PHP strtotime获取给定年份的最后一天

Question

How can i get the last day of a year passed as parameter using strtotime?

I try this but the results are not the correct last days :

$yearEnd2015 = date('Y-m-d', strtotime('last day of december this year -3'));
    $yearEnd2016 = date('Y-m-d', strtotime('last day of december this year -2'));
    $yearEnd2017 = date('Y-m-d', strtotime('last day of december this year -1'));
    $yearEnd     = date('Y-m-d', strtotime('last day of december this year'));

    echo " <br>2015 : ".$yearEnd2015;
    echo " <br>2016 : ".$yearEnd2016;
    echo " <br>2017 : ".$yearEnd2017;
    echo " <br>2018 : ".$yearEnd;

Answer 1

You can do it like this, just send the last day in the year(31 Dec) to the strtotime function, for example to get the date name of the last day in 2015:

$yearEnd2015 = date('l', strtotime('2015-12-31'));
echo $yearEnd2015;

Or you can store the desired year in a variable(if you want to use it in a function or the year changes dynamically): for example to get all the last date names from 2000 till now:

$year = 2000;
while($year++ <= 2018) {
  $yearEnd = date('l', strtotime($year . '-12-31'));
  echo $yearEnd . '<br />';
}

Hope I pushed you further.