PHP问题：如何修复这些if / elseif语句

我正在尝试使用这些if / else if语句来显示这些PHP页面. if / elseif语句允许显示PHP页面.数据存储在mysql中.我们如何获得它,以便如果它已经显示过,它只能输入一次？谢谢.希望您能提供帮助.抱歉,这有点令人困惑.我最近才学英语.
谢谢.

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";
                                                            }

解决方法:

就像是：

$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.";
   }
}

如果找不到该页面,我不确定要做什么.或后续记录重复.