EvilNabster EvilNabster - 5 months ago 12
PHP Question

Easier menu generation depended on session

I am making a menu for website language and it works but i think the way i am doing it is not the best way in case i add more languages in future.
Maybe someone can show a better example.
Here is a piece of what i've got.

if(isSet($_SESSION['lang'])) {
$lang = $_SESSION['lang'];
if ($lang == "en") {
echo '<li class="">EN
<ul>
<li class="submenu">RU</li>
<li class="submenu">ES</li>
<li class="submenu">GR</li>
<li class="submenu">DE</li>
</ul>
</li>';
}
if ($lang == "de") {
echo '<li class="">DE
<ul>
<li class="submenu">RU</li>
<li class="submenu">ES</li>
<li class="submenu">GR</li>
<li class="submenu">EN</li>
</ul>
</li>';
}
// and so on for every language..
}


Probably would be better doing some array and than foreach function, but i have no idea how to do that x,x way too confusing

Answer

If you want an array, it's possible to create a function, this function will get the language as parameter and display the options according to it :

<?php

function display ( $lang )
{ $arr = array( "en" => array( "RU","ES","GR","DE" ),
                "de" => array( "RU","ES","GR","DE" )
              );
  echo '<li class="">' .
       '<ul>';
  foreach ( $arr[ $lang ] as $item ) // DISPLAY THE OPTIONS.
    echo '<li class="submenu">' . $item . '</li>';
  echo '</ul>
        </li>';
}

display( "en" ); // DISPLAY OPTIONS FOR "EN".
display( "de" ); // DISPLAY OPTIONS FOR "DE".

?>

In the future, you may add more languages to the array.

To test it, copy-paste previous code in a file, save it as PHP and open it in your browser.