So I have the date like this in wordpress, I get it from a custom metabox where is stored like this, 23/02/2012, now how can I set wp_locale in WP or something like that, and I need to convert the date to: Monday 23 February 2012, but I need also to set the language, thats why I need that wp_locale because that output will be in Dutch.Thank you
The date format can be converted with the code below as a guide.
date('l j F Y', strtotime(str_replace('/', '-', '23/02/2012')))
The str_replace is necessary because with / PHP assumes m/d/y American date format, not the European d/m/y.
As for doing this in Wordpress with i18n support, you might consider http://codex.wordpress.org/Function_Reference/date_i18n
date_i18n('l j F Y', strtotime(str_replace('/', '-', '23/02/2012')))
I guess there are other ways to do this, but I'd do it as follows:
Upon 'save_post':
$date = explode ('/',$_POST["date_field"]);
$date = $date[1].'/'.$date[0].'/'.$date[2]; // dd/mm/yyyy to mm/dd/yyyy
$s = strtotime($date); /* UNIX TIMESTAMP */
and then store the Unix timestamp in the database.
Upon 'amdin_init' you'll need to do the exact opposite to load the date in the right format in the metabox:
global $post;
$custom = get_post_custom($post->ID);
if ($custom["date_field"][0]) {
$d = date("d/m/Y",$custom["date_field"][0]); // convert unix timestamp
} else {
$d = "";
}
And in your template file, use php functions setlocale and strftime to display the date:
$custom = get_post_custom($post->ID);
$d = $custom["date_field"][0];
setlocale(LC_TIME, 'nl_NL');
$s = strftime('%#d %B %Y',$s);
This should output something like 13 January 2012.
Related
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", $date);
It returns an array with very useful information like month, year etc.
I have a function that reads out the date in a file on the first line. This date is formatted in dutch like this 2 mei 2013 or 28 jun. 2013
It needs to convert the date string into a timestamp, but whatever i try it won't work for the mei moths or any other dutch named month. Here is the code I currently have (the original function is a bit more code, but this is where it goes wrong)
function getTimestamp($date){
date_default_timezone_set('Europe/Amsterdam');
setlocale(LC_ALL, 'nl_NL');
$timestamp = strtotime($date);
return $timestamp;
}
Now, here are some results when using this function:
$timestamp = getTimestamp('28 jun. 2013') //1372370400
$timestamp2 = getTimestamp('2 mei 2013') // false
but, when i put this code in the function
echo strftime('%e %b %Y', 1367445600)."\n";
it prints '2 mei 2013'
How can I tell php not only format the date-time string in Dutch, but also read it in Dutch?
=======================
Thanks to some explanation below I now have the code working (this is the full function)
public function getReportDate(){
$mothsTranslated = array('mrt'=> 'mar','mei'=>'may', 'okt'=>'oct');
$content = file($this->file);
$line = $content[0];
$header = str_getcsv($line, $this->delimiter);
$date = str_replace('.', '', $header[1]);
foreach ($mothsTranslated as $dutch => $eng) {
if(strpos($date, $dutch) !== false){
$date = str_replace($dutch, $eng, $date);
}
}
$timestamp = strtotime($date);
return $timestamp;
}
Without creating your own date parser, the native PHP functions only use English dates.
However, there is an international dateformatter extension available for PHP. You can install this plugin and then would be able to parse non-english dates.
http://www.php.net/manual/en/intldateformatter.parse.php
As others found out, strtotime does not respect the set locale.
Indeed, it's description in the manual states: "Parse about any English textual datetime description into a Unix timestamp"
Solutions
You can use strptime() since PHP5 that does respect the locale (like strftime), but there are some warnings about using it on the php website.
You could write a function that replaces the Dutch month names to English month names and then calls strtotime.
I want to convert 1373892900000 to Monday 2013/07/15 8:55 AM in Codeigniter.
However, I keep receiving a totally different result by converting the timestamp using the function i have written, please note:I need to change the dates according to different timezones, that is why I want to write it this way:
public function time_convert($timestamp){
$this->load->helper('date');
date_default_timezone_set('UTC');
$daylight_saving = TRUE;
$timezone = "UM4"; //toronto or new york timezone
$time = gmt_to_local($timestamp, $timezone, $daylight_saving);
$final_time = standard_date('DATE_RFC822', $time);
return $final_time;
}
Result from the above function is: Sat, 08 Dec 06 01:40:00 +0000
And if I don't put date_default_timezone_set('UTC'); in the above function, I get this date instead Sat, 08 Dec 06 02:40:00 +0100. My codeigniter seems to default the timezone to Europe/Berlin.
Can anyone please help me correct any of the mistakes I might have made?
Why not just use PHP's date function?
public function time_convert($timestamp){
return date('l Y/m/d H:i', $timestamp);
}
For different timezones use a DateTime object:
public function time_convert($timestamp, $timezone = 'UTC'){
$datetime = new DateTime($timestamp, new DateTimeZone($timezone));
return $datetime->format('l Y/m/d H:i');
}
Think that should work. Note: I tihnk you need at least PHP version 5.20 for the TimeZone class.
<?php
$time_str=1373892900000;
echo gmdate("fill with your format", $time_str);
?>
your format = format your time in php, reading this page for details.
http://php.net/manual/en/function.date.php
http://php.net/manual/en/function.gmdate.php
Appears as though an invocation of standard_date with the DATE_ATOM format may sort you:
echo unix_to_human(time(), true, 'us'); # returns 2013-07-12 08:01:02 AM, for example
There are a whole host of other options for the format, enumerated on the linked page.
This how to covert timestamp to date very simple:
echo date('m/d/Y', 1299446702);
to convert timestamp to human readable format try this:
function unix_timestamp_to_human ($timestamp = "", $format = 'D d M Y - H:i:s')
{
if (empty($timestamp) || ! is_numeric($timestamp)) $timestamp = time();
return ($timestamp) ? date($format, $timestamp) : date($format, $timestamp);
}
$unix_time = "1251208071";
echo unix_timestamp_to_human($unix_time); //Return: Tue 25 Aug 2009 - 14:47:51
if you want to convert it to a format like this: 2008-07-17T09:24:17Z than use this method
<?php
$timestamp=1333699439;
echo gmdate("Y-m-d\TH:i:s\Z", $timestamp);
?>
for details about date:
http://php.net/manual/en/function.date.php
Your timestamp is coming from javascript on the client, I would guess, because it appears to be in milliseconds. php timestamps are in seconds. So to get the answer you want, first divide by 1000.
Showing the full year would have made the issue more obvious, as you would have seen the year as 45,506.
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');