PHP Question: How to fix these if/elseif statements

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

Answer 1

You need to use == instead of = I don't know what the right side of the statement is, for example Politics. If its not a variable then you should put it in quotes.

Something like this

if ($result_array[0] == "Politics") {
        require 'news/political.php';
    }else ...

Answer 2

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.

Answer 3

How about using a switch statement?

switch $result_array[0] {
     case 'Politics':     include('politics.php'); break;
     case 'This': include(...); break;
     case 'That': include(....); break;
     default: include('default.php'); break;
}

for a loooong set of if/then/else tests with simple "todo" sections, a switch statement is ideal.

Answer 4

It looks like you're iterating over an array of options, and including a bunch of set files?

I would try something like the following:

switch( TRUE )
{
    case in_array("Politics", $result_array):
        require 'news/political.php';
        break; 

    case in_array("Gossip", $result_array):
        require 'news/celebgossib';
        break;

    // etc.
}