I have a date 01/31/2014, and I need to add a year to it and make it 01/31/2015. I am using
$xdate1 = 01/31/2014;
$xpire = strtotime(date("m/d/Y", strtotime($xdate1)) . " +1 year");
But it is returning 31474800.
Waaaay too complicated. You're doing multiple date<->string conversions, when
php > $x = strtotime('01/31/2014 +1 year');
php > echo date('m/d/Y', $x);
01/31/2015
would do the trick.
Another way is below:
<?php
$date = new DateTime('2014-01-31');
$date->add(new DateInterval('P01Y'));
echo $date->getTimestamp();
?>
There are 2 mistakes here.
You are missing the quote sign " when assigning to $xdate1. It should be
$xdate1 = "01/31/2014";
And the second, to get "01/31/2015", use the date function. strtotime returns a timestamp, not a date format. Therefore, use
$xpire = date("m/d/Y", strtotime(date("m/d/Y", strtotime($xdate1)) . " +1 year"));
May I introduce a simple API extension for DateTime with PHP 5.3+:
$xdate1 = Carbon::createFromFormat('m/d/Y', '01/31/2014');
$xpire = $xdate1->addYear(1);
First make $xdate1 as string value like
$xdate1 = '01/31/2014';
then apply date function at it like bellow
$xpire = date('m/d/Y', strtotime($xdate1.' +1 year')); // 01/31/2015
Related
I have a variable is which the value coming is Date along with time in php. How do I convert it into a variable to get only the year? I do not need automatic updation but the format change is needed. Normal answers are giving it about date but my variable is containing time as well.
The format coming by now is 2017-12-11 4:06:37 and i need only 2017
Use like this:
<?php echo date('Y',strtotime('now'));?>
You can you simple DateTime function and date_formate() function for displaying separate year, month and date.
For that you have to first convert in Object of your current Date time string by using :
$date = new \DateTime('2017-12-11 4:06:37');
And then you can use date format function by using below code:
echo date_format($date, "Y"); //for Display Year
echo date_format($date, "m"); //for Display Month
echo date_format($date, "d"); //for Display Date
You can code like this (working perfectly):
$format = 'Y-m-d H:i:s';
$date = DateTime::createFromFormat($format, '2009-02-15 15:16:17');
echo "Format: $format; " . $date->format('Y') . "\n";
As mentioned by Himanshu Upadhyay, this is correct and the easiest way.
<?php
echo date('Y',strtotime('now'));
?>
But i would recommend you to read this here. You should really do actually!
By using DateTime class
$date = new \DateTime('2017-12-11 4:06:37');
echo $date->format('Y');
I'm upload the excel data into mysql. in there date save like 16.5.59(dd.mm.yy) formate. I try to change this using this code
$date = '16.5.59';
echo date('d-m-y',strtotime($date));
it always show current time like 04-02-15.
Pls change this to dd-mm-yyyy formate.
Thanks in Advance.
Timestamp was limited from 01-01-1970 to 19-01-2038 on some systems (e.g. Windows).
If it not cover the between given range then it will take current date.
$date = '16.5.59';
$date_array = explode(".",$date);
$var_day = $date_array[0];
$var_month = $date_array[1];
> Blockquote
$var_year = $date_array[2];
echo $new_date_format = "$var_day-$var_month-$var_year";
Try to make the date yourself using explode and strtotime:
date_default_timezone_set('America/Los_Angeles');
$date = '16.5.59';
$date_time = strtotime(implode('-', array_reverse(explode('.', $date))))
$date_str = date('d-m-Y', $date_time);
echo $date_str;
Output: 16-5-2059
You ARE required to specify that the year is 1959 instead of 2059 (or are you sure that it's really 2059?)
You can make change from 2059 to 1959 like this:
date_default_timezone_set('America/Los_Angeles');
$date = '16.5.59';
$date_arr = array_reverse(explode('.', $date))
# This line does the job. Make sure all years are between 1900 - 1999.
$date_arr[0] = '19' . $date_arr[0];
$date_time = strtotime(implode('-', $date_arr))
$date_str = date('d-m-Y', $date_time);
echo $date_str;
I hope this answer can help you.
