PHP date() print 24:00 instead of 00:00

Question

PHP date("H:i (d.m.Y)",$timestamp) function represents exact midnight as 00:00 of following day. But I need it to represent it as 24:00 of preceding day. Is it possible to be done without writing completely new date() parser function?

edit: Why do I need such 'weird' format? Because my client demands it. In my country (CZ), 24:00 is sometimes used when referring to exact midnight.

edit2: My current 'dirty' solution is: (does not work with all possible format strings)

function date_24midnight($format,$ts)
{
   if(date("Hi",$ts)=="0000")
      return preg_replace('/23:59/',"24:00",date($format,$ts-1));
   else
      return date($format,$ts);
}

Answer 1

function date_24midnight($format,$ts)
{
   if(date("Hi",$ts)=="0000") {
      $replace = array(
        "H" => "24",
        "G" => "24",
        "i" => "00",
      );

      return date(
        str_replace(
          array_keys($replace),
          $replace, 
          $format
        ),
        $ts-60 // take a full minute off, not just 1 second
      );
   } else {
      return date($format,$ts);
   }
}

This function is based on yours, But works for all formats (that I tested)

So H:i:s works for example

It will need some extra work to for for formats like "\H\i Y-m-d"