How to convert date from yyyy:mm:dd to yyyy-m-dd and yyyy:mm:dd (without leading zero for the month) to yyyy-m-dd (with leading zero to month)?
You could use DateTime::createFromFormat and then use DateTime::format.
Example:
$date = DateTime::createFromFormat('Y:m:d', '2012:08:02');
echo $date->format('Y-m-d');
// without leading zero for month
$date = DateTime::createFromFormat('Y:n:d', '2012:8:02');
echo $date->format('Y-m-d');
try this:
$dateFrom ="2012:8:2";
$dateTo = str_replace(":","-",$dateFrom);
$dateTo = date("Y-m-d", strtotime($dateTo));
echo $dateTo;
Use the $date=strtotime($date) function to get the date in unix timestamp. After that you can use the date("Y-m-d",$date) function to convert it to the format you want.Here's an example:
$date=strtotime($olddate);
$date=date("Y-m-d",$date);
echo $date; // Now this will show you the date in the format you wanted :)
Use the date function.
date will not understand : as a separator, so you will need to replace that with a separator that it understands, like / or -, with str_replace.
Code:
$orig_date = '2012:8:2';
$final_date = date('Y-n-d', str_replace(':', '/', $orig_date));
echo $final_date; // Result: 2012-8-02
Use date() function
echo date('Y-m-d'); // for 1st case (replacing ':' with '-')
echo date('Y-j-d'); // for 2nd case (without leading zero)
Related
I am dealing with a problem of time conversion from 12 hr format to 24 hour format.
Is there any single function in php to replace the first two characters of a string?
str_replace can be used only when I know the substring content to be replaced.
$str_to_replace = '12';
$input_str = 'ab345678';
$output_str = $str_to_replace . substr($input_str, 2);
echo $output_str;
"12345678"
If the date is always given in a specific format you could try to convert it to a DateTime object and format the output.
$dateString = '15-Feb-2009 2:24 PM';
$date = DateTime::createFromFormat('j-M-Y g:i A', $dateString);
echo $date->format('Y-m-d G:i'); // will show "2009-02-15 14:24"
In general you should try in avoid holding a date in a string. Convert it to a DateTime -- this makes it also easier for you to manipulate the object (e.g. move date +1 day)
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 am building a timestamp from the date, month and year values entered by users.
Suppose that the user inputs some wrong values and the date is "31-02-2012" which does not exist, then I have to get a false return. But here its converting it to another date nearby. Precisely to: "02-03-2012"..
I dont want this to happen..
$str = "31-02-2012";
echo date("d-m-Y",strtotime($str)); // Outputs 02-03-2012
Can anyone help? I dont want a timestamp to be returned if the date is not original.
You might look into checkdate.
That's because strtotime() has troubles with - since they are used to denote phrase like -1 week, etc...
Try
$str = '31-02-2012';
echo date('d-m-Y', strtotime(str_replace('-', '/', $str)));
However 31-02-2012 is not a valid English format, it should be 02-31-2012.
If you have PHP >= 5.3, you can use createFromFormat:
$str = '31-02-2012';
$d = DateTime::createFromFormat('d-m-Y', $str);
echo $d->format('d-m-Y');
You'll have to check if the date is possible before using strtotime. Strtotime will convert it to unix date meaning it will use seconds since... This means it will always be a date.
You can workaround this behavior
<?php
$str = "31-02-2012";
$unix = strtotime($str);
echo date('d-m-Y', $unix);
if (date('d-m-Y', $unix) != $str){
echo "wrong";
}
else{
echo date("d-m-Y", $unx);
}
or just use checkdate()
Use the checkdate function.
$str = "31-02-2012";
$years = explode("-", $str);
$valid_date = checkdate($years[1], $years[0], $years[2]);
Checkdate Function - PHP Manual & Explode Function - PHP Manual
Combine date_parse and checkdate to check if it's a valid time.
<?php
date_default_timezone_set('America/Chicago');
function is_valid_date($str) {
$date = date_parse($str);
return checkdate($date['month'], $date['day'], $date['year']);
}
print is_valid_date('31-02-2012') ? 'Yes' : 'No';
print "\n";
print is_valid_date('28-02-2012') ? 'Yes' : 'No';
print "\n";
Even though that date format is acceptable according to PHP date formats, it may still cause issues for date parsers because it's easy to confuse the month and day. For example, 02-03-2012, it's hard to tell if 02 is the month or the day. It's better to use the other more specific date parser examples here to first parse the date then check it with checkdate.
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";