convert date in php day/month name, year - php

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

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));
?>

Display reformatted date from MySQL in PHP

I have a field within a MySQL database that has a date format of the following:
Mon, 17 Aug 2015 19:14:22 +0100
I want to display this date using PHP in the following format:
YYYY-MM-DD HH:MM:SS
I have no idea how to pick out the different elements and reformat them.
Can somebody help?
Many thanks,
John
You can try these :-
$date = "Mon, 17 Aug 2015 19:14:22 +0100";
$newDate = date('Y-m-d H:i:s', strtotime($date));
echo $newDate;
You can do it in mysql
SELECT
DATE_FORMAT(test.dateFrom, '%Y-%M-%d %H:%i:%s') as date,
FROM test
Or in php
$date = date('Y-m-d H:i:s', strtotime($datefrommysql) );

Get current datetime using PHP

I want to add content to a MySQL table with current date and time.
When I insert the content to database it shows the correct date but the wrong time.
$date1 = date("Y-m-d H:i:s");
date_default_timezone_set('Asia/Kolkata');
$date1 = date("Y-m-d H:i:s");
Asia/calcutta has changed to Asia/Kolkata
[a] strftime(): Format a local time/date according to locale settings
[b] date : Format a local time/date
Examples
Consider following simple example:
<?php
print strftime('%c');
?>
Output:
Mon Apr 23 01:22:58 2007
You need to pass format such as %c to strftime() to print date and time representation for the current system. You can use following format characters:
%m - month as a decimal number (range 01 to 12)
%d - day of the month as a decimal number (range 01 to 31)
%Y - year as a decimal number including the century
You can see the complete format conversion specifiers online here
You can also use date() as follows:
<?php
print date('r');
print "\\n";
print date('D, d M Y H:i:s T');
print "\\n";
?>
Output:
Mon, 23 Apr 2007 01:29:56 +0530
Mon, 23 Apr 2007 01:35:14 IST

how to php convert this time format?

i was fetching this date from table in the database like this format
Sunday 16th of January 2011 06:55:41 PM
and i want to convert it to be like this format
11-05-2012
how to do that with date function or any function
when i use date function
<td><?php echo date('d-m-Y', $v['v_created']); ?></td>
i get error message
'Severity: Warning
Message: date() expects parameter 2 to be long, string given'
This works for me (just tested on local web server)
<?php
date_default_timezone_set ('Europe/Rome');
$date = "Sunday 16th of January 2011 06:55:41 PM";
//.Strip "of" messing with php strtotime
$date = str_replace('of', '', $date);
$sql_friendly_date = date('y-m-d H:i', strtotime($date));
echo $sql_friendly_date;
?>
You can format the date as you prefer changing the first parameter of Date function according to: http://it2.php.net/manual/en/function.date.php
You have the following format:
Sunday 16th of January 2011 06:55:41 PM
that is a string based format, so the date information is more or less encoded in a human readable format. Luckily in english language. Let's see, that are multiple things all separated by a space:
Sunday - Weekdayname
16th - Date of Month, numeric, st/nd/th form
of - The string "of".
January - Monthname
2011 - Year, 4 digits
06:55:41 - Hour 2 digits 12 hours; Colon; Minute 2 digits; Colon; Seconds 2 digits
PM - AM/PM
So you could separate each node by space and then analyze the data. All you need is all Monthnames and the sscanf function because you only need to have the month, date of month and year:
$input = 'Sunday 16th of January 2011 06:55:41 PM';
$r = sscanf($input, "%*s %d%*s of %s %d", $day, $monthname, $year);
Which will already give you the following variables:
$monthname - string(7) "January"
$day - int(16)
$year - int(2011)
So all left to do is to transpose the monthname to a number which can be done with a map (in the form of an array in PHP) and some formatted output:
$monthnames = array(
'January' => 1,
# ...
);
printf("%02d-%02d-%04d", $day, $monthnames[$monthname], $year);
So regardless of which problem, as long as the input is somewhat consistently formatted you can pull it apart, process the gained data and do the output according to your needs. That is how it works.
try this. always worked for me
$date = Sunday 16th of January 2011 06:55:41 PM
$new_date = date('d-M-Y', strtotime($date));
echo $new_date;
The format you are using Sunday 16th of January 2011 06:55:41 PM is a wrong format.from the form you are inserted this date in database should be in date(Y-m-d) than the value of date inserted in database like:- 11-05-2012. and you can fetch this and get the format what you want.
<?php
$old_date = date('l, F d y h:i:s'); // returns Saturday, January 30 10 02:06:34
$new_date = date('d-M-Y', strtotime($old_date));
echo $new_date
?>
more details about date plz visit this url
http://www.php.net/manual/en/datetime.createfromformat.php

Convert data from email header

Does anyone could help me how to convert data from email header?
I have the next date format from email header:
Wed, 28 Apr 2010 21:59:49 -0400
I need to convert them into mysql Date, or timestamp. Thanks!
You should be using DateTime for this, specifically DateTime::createFromFormat():
$str = 'Wed, 28 Apr 2010 21:59:49 -0400';
$date = DateTime::createFromFormat( 'D, d M Y H:i:s O', $str);
Now, you have a Date object in $date, and you can grab the unix timestamp (if that's what you want), or you can format it into a date for MySQL.
echo $date->getTimestamp(); // Outputs: 1272506389
echo $date->format( 'Y-m-d H:i:s'); // For MySQL column, 2010-04-28 21:59:49
You can see it working in the demo.

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