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syntax – PHP "unexpected $end"

Posted by: admin July 12, 2020 Leave a comment

Questions:

i actually checked my code like a hundred times, but i cant find the error:

The error i get:
Parse error: syntax error, unexpected $end in /home/tennox/public_html/php/kalender.php on line 46

I tried to do a function that calculates the easter sunday, in another script without all the other it just works, but in this one not. The other way round the same thing!

<?php
$year = isset($_POST['year']) ? intval($_POST['year']) : date('Y');
$months = array("", "Januar", "Februar", "März", "April", "Mai", "Juni", "Juli", "August", "September", "Oktober", "November", "Dezember");
$days = array('So', 'Mo', 'Di', 'Mi', 'Do', 'Fr', 'Sa');
$ostern = getOsterSonntag($year);
?>
<form action="kalender.php" method="POST" target="_self">
<input type="text" name="year" value="<?php echo $year; ?>">
</form>
<table border="1" cellspacing="0">
<?php
for($y=0; $y<32; $y++) {
   echo "<tr height=\"20\">";
   for($x=1; $x<=12; $x++) {
      echo "<td width=\"5%\">";
      $date = strtotime("$y.$x.$year");
      $day = date("d", $date);
      $dayname = $days[date("w", $date)];

      if ($y == 0) {
         echo "<div align=\"center\"><b>$months[$x]</b></div>";
      } elseif ($y < date("t",$date) && !($y == 30 && $x == 2)) {
         if (date("w", $date) == 6 || date("w", $date) == 0)
            echo "<b>";
         echo "$day $dayname";
         if ($date == $ostern)
            echo "(Ostersonntag)";
         if (date("w", $date) == 6 || date("w", $date) == 0)
            echo "</b>";
      } else {
         echo "   -   ";
      }
      echo "</td>";
   }
   echo "</tr>";
}
?>
</table>

<?php
function getOsterSonntag($year) {
   $a = $year % 19;
   $b = $year % 4;
   $c = $year % 7;
   $k = floor($year / 100);
   $p = floor((8*$k + 13) / 25);
   $q = ($k / 4);
   $d = (19*$a + ((15 + $k - $p - $q) % 30)) % 30;
   $e = (2*$b + 4*$c + 6*$d + ((4 + $k - $q) % 7)) % 7;

   $ostern = 22 + $d + $e;
   if ($ostern > 31){
      $ostern -= 31;
      return strtotime("$ostern.4.$year");
   } else
      return strtotime("$ostern.3.$year");
}
?>
How to&Answers:

That error means that PHP has finished analyzing your code, but you forgot to close a symbol somewhere in your page.Its either you forgot to close a quote, bracket, parenthesis or comma.

Hope this helps.

Answer:

This error usually means that you missed a }. Check all your braces and make sure you have the same number of { as } – a code editor with bracket matching (such as Notepad++) can make this a lot easier.

Answer:

In this part of your code, i think you should add { } for else.

 $ostern = 22 + $d + $e;
       if ($ostern > 31){
          $ostern -= 31;
          return strtotime("$ostern.4.$year");
       } else
          return strtotime("$ostern.3.$year");

also in this part put { } in this part like this :

if (date("w", $date) == 6 || date("w", $date) == 0) {
      echo "<b>";
      echo "$day $dayname";
}