Sir,
$datfrs=$_REQUEST["datfrs"];
This is my php code here the $datfrs is the date and its format is (01 / Nov / 2013 ),I need to convert it into (2013-11-01) to fetch the value from data base.I looked through date function but the date format is changing to (1970-01-01) but I need it as(2013-11-01) .Please help me.
Something like this perhaps...
$dt = DateTime::createFromFormat('d/M/Y', $datefrs);
$formatted = $dt->format('Y-m-d');
The first argument to DateTime::createFromFormat should match the incoming format. I can't quite tell if the parentheses and spaces are included. Taken literally, the format would be something like
'(d / M / Y)'
try this
date("Y-m-d", strtotime(str_replace("/", " ",$_REQUEST["datfrs"])));
Try:
date("Y-m-d", strtotime($datfrs))
Use this function to get your required date. This will surely work.
$datfrs=$_REQUEST["datfrs"];
$newdate=MyDateFormat($datfrs);
function MyDateFormat($date) // format is (01 / Nov / 2013 )
{
$exp=explode("/",$date);
$mon=date("m",strtotime($exp[1]))
$ndate=$exp[2]."-".$mon."-".$exp[0];
return $ndate;
}
Related
I have much data with several timestamps and I just recognized that some are in "dd.mm.YYYY" which works very well with date("Y-m-d", strtotime($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.
a line of my XML looks like this:
<observation_time_rfc822>Thu, 09 Oct 2014 22:59:16 +0200</observation_time_rfc822>
I grab it and give it out:
$ob_time= $xml->observation_time_rfc822;
echo $ob_time;
The output looks like this:
Thu, 09 Oct 2014 22:59:16 +0200
But what I need should look like this (yes, the funny '%3A' replaces ':')
2014-10-09+22%3A59%3A16
I think string replace can do this, please someone can help me to find out!
Thank you!
Edit: Use #Ghost's solution, it correctly handles the timezone offset.
First you need to reformat your date. You do this by parsing it with strtotime and formatting it with the date function. Those "funny %3A replaces" are actually URL-encoded characters:
$date = date('Y-m-d H:i:s', strtotime($ob_time));
$date = urlencode($date); // 2014-10-09+20%3A59%3A16
You could use DateTime class in this case, then use urlencode():
Example:
$ob_time = (string) $xml->observation_time_rfc822;
$date = DateTime::createFromFormat('D, d M Y H:i:s O', $ob_time);
$real_date = $date->format('Y-m-d H:i:s');
echo urlencode($real_date); // 2014-10-09+22%3A59%3A16
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.
I am passing a date in a URL in a UK format as per the following:
http://www.website.com/index.php?from_date=01/04/2013&to_date=12/04/2013
The date range is 1st April 2013 to 12th April 2013.
Then I am using the following PHP to convert it to the YYYY-MM-DD format.
<?php
$rpt_from_date = date("Y-m-d", strtotime($_GET["from_date"]) );
$rpt_to_date = date("Y-m-d", strtotime($_GET["to_date"]) );
echo $rpt_from_date;
echo "</br>";
echo $rpt_to_date;
?>
For some reason this is not working. The above returns the following:
2013-01-04
2013-12-04
It's switching the month and day around. I want it to return:
2013-04-01
2013-04-12
Does anyone have any ideas?
Use DateTime object, to get php understand in which format you passing date to it.
$rpt_from_date = DateTime::createFromFormat('d/m/Y', $_GET["from_date"]);
echo $rpt_from_date->format('Y-m-d');
PHP is reading your time string in the US format (MM/DD/YYYY) because you are using slashes. You could use dashes to give the time: index.php?from_date=01-04-2013&to_date=12-04-2013, or convert it yourself:
$uktime = implode("-", explode("/", $_GET['from_date']));
I will provide the solution but it is not using the date function:
$arr = explode("/",$_GET["from_date"]);
$from_date = $arr[2]."-".$arr[1]."-"$arr[0];
Second solution is as following:
$from_date = implode(array_reverse(explode("/",$_GET["from_date"])));
I have the following date format in a xml sheet:
Jan 30
and I want to display as:
2011-01-30
2011 being the current year
Can someone help me with that?
strtotime will use the current year if none is specified, so this would work
$t=strtotime("Jan 30");
echo strftime("%Y-%m-%d", $t);
strtotime + date gives you:
echo date("Y-m-d", strtotime('30 Jan')); //echoes '2011-01-30'
care to post the xml?
in anycase, take the string
$x = "Jan 30"
add 2011
$x = $x . ' ' . date('Y');
convert it to date time and format it
$y = date_format('Y-m-d', strtotime($x));
echo $y;`
strtotime is your key here, it converts most any date format into seconds since the epoch, then all you need do is use the date_format function
ps... if you like my answer, accept it as the answer to bring your accepted question ratio, otherwise people will be less likely to answer your questions