I have a date in this format 2068-06-15. I want to get the year from the date, using php functions. Could someone please suggest how this could be done.
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Yes, I have maximum input 2069Shakti Singh– Shakti Singh2010-12-25 07:54:56 +00:00Commented Dec 25, 2010 at 7:54
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10 Answers
$date = DateTime::createFromFormat("Y-m-d", "2068-06-15");
echo $date->format("Y");
The DateTime class does not use an unix timestamp internally, so it han handle dates before 1970 or after 2038.
Comments
You can use the strtotime and date functions like this:
echo date('Y', strtotime('2068-06-15'));
Note however that PHP can handle year upto 2038
If your date is always in that format, you can also get the year like this:
$parts = explode('-', '2068-06-15');
echo $parts[0];
6 Comments
Shakti Singh
I already tried this but it is giving me 1970 that it is why php can't read the year above the 2038. right?
Sarfraz
@Shakti: Because as i have said in my answer, PHP can handle year up to 2038 and you have specified 2068. You can test it here: codepad.org/iwqtML6s
Shakti Singh
ya, I agree with you but I have maximum input 2069. so I think I have to use substr or similar function which can split it but the problem is input date fromat can be changed now it is yyyy-mm-dd may be later changed to mm-dd-yyyy
Sarfraz
@Shakti: As i have shown in my answre later, you can get the date with
explode function too: $parts = explode('-', '2068-06-15'); echo $parts[0];. You can test it here: codepad.org/HbS63y2n
Maerlyn
I do not understand why you're using unix timestamps when the OP has given an example outside of it's range.
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You can try strtotime() and date() functions for output in minimum code and using standard way.
echo date('Y', strtotime('2068-06-15'));
output:
2068
echo date('y', strtotime('2068-06-15'));
output:
68
Comments
I would use this:
$parts = explode('-', '2068-06-15');
echo $parts[0];
It appears the date is coming from a source where it is always the same, much quicker this way using explode.
Comments
public function getYear($pdate) {
$date = DateTime::createFromFormat("Y-m-d", $pdate);
return $date->format("Y");
}
public function getMonth($pdate) {
$date = DateTime::createFromFormat("Y-m-d", $pdate);
return $date->format("m");
}
public function getDay($pdate) {
$date = DateTime::createFromFormat("Y-m-d", $pdate);
return $date->format("d");
}
Comments
<?php
list($year) = explode("-", "2068-06-15");
echo $year;
?>
Comments
You can achieve your goal by using php date() & explode() functions:
$date = date("2068-06-15");
$date_arr = explode("-", $date);
$yr = $date_arr[0];
echo $yr;
That is it. Happy coding :)
Comments
Assuming you have the date as a string (sorry it was unclear from your question if that is the case) could split the string on the - characters like so:
$date = "2068-06-15";
$split_date = split("-", $date);
$year = $split_date[0];
1 Comment
Sarfraz
Note that
split is deprecated.$Y_date = split("-","2068-06-15");
$year = $Y_date[0];
You can use explode also
Comments
You wrote that format can change from YYYY-mm-dd to dd-mm-YYYY you can try to find year there
$parts = explode("-","2068-06-15");
for ($i = 0; $i < count($parts); $i++)
{
if(strlen($parts[$i]) == 4)
{
$year = $parts[$i];
break;
}
}
