1.2 and need to convert a date from dd/mm/yyyy to yyyy-mm-dd
For example if the date is in format 07/08/2014, it should appear as 2014-08-07
How can this be done? I know strtotime returns unix timestamp but it doesn't seem to work with dates with Slashes (/) in it. SInce I'm using 5.1, a lot of DateTime functions are not supported in it.
Please help.
Use DateTime class, strtotime function would create issue when date less then 1901 with PHP 5.3.0
Try this way
$date = DateTime::createFromFormat('d/m/Y', "07/08/2014");
$new_date_format = $date->format('Y-m-d');
Need to pass a correct format with -(date string separation with dash) in date() try
$d = str_replace('/', '-','07/08/2014');
echo date('Y-m-d', strtotime($d)); //2014-08-07
with DateTime
$objDateTime = new DateTime($d);
echo $objDateTime->format('Y-m-d'); //2014-08-07
You can do it by date('Y-m-d',strtotime($date))
Where $date is in any format that you want to convert to YYYY-MM-DD format.
By using date() function yuo can try this
echo date('Y-d-m',strtotime('07/08/2014'));
Check the documentation for more
Method : 1 demo
$date1 = "07/08/2014";
$arr = explode("/", $date1);
$date2 = $arr[2]."-".$arr[1]."-".$arr[0];
echo $date2;
Method : 2 demo
$date1 = "07/08/2014";
list($day, $month, $year) = explode("/", $date1);
$date2 = $year."-".$month."-".$day;
echo $date2;
Method 3 : with strtotime Demo
$date1 = "07/08/2014";
$date1 = str_replace("/", "-", $date1);
$date2 = date('Y-m-d', strtotime($date1));
echo $date2;
Related
I trying to format string to date. String looks like that: 20200219 and I want to format it like 2020-02-19. Can some help me with this?
You have multiple options:
Use strtotime() on your first date then date('Y-m-d') to convert it back:
$changed_date = "20200219";
echo date("Y-m-d", strtotime($changed_date ) );
$time = strtotime('03/05/2020');
$newformat = date('Y-m-d',$time);
echo $newformat;
// 2020-03-05
You need to be careful with m/d/Y and m-d-Y formats. PHP considers / to mean m/d/Y and - to mean d-m-Y. I would explicitly describe the input format in this case:
$ymd = DateTime::createFromFormat('m-d-Y', '03/05/2020')->format('Y-m-d');
Another Option:
$d = new DateTime('03/05/2020');
$timestamp = $d->getTimestamp(); // Unix timestamp
$formatted_date = $d->format('Y-m-d'); // 2020-03-05
you can do like that
$s = '20200219';
$date = strtotime($s);
echo date('Y-m-d', $date);
if it's a string you could do like this:
$date=date_create("20200219");
return date_format($date,"Y-m-d");
Hope it helped.
Try this one:-
$var = "20200219";
echo date("Y-m-d", strtotime($var) );
You can try php way:
date("Y-m-d", strtotime("20200219") );
Date Format:
$date1 = '16-MAR-2015';
$date2 = '04-FEB-15';
How can I check if
$date1 <= $date2 || $date1 => $date2
Do I need to convert date format in
$date1 = '16-3-2015';
$date2 = '04-2-15';
Use strtotime():
<?php
$date1 = strtotime("16-MAR-2015");
$date2 = strtotime("04-FEB-15");
?>
and compair
Try this way, it works. DEMO
$date1 = DateTime::createFromFormat('j-M-Y', '16-MAR-2015');
$date2 = DateTime::createFromFormat('j-M-y', '04-FEB-15');
$date1=$date1->format('Y-m-d');
$date2=$date2->format('Y-m-d');
var_dump($date1 <= $date2);
var_dump($date1 >= $date2);
You don't need to convert anything with the given formats, DateTime will do what you need.
Just turn them into DateTime objects and compare them :
$date1 = '16-MAR-2015';
$date2 = '04-FEB-15';
$date1 = date_create($date1);
$date2 = date_create($date2);
// $date1 > $date2 -> true
// $date1 < $date2 -> false
Or
if (date_create($date1) > date_create($date2)) {
// Use $date1 unchanged
}
First, you need to turn them into a DateTime object.
DEMO
$date1 = '16-MAR-2015';
$date2 = '04-FEB-15';
$dateComp1 = new DateTime($date1);
$dateComp2 = new DateTime($date2);
Then you can compare them, DateTime is smart enough to convert the format automatically.
$myDateTime = DateTime::createFromFormat('Y-m-d', $dateString); $newDateString = $myDateTime->format('m/d/Y'); There is a magical function called strtotime()
If you need to print the value in a readable format you can use the function DATE
date("d-m-Y h:i:s",$date1);
In your case.
You want to use
Date.parse()
try
date1= new DateTime();
date1= new Datetime->format('d-m-Y'); // this will print year 4 digits ie; 2017
date2= new DateTime();
date2= new DateTime->format('d-m-y'); //this will print 2 digit year ie; 17
// to compare
$differance=$date1->diff($date2);
echo $differance->format('d-m-y'); //or change to upper case Y for 4 digit year
note... in mysql 5.6 it seems the input format must be Y first: if you insert it with Y-first format first, it seems to compare correctly no matter what the second format is.
Im trying to add a certain amount of days to a timestmp using this in PHP:
$capturedDate = '2008-06-20';
$endDate = strtotime($capturedDate);
$endDate2 = strtotime('+1 day',$endDate);
echo $endDate2;
but its displaying: 1216526400
any ideas?
Try:
echo date("Y-m-d H:i:s",$endDate2);
Or (for just the date):
echo date("Y-m-d",$endDate2);
You can find documentation about how to format your string here: http://php.net/manual/en/function.date.php
You should be using DateTime for working with dates. It's timezone friendly.
