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];
Related
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
I have a problem by converting a date in proper format.
I get the time the from Facebook API in this format: 2013-08-23T09:00:00
I then $fbdate = date('2013-08-23T09:00:00');
When I echo $fbdate, it retuns 2013-08-25UTC05:00:00.
Then I tried:
$datum = date("d.m.Y",$fbdate);
$uhrzeit = date("H:i",$fbdate);
To extract the date and the time but it always returns:
01.01.1970 for $datum and 00:33 for $uhrzeit.
You should use strtotime() to parse a date string into a UNIX timestamp:
$fbdate = strtotime('2013-08-23T09:00:00');
$datum = date('d.m.Y', $fbdate);
$uhrzeit = date('H:i', $fbdate);
Try using the DateTime class:
$fbdate = '2013-08-23T09:00:00';
$date = DateTime::createFromFormat('Y-m-d\TH:i:s', $fbdate);
$datum = $date->format('d.m.Y');
$uhrzeit = $date->format('H:i');
echo $datum;
echo $uhrzeit;
$datum = date("d.m.Y",strtotime($fbdate));
$uhrzeit = date("H:i",strtotime($fbdate));
The date function in PHP is meant to convert a microtime into a date, so you need to convert your string dates to microtimes first.
Have you tried this? You should use strtotime() when using the date function.
$datum = date("d.m.Y", strtotime($fbdate));
$uhrzeit = date("H:i", strtotime($fbdate));
Use strtotime()
$a = date("Y-M-d", strtotime($datum));
echo $a.$uhrzeit;
More info http://php.net/manual/en/function.strtotime.php
I'm not a huge fan of using timestamp.
What I did to solve this issue was:
$updatedDate = new DateTime(
preg_replace('/^(.*)\+0000$/', '$1', $fbUser->getProperty("updated_time")),
new DateTimeZone("UTC")
);
It chops the +0000 part and creates the DateTime object with an explicit UTC timezone.
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');
How can I convert this string 05/Feb/2010:14:00:01 to unixtime ?
Use the strtotime function:
Example:
$date = "25 december 2009";
$my_date = date('m/d/y', strtotime($date));
echo $my_date;
For PHP 5.3 this should work. You may need to fiddle with passing $dateInfo['is_dst'], wasn't working for me anyhow.
$date = '05/Feb/2010:14:00:01';
$dateInfo = date_parse_from_format('d/M/Y:H:i:s', $date);
$unixTimestamp = mktime(
$dateInfo['hour'], $dateInfo['minute'], $dateInfo['second'],
$dateInfo['month'], $dateInfo['day'], $dateInfo['year'],
$dateInfo['is_dst']
);
Versions prior, this should work.
$date = '05/Feb/2010:14:00:01';
$format = '#^(?P<day>\d{2})/(?P<month>[A-Z][a-z]{2})/(?P<year>\d{4}):(?P<hour>\d{2}):(?P<minute>\d{2}):(?P<second>\d{2})$#';
preg_match($format, $date, $dateInfo);
$unixTimestamp = mktime(
$dateInfo['hour'], $dateInfo['minute'], $dateInfo['second'],
date('n', strtotime($dateInfo['month'])), $dateInfo['day'], $dateInfo['year'],
date('I')
);
You may not like regular expressions. You could annotate it, of course, but not everyone likes that either. So, this is an alternative.
$day = $date[0].$date[1];
$month = date('n', strtotime($date[3].$date[4].$date[5]));
$year = $date[7].$date[8].$date[9].$date[10];
$hour = $date[12].$date[13];
$minute = $date[15].$date[16];
$second = $date[18].$date[19];
Or substr, or explode, whatever you wish to parse that string.
You should look into the strtotime() function.
http://www.php.net/date_parse_from_format
$d="05/Feb/2010:14:00:01";
$dr= date_create_from_format('d/M/Y:H:i:s', $d);
echo $dr->format('Y-m-d H:i:s');
here you get date string, give format specifier in ->format() according to format needed
Simple exploding should do the trick:
$monthNamesToInt = array('Jan'=>1,'Feb'=>2, 'Mar'=>3 /*, [...]*/ );
$datetime = '05/Feb/2010:14:00:01';
list($date,$hour,$minute,$second) = explode(':',$datetime);
list($day,$month,$year) = explode('/',$date);
$unixtime = mktime((int)$hour, (int)$minute, (int)$second, $monthNamesToInt[$month], (int)$day, (int)$year);
If you're up for it, use the DateTime class
Try this:
$new_date=date('d-m-Y', strtotime($date));
If it's a string that you trust meaning that you have checked it before hand then the following would also work.
$date = new DateTime('2015-03-27');