I am passing a date in a URL in a UK format as per the following:
http://www.website.com/index.php?from_date=01/04/2013&to_date=12/04/2013
The date range is 1st April 2013 to 12th April 2013.
Then I am using the following PHP to convert it to the YYYY-MM-DD format.
<?php
$rpt_from_date = date("Y-m-d", strtotime($_GET["from_date"]) );
$rpt_to_date = date("Y-m-d", strtotime($_GET["to_date"]) );
echo $rpt_from_date;
echo "</br>";
echo $rpt_to_date;
?>
For some reason this is not working. The above returns the following:
2013-01-04
2013-12-04
It's switching the month and day around. I want it to return:
2013-04-01
2013-04-12
Does anyone have any ideas?
Use DateTime object, to get php understand in which format you passing date to it.
$rpt_from_date = DateTime::createFromFormat('d/m/Y', $_GET["from_date"]);
echo $rpt_from_date->format('Y-m-d');
PHP is reading your time string in the US format (MM/DD/YYYY) because you are using slashes. You could use dashes to give the time: index.php?from_date=01-04-2013&to_date=12-04-2013, or convert it yourself:
$uktime = implode("-", explode("/", $_GET['from_date']));
I will provide the solution but it is not using the date function:
$arr = explode("/",$_GET["from_date"]);
$from_date = $arr[2]."-".$arr[1]."-"$arr[0];
Second solution is as following:
$from_date = implode(array_reverse(explode("/",$_GET["from_date"])));
Related
I have a string: 30/06/18 (30th June 2018)
I am converting to a date:
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18')->format('d-m-Y');
echo $calcFieldDate;
Result: 18-06-2018
Now I want to add 20 days to the date:
$expiryDate = date("d-m-Y", strtotime("+20 days", $calcFieldDate));
echo $expiryDate;
Expected Result: 08-07-2018
Actual Result: 31-01-1970
I am obviously creating a date format which is then subsequently being treated as a string...
Every time I try a conversion, I just hit another road block - is there anyway to create a date that is then treated like a date?
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18')->format('d-m-Y');
echo $calcFieldDate;
Result:30-06-2018
$expiryDate = date("d-m-Y", strtotime("+20 days", strtotime($calcFieldDate)));
echo $expiryDate;
Result:20-07-2018
Strtotime() The second parameter is the timestamp
You actually don't need to revert using strtotime and date functions, you can actually use DateTime to simply add dates into it:
$calcFieldDate = date_create_from_format('d/m/y', '30/06/18');
echo $calcFieldDate->format('d-m-Y'); // get inputted date
$expiryDate = clone $calcFieldDate; // clone the original date object
$expiryDate->modify('+20 days'); // adjust the cloned date
echo $expiryDate->format('d-m-Y'); // show the adjusted date
This will sort your problem.
$str="30/06/18 (30th June 2018)";
$arr_temp=explode(" ",$str);
$str_date=str_replace("/","-",$arr_temp[0]);
$dt = DateTime::createFromFormat('d-m-y',$str_date);
$date=$dt->format('d-m-Y');
$new_date=date('d-m-Y',strtotime("+20 days",strtotime($date)));
echo $new_date;
I have much data with several timestamps and I just recognized that some are in "dd.mm.YYYY" which works very well with date("Y-m-d", strtotime($input)); but some are in "dd.mm.YY" and this does not work anymore - it always returns the current date.
My problem is that my data is too huge to fix this problem manually by editting. Is there any way to get the YYYY-mm-dd out of mm.dd.YY ?
Here you go...
$date = "20.02.71"; // sample date... (common German format)
$date = DateTime::createFromFormat('d.m.y', $date);
echo $date->format('Y-m-d');
will result in:
1971-02-20
Create a DateTime object, then format it to anything you want...
Well you can replace the . by -, you could do something like the following:
$date = str_replace(".", "-", "mm.dd.YY")
This would return
mm-dd-YY
You could use date_parse_from_format which would convert any formate into the formate you specify.
date_parse_from_format("y-m-d", $date);
It returns an array with very useful information like month, year etc.
Suppose the current year is 2014; then I want get the date 2014-06-30, if the current year is 2015 then I want 2015-06-30
I tried date('Y'); but it is just giving current year.