$datetime = new DateTime('2008-06-20');
$datetime->modify('+1 day');
echo $datetime->getTimestamp();
strtotime() converts the date into a unix timestamp which is the number of seconds since January 1st 1970. If you want a date output you have to run the finished timestamp through date() first.
$capturedDate = '2008-06-20';
$endDate = strtotime($capturedDate.' +1 day');
echo date("Y-m-d", $endDate);
strtotime creates a Unix timestamp so if you want to be presented with a formatted date, you need to pass the timestamp as an argument to the date function as follows:
$capturedDate = '2008-06-20';
$endDate = strtotime($capturedDate);
$endDate2 = strtotime('+1 day',$endDate);
echo date('Y-m-d', $endDate2);
Additionally, there are a wide variety of parameters you can use in the date function if you want to display additional information.
e.g.: echo date('Y-m-d H:i:s', $endDate2); or echo date('Y-m-d h:i:s a', $endDate2);, etc.
Sooooo close, just take your timestamp and convert it back into date format using date("desired format",$endDate2);
DateTime is a very nice way to deal with dates. You can try like this:
$capturedDate = '2008-06-20';
$date = DateTime::createFromFormat('Y-m-d', $capturedDate)->modify('+1 day');
echo $date->getTimestamp();
<?php
$date = "04-15-2013";
$date = strtotime($date);
$date = strtotime("+1 day", $date);
echo date('m-d-Y', $date);
?>
This is driving me crazy and seems so simple. I'm pretty new to PHP, but I can't figure this out. The echo returns 01-01-1970.
The $date will be coming from a POST in the format m-d-Y, I need to add one day and have it as a new variable to be used later.
Do I have to convert $date to Y-m-d, add 1 day, then convert back to m-d-Y?
Would I be better off learning how to use DateTime?
there you go
$date = "04-15-2013";
$date1 = str_replace('-', '/', $date);
$tomorrow = date('m-d-Y',strtotime($date1 . "+1 days"));
echo $tomorrow;
this will output
04-16-2013
Documentation for both function
date
strtotime
$date = DateTime::createFromFormat('m-d-Y', '04-15-2013');
$date->modify('+1 day');
echo $date->format('m-d-Y');
See it in action
Or in PHP 5.4+
echo (DateTime::createFromFormat('m-d-Y', '04-15-2013'))->modify('+1 day')->format('m-d-Y');
reference
DateTime::createFromFormat()
$date = strtotime("+1 day");
echo date('m-d-y',$date);
use http://www.php.net/manual/en/datetime.add.php like
$date = date_create('2000-01-01');
date_add($date, date_interval_create_from_date_string('1 days'));
echo date_format($date, 'Y-m-d');
output
2000-01-2
The format you've used is not recognized by strtotime(). Replace
$date = "04-15-2013";
by
$date = "04/15/2013";
Or if you want to use - then use the following line with the year in front:
$date = "2013-04-15";
Actually I wanted same alike thing,
To get one year backward date, for a given date! :-)
With the hint of above answer from #mohammad mohsenipur
I got to the following link, via his given link!
Luckily, there is a method same as date_add method, named date_sub method! :-)
I do the following to get done what I wanted!
$date = date_create('2000-01-01');
date_sub($date, date_interval_create_from_date_string('1 years'));
echo date_format($date, 'Y-m-d');
Hopes this answer will help somebody too! :-)
Good luck guys!
I've got a date in this format:
2009-01-01
How do I return the same date but 1 year earlier?
You can use strtotime:
$date = strtotime('2010-01-01 -1 year');
The strtotime function returns a unix timestamp, to get a formatted string you can use date:
echo date('Y-m-d', $date); // echoes '2009-01-01'
Use strtotime() function:
$time = strtotime("-1 year", time());
$date = date("Y-m-d", $time);
Using the DateTime object...
$time = new DateTime('2099-01-01');
$newtime = $time->modify('-1 year')->format('Y-m-d');
Or using now for today
$time = new DateTime('now');
$newtime = $time->modify('-1 year')->format('Y-m-d');
an easiest way which i used and worked well
date('Y-m-d', strtotime('-1 year'));
this worked perfect.. hope this will help someone else too.. :)
On my website, to check if registering people is 18 years old, I simply used the following :
$legalAge = date('Y-m-d', strtotime('-18 year'));
After, only compare the the two dates.
Hope it could help someone.
// set your date here
$mydate = "2009-01-01";
/* strtotime accepts two parameters.
The first parameter tells what it should compute.
The second parameter defines what source date it should use. */
$lastyear = strtotime("-1 year", strtotime($mydate));
// format and display the computed date
echo date("Y-m-d", $lastyear);
Although there are many acceptable answers in response to this question, I don't see any examples of the sub method using the \Datetime object: https://www.php.net/manual/en/datetime.sub.php
So, for reference, you can also use a \DateInterval to modify a \Datetime object:
$date = new \DateTime('2009-01-01');
$date->sub(new \DateInterval('P1Y'));
echo $date->format('Y-m-d');
Which returns:
2008-01-01
For more information about \DateInterval, refer to the documentation: https://www.php.net/manual/en/class.dateinterval.php
You can use the following function to subtract 1 or any years from a date.
function yearstodate($years) {
$now = date("Y-m-d");
$now = explode('-', $now);
$year = $now[0];
$month = $now[1];
$day = $now[2];
$converted_year = $year - $years;
echo $now = $converted_year."-".$month."-".$day;
}
$number_to_subtract = "1";
echo yearstodate($number_to_subtract);
And looking at above examples you can also use the following
$user_age_min = "-"."1";
echo date('Y-m-d', strtotime($user_age_min.'year'));