This is by no means the best solution but using php's date_parse_from_format is possible. You could supply a format and use the following
$date = "16.5.59";
$dateObj = date_parse_from_format("j.n.yy", $date)
with the $dateObj you can work what you need such as :
echo $dateObj['year'];
IDE Running Example
Worth noting you require PHP >=v5.3
You can do this if it is for simpler task. It will consume time.
$date = '16.5.59';
$dtSplit=explode(".",$date);
echo "<br>".$dtSplit[0].".".$dtSplit[1].".".$dtSplit[2];
echo "<br>".$dtSplit[0].".".$dtSplit[2].".".$dtSplit[1];
echo "<br>".$dtSplit[1].".".$dtSplit[0].".".$dtSplit[2];
echo "<br>".$dtSplit[1].".".$dtSplit[2].".".$dtSplit[0];
echo "<br>".$dtSplit[2].".".$dtSplit[0].".".$dtSplit[1];
echo "<br>".$dtSplit[2].".".$dtSplit[1].".".$dtSplit[0];
Hi I know this question is a bit trivial. I googled it but could not find the exact problem anywhere.
I have a string which contains a application filling date how can I convert this string into date
$appln_filling_date = '20020315';
The type of appln_filling_date is string I want to convert its type to date and want the data remain the same. The field in the database has a type date.
EDIT
This is what I am trying to do
$appln_filling_date = strtotime($appln_data['bibliographic-data']['application-reference']['document-id']['1']['date']['$']);
$appln_filling_date = date('Y-m-d', $appln_filling_date);
If your date is in some standard date format use this example:
$d = new DateTime('20020315');
echo $d->format('Y-m-d');
# or [if you do not have DateTime, which is in php >= 5.2.0]
$t = strtotime('20020315');
echo date('Y-m-d', $t);
If your date is not in some standard format use this example:
$d = DateTime::createFromFormat('d . Y - m', '15 . 2002 - 03');
echo $d->format('Y-m-d');
$date = DateTime::createFromFormat('Ymd', $appln_filling_date);
This worked for me like a charm
$appln_filling_date = $appln_data['bibliographic-data']['application-reference']['document-id']['1']['date']['$'];
$appln_filling_date = date_create(date('Y-m-d', $appln_filling_date));
Thanks for the down voting
I'm having date 20/12/2001 in this formate . i need to convert in following format 2001/12/20 using php .
$var = explode('/',$date);
$var = array_reverse($var);
$final = implode('/',$var);
Your safest bet
<?php
$input = '20/12/2001';
list($day, $month, $year) = explode('/',$input);
$output= "$year/$month/$day";
echo $output."\n";
Add validation as needed/desired. You input date isn't a known valid date format, so strToTime won't work.
Alternately, you could use mktime to create a date once you had the day, month, and year, and then use date to format it.
If you're getting the date string from somewhere else (as opposed to generating it yourself) and need to reformat it:
$date = '20/12/2001';
preg_replace('!(\d+)/(\d+)/(\d+)!', '$3/$2/$1', $date);
If you need the date for other purposes and are running PHP >= 5.3.0:
$when = DateTime::createFromFormat('d/m/Y', $date);
$when->format('Y/m/d');
// $when can be used for all sorts of things
You will need to manually parse it.
Split/explode text on "/".
Check you have three elements.
Do other basic checks that you have day in [0], month in [1] and year in [2] (that mostly means checking they're numbers and int he correct range)
Put them together again.
$today = date("Y/m/d");
I believe that should work... Someone correct me if I am wrong.
You can use sscanf in order to parse and reorder the parts of the date:
$theDate = '20/12/2001';
$newDate = join(sscanf($theDate, '%3$2s/%2$2s/%1$4s'), '/');
assert($newDate == '2001/12/20');
Or, if you are using PHP 5.3, you can use the DateTime object to do the converting:
$theDate = '20/12/2001';
$date = DateTime::createFromFormat('d/m/Y', $theDate);
$newDate = $date->format('Y/m/d');
assert($newDate == '2001/12/20');
$date = Date::CreateFromFormat('20/12/2001', 'd/m/Y');
$newdate = $date->format('Y/m/d');
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";