How to Display date as day/month/year - php

I want to display my date as day/month/year.
example 31st October 2011
echo '<h4>Date Of Birth: '.$row['dob'].'</br></h4>';


You can try below code ::-
echo '<h4>Date Of Birth: '.date('dS F Y', strtotime($row['dob'])).'</br></h4>';
// ^^^^^^^ Here the is format of date, in which format you want.
Output will be ::-
31st October 2011
For more you can refer click here


<?php
$mydate = "2010-03-3";
$newDate = date("d M Y", strtotime($mydate));
$new_date = date('dS F Y', strtotime($newDate));
echo $new_date;
?>
/*Out Put*/
03rd March 2010


So if you have that data stored as or you can convert it to 3 variables representing DD MM YYYY then I'd suggest creating array of months and an array of post-fixes(don't know how to call it, but in your example it's 31st) and getting DD % 10 to use reminder for post-fix array
$month = ['January','February',....,'December']; //remember it's $month[MM-1];
$post = ['th','st','nd',...,'th'] //not sure if this on is correct but hope you get the idea


echo date('dS F Y', strtotime($row['dob']));


Refer to DateTime. I normally avoid strtotime() conversions.
Heres a working solution:
<?php
$date = '2016-11-12';
$newDate = new DateTime($date);
echo $newDate->format('jS F Y');
?>
OUTPUT:
12th November 2016

Related

date conversion in CodeIgniter / PHP does not gives output

I have date in this type of format: April 1st 2017 and I want to convert it into this type of format: 2017/04/01 in my CodeIgniter code using php. I have used below posted piece of code but it is not working. Please solve the issue.
Code:
$date = DateTime::createFromFormat('m/d/Y', "April 1st 2017");
echo "Date = ".$date->format('Y-m-d');

You can use strtotime() and date() php functions as
$newDate = date("m/d/Y", strtotime("April 1st 2017"));
Or in CodeIgniter
$date = DateTime::createFromFormat('j F Y - H:i', 'April 1st 2017');
echo $date->format('m/d/Y H:i:s');

Your format can be used in the constructor of DateTime. See accepted formats.
$date = new DateTime("April 1st 2017");
echo "Date = ".$date->format('Y-m-d');
Outputs:
Date = 2017-04-01
If you want to use DateTime::createFromFormat(), you have to use the proper format
"F jS Y"

The format you specified for your date is incorrect.
It would convert '04/01/2017' but it does not suit
April 1st 2017.
Try instead: createFromFormat('F dS Y')
Explanation:
F - full textual representation of a month, such as January.
d - day
S - English ordinal suffix for the day of the month
Y - 4-digit representation of year

you can try this also
<?php
$date='22 march 2018';
echo date('m/d/Y', strtotime($date));
?>

Convert date to month name & year

I am trying to convert a date to month name and year.
$date = '2017-07-00';
$date = date('m/y', strtotime($date));
echo DATE_FORMAT($date, '%M %Y');
I am expecting output like
July, 2017
Here is error i am getting
Warning: date_format() expects parameter 1 to be DateTimeInterface, string given

No Need of DATE_FORMAT() function.
Example-1: If 00 used in day. Then, output will be June, 2017
<?php
$date = '2017-07-00';
echo date('F, Y', strtotime($date)); //June, 2017
?>
Example-2: If 01 or valid day used in day. Then, output will be July, 2017
<?php
$date = '2017-07-01';
echo date('F, Y', strtotime($date)); //July, 2017
?>

You are not using correct parameters, use F for moth and Y for year
Full code:
$date = '2017-07-00';
$date = date('F, Y ', strtotime($date));
echo $date;

Try This.
$date = '2017-07-01';
$date = date('F, Y', strtotime($date));
echo $date;

You can use this
echo date('F,Y',strtotime($date));

PHP date increase

I've written a piece of code which converts a date to the specific format and increases it by 1 day.
<?php
date_default_timezone_set('Europe/Moscow');
$mehdate = "2011-11-25";
$mehdate = date ('d m Y', strtotime ('+1 day', strtotime($mehdate)));
echo $mehdate, "\n";
?>
But then I have to increase $mehdate by 1 day one more time.
And I cannot understand how to do that. I already tried
$mehdate = date ('d m Y', strtotime ("+1 day", $mehdate));
and
$mehdate = date ('d m Y', strtotime ('+1 day', strtotime($mehdate)));
again but it won't work because
strtotime($mehdate)
returns FALSE.
So, how can I increase the $mehdate which was already formatted?

Your issue can easily be resolved if you use DateTime class.
Try this:
$mehdate = new DateTime('2011-11-25');
$mehdate->modify('+1 day');
echo $mehdate->format('d m Y')."\n"; // Gives 26 11 2011
$mehdate->modify('+1 day');
echo $mehdate->format('d m Y'); // Gives 27 11 2011

date_default_timezone_set('Europe/Moscow');
$mehdate = "2011-11-25";
$mehdate = strtotime ('+1 day', strtotime($mehdate));
$mehdate = date ('d m Y', $mehdate);
echo $mehdate, "\n";
Result
26 11 2011

For all newbies like me there is a simple advice: don't use 'd m Y' format, you'd better operate with 'd-m-Y'.
Or you have to use DateTime class, as Object Manipulator advised.

PHP date conversion dd [th/st/rd] / month / yyyy

I have my users entering the date in this format :- mm/dd/yyyy (11/21/2012)
My PHP script converts the date into the following format :- dd-Month-yyyy (21-November-2012)
I do this using :-
$new_date = date('d-F-Y', strtotime($user_date));
How can I have the date in this format :- 21st November 2012?
Thanks

You can use S letter as following:
$new_date = date('jS F Y', strtotime($user_date));
Check manual.

It will output as you expect
$my_date = '2016-01-01';
echo date('F jS, Y', strtotime($my_date));
# January 1st, 2016
while dS will also prepends 0
echo date('F dS, Y', strtotime($my_date));
# January 01st, 2016

$new_date = date('jS F Y', strtotime($date));
S - English ordinal suffix for the day of the month, 2 characters
(st, nd, rd or th. Works well with j)

**My Date = 22-12-1992**
<?php
$mydate = "22-12-1992";
$newDate = date("d M Y", strtotime($mydate));
$new_date = date('dS F Y', strtotime($newDate));
echo $new_date;
?>
**OutPut = 22nd December 1992**

You can use something like:
echo date('l, F jS');
Or even get a bit fancy with the HTML:
echo date('l, F j<\s\u\p>S</\s\u\p>');

You can do that :
<?php echo date('dS M. Y');?>

$date = date_create('09-22-2012');
echo $date->format('d S F Y');
by this code you will get your desire
for more you can also visit http://php.net/manual/en/datetime.formats.date.php

Convert date integer to text?

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

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