Question

I am trying to use these if/else if statements to display these php pages. The if/elseif statements allow for the php page to show up. The data is stored in the mysql. How do we get it so that if its already being displayed it only enters once? Thank you. I hope you can help. Sorry this is a little confusing. I just learned English recently.
Thank you.

if ($result_array[0] = Politics) {
        require 'news/political.php';
    } elseif ($result_array[0] = Gossip) {
        require 'news/celebgossib';
    }   elseif ($result_array[0] = Entertainment) {
            require 'news/entertainment.php';
        }   elseif ($result_array[0] = Finance) {
                require 'news/finance.php';

            }   elseif ($result_array[0] = Health) {
                    require 'news/health.php';
                }   elseif ($result_array[0] = Leisure) {
                        require 'news/leisure.php';
                    }   elseif ($result_array[0] = Sports) {
                            require 'news/sports.php';
                        }   elseif ($result_array[0] = Tech) {
                                require 'news/tech.php';
                            }   elseif ($result_array[0] = World) {
                                    require 'news/world.php';
                                } else {
                                    echo "There is no interests in your database";
                                }

                                if ($result_array[1] = Politics) {
                                    require 'news/political.php';
                                } elseif ($result_array[1] = Gossip) {
                                    require 'news/celebgossib';
                                }   elseif ($result_array[1] = Entertainment) {
                                        require 'news/entertainment.php';
                                    }   elseif ($result_array[1] = Finance) {
                                            require 'news/finance.php';

                                        }   elseif ($result_array[1] = Health) {
                                                require 'news/health.php';
                                            }   elseif ($result_array[1] = Leisure) {
                                                    require 'news/leisure.php';
                                                }   elseif ($result_array[1] = Sports) {
                                                        require 'news/sports.php';
                                                    }   elseif ($result_array[1] = Tech) {
                                                            require 'news/tech.php';
                                                        }   elseif ($result_array[1] = World) {
                                                                require 'news/world.php';
                                                            } else {
                                                                echo "There is no interests in your database";
                                                            }

Answer 1

Something like:

$pages = array(
  'Politics' => 'political',
  'Gossip' => 'celebgossib',
   ...
);

$used = array();

for ($i = 0; $i < 2; ++$i)
{
   if (array_key_exists($result_array[$i], $pages)
   {
      if (!array_key_exists($result_array[$i], $used))
      {
         # only display this section once
         include 'news/'.$pages[$result_array[$i]].'.php';
         $used[$result_array[$i]] = true;
      }
   }
   else
   {
      echo "Nothing to see here.";
   }
}

I’m not sure exactly what you want to do if the page isn’t found; or if subsequent records are duplicates.