Why not just use unix time, subtract 7 * 24 * 3600 * 16 from that and then convert that back into the date that you need.
This will help with the conversion back:
http://php.net/manual/en/function.date.php