Error with PHP 8

Question

Hi i use a widget with php code

on php7.4 everything ok

with php 8 i have an error....

Warning: Undefined variable $post in /xxxxxxx/wp-content/plugins/php-code-widget/execphp.php(27) : eval()'d code on line 2

Warning: Attempt to read property "ID" on null in /xxxxxxx/wp-content/plugins/php-code-widget/execphp.php(27) : eval()'d code on line 2

Warning: Undefined variable $post in /xxxxxxx/wp-content/plugins/php-code-widget/execphp.php(27) : eval()'d code on line 3

Warning: Attempt to read property "ID" on null in /xxxxxxx/wp-content/plugins/php-code-widget/execphp.php(27) : eval()'d code on line 3

My code:

<?php
if (get_the_terms($post->ID, 'ausstattung')) {
  $taxonomy_ar = get_the_terms($post->ID, 'ausstattung');
echo "";
  $output = '<ul>';
  foreach ($taxonomy_ar as $taxonomy_term) {
    $output .= '<li>'. $taxonomy_term->name .'</li>';
  }
  $output .= '</ul>';

  echo $output;
}
?>

What is wrong?

Answer 1

You are using a global variable called $post in your code, but this variable is not defined.

To fix this issue, you can try defining the $post variable at the top of your script:

 <?php
    global $post;
    
    if (get_the_terms($post->ID, 'ausstattung')) {
      $taxonomy_ar = get_the_terms($post->ID, 'ausstattung');
    echo "";
      $output = '<ul>';
      foreach ($taxonomy_ar as $taxonomy_term) {
        $output .= '<li>'. $taxonomy_term->name .'</li>';
      }
      $output .= '</ul>';
    
      echo $output;
    }
    ?>