You can just use:
date("Y-06-30")
Demo : https://eval.in/103028
try:
echo date("Y-06-30");
OUTPUT:
2014-06-30
See date documentation:
http://php.net/manual/en/function.date.php
You can concatenate the rest of your required date to your date() function as follow
echo date('Y') . '-06-30';
//output 2014-06-30
Use this :
date("Y-06-30");
Y will be the your year and rest will same.
Where y is A full numeric representation of a year, 4 digits for more Info : Date()
You need to put like below code:
$d = date('Y');
$data = $d."-06-30";
echo $data;
I have a PHP date in a database, for example 8th August 2011. I have this date in a strtotime() format so I can display it as I please.
I need to adjust this date to make it 8th August 2013 (current year). What is the best way of doing this? So far, I've been racking my brains but to no avail.
Some of the answers you have so far have missed the point that you want to update any given date to the current year and have concentrated on turning 2011 into 2013, excluding the accepted answer. However, I feel that examples using the DateTime classes are always of use in these cases.
The accepted answer will result in a Notice:-
Notice: A non well formed numeric value encountered......
if your supplied date is the 29th February on a Leapyear, although it should still give the correct result.
Here is a generic function that will take any valid date and return the same date in the current year:-
/**
* #param String $dateString
* #return DateTime
*/
function updateDate($dateString){
$suppliedDate = new \DateTime($dateString);
$currentYear = (int)(new \DateTime())->format('Y');
return (new \DateTime())->setDate($currentYear, (int)$suppliedDate->format('m'), (int)$suppliedDate->format('d'));
}
For example:-
var_dump(updateDate('8th August 2011'));
See it working here and see the PHP manual for more information on the DateTime classes.
You don't say how you want to use the updated date, but DateTime is flexible enough to allow you to do with it as you wish. I would draw your attention to the DateTime::format() method as being particularly useful.
strtotime( date( 'd M ', $originaleDate ) . date( 'Y' ) );
This takes the day and month of the original time, adds the current year, and converts it to the new date.
You can also add the amount of seconds you want to add to the original timestamp. For 2 years this would be 63 113 852 seconds.
You could retrieve the timestamp of the same date two years later with strtotime() first parameter and then convert it in the format you want to display.
<?php
$date = "11/08/2011";
$time = strtotime($date);
$time_future = strtotime("+2 years", $time);
$future = date("d/m/Y", $time_future);
echo "NEW DATE : " . $future;
?>
You can for instance output it like this:
date('2013-m-d', strtotime($myTime))
Just like that... or use
$year = date('Y');
$myMonthDay = date('m-d', strtotime($myTime));
echo $year . '-' . $myMonthDay;
Use the date modify function Like this
$date = new DateTime('2011-08-08');
$date->modify('+2 years');
echo $date->format('Y-m-d') . "\n";
//will give "2013-08-08"
I have a date/time string like this: 180510_112440 in this format ddmmyy_hhmmss
I need a snippet for having a string formatted like this way: 2010-05-18 11:24:40
Thanks for help.
another possible answer is the common use of strptime to parse your date and the mktime function:
<?php
$orig_date = "180510_112440";
// Parse our date in order to retrieve in an array date's day, month, etc.
$parsed_date = strptime($orig_date, "%d%m%y_%H%M%S");
// Make a unix timestamp of this parsed date:
$nice_date = mktime($parsed_date['tm_hour'],
$parsed_date['tm_min'],
$parsed_date['tm_sec'],
$parsed_date['tm_mon'] + 1,
$parsed_date['tm_mday'],
$parsed_date['tm_year'] + 1900);
// Verify the conversion:
echo $orig_date . "\n";
echo date('d/m/y H:i:s', $nice_date);
$inDate = '180510_112440';
$date = strtotime('20'.substr($inDate,4,2).'-'.
substr($inDate,2,2).'-'.
substr($inDate,0,2).' '.
substr($inDate,7,2).':'.
substr($inDate,9,2).':'.
substr($inDate,11,2));
echo date('d-M-Y H:i:s',$date);
Assumes date will always be in exactly the same format, and always 21st century
list($d,$m,$y,$h,$i,$s)=sscanf("180510_112440","%2c%2c%2c_%2c%2c%2c");
echo "20$y-$m-$d $h:$i:$s";