Get year from specified php date

Question

I have a date in this format 2068-06-15 . I want to get the year from the date using php functions. Can anyone suggest how this can be done.

Answer 1

 $date = DateTime::createFromFormat("Ymd", "2068-06-15"); echo $date->format("Y"); 

The DateTime class does not use the unix timestamp inside, so it processes dates before 1970 or after 2038.

Answer 2

You can use strtotime and date acts as follows:

 echo date('Y', strtotime('2068-06-15')); 

Please note that PHP can handle the year until 2038

You can check it out here.

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If your date is always in this format, you can also get the year as follows:

 $parts = explode('-', '2068-06-15'); echo $parts[0]; 

Answer 3

 <?php list ($ year) = explode ( "-", "2068-06-15" ); echo $year; ?> > 

Answer 4

I would use this:

 $parts = explode('-', '2068-06-15'); echo $parts[0]; 

It seems that the date comes from the source, where it is always the same, much faster using explode.

Answer 5

  public function getYear($pdate) { $date = DateTime::createFromFormat("Ymd", $pdate); return $date->format("Y"); } public function getMonth($pdate) { $date = DateTime::createFromFormat("Ymd", $pdate); return $date->format("m"); } public function getDay($pdate) { $date = DateTime::createFromFormat("Ymd", $pdate); return $date->format("d"); } 

Answer 6

Assuming you have a date as a string (sorry, it was not clear from your question, if so), you can break the string into-characters as follows:

 $date = "2068-06-15"; $split_date = split("-", $date); $year = $split_date[0]; 

Answer 7

 $Y_date = split("-","2068-06-15"); $year = $Y_date[0]; 

You can also use explode

Answer 8

You wrote that the format can vary from YYYY-mm-dd to dd-mm-YYYY, you can try to find the year there

 $parts = explode("-","2068-06-15"); for ($i = 0; $i < count($parts); $i++) { if(strlen($parts[$i]) == 4) { $year = $parts[$i]; break; } }