[Resolved] Question about "switch" command

Tomz

Right now I am using the swich command in one of my scripts but am wondering if it's smart to use that command in this particular situation. Maybe some of you have a better/smarter idea...

I have a file (menu.php) that loads a menu with links for the user to click on. The menu choice that is the page that the user is browsing right now should be in a different colour.

The complete page, including the menu is reloaded when the user clicks a link. I am not using frames.
In a stylesheet I have defined two classes; one for a normal link and one for an active link in the menu.

There are four menu links. In menu.php I have assigned a variable to each choice. So

Link 1 = $link1
Link 2 = $link2, and so on

Lets say the user clicks on link 2, that link should be highlighted. To do this variable $link2 is assigned the value of "menu_active" (menu_active being the name of the stylesheet class that displays an active link). The other link-variables are assigned the value of the not-active-link.
This way the right class is used for the active link.

The menu looks like this:


<?php
  echo "<td class=\"$link1\">";
  echo "Choice 1";
  echo "</td>";

  echo "<td class=\"$link2\">";  // should be "menu_active" in this example
  echo "Choice 2";
  echo "</td>";

  echo "<td class=\"$link3\">";
  echo "Choice 3";
  echo "</td>";

  echo "<td class=\"$link4\">";
  echo "Choice 4";
  echo "</td>";



?>

When the user clicks a link the link number is stored in a cookie $menuC, which I use when the menu is reloaded after the click.
The link number that corresponds with $menuC is then given the value of "menu_active".

To evaluate $menuC to see what choice was clicked I use the switch command below:


switch ($menuC){
    case 1:
    	$link1 = menu_active;
    case 2:
    	$link2 = menu_active;
    case 3:
    	$link3 = menu_active;
    case 4:
    	$link4 = menu_active;
    break;

With only four menu choices it's easy but what if i have to add many more later...?

I think there must be a much easier and smarter way to do this...
Anyone? 🙂

slick

You could use an array to store the link classes (you could even expand it to link URLs and names). On each page, keep a variable denoting the page number. Then use a function to generate the menus.

In fact, I wouldn't store the classes at all (because you've only two of them). I would store the link URLs and names in two arrays in an include file.

e.g.

$linkNamesArray = array("Link1", "Link2", "Link3", "Link4");
$linkURLsArray = array("page1.php", "page2.php", "page3.php", "page4.php");

Your function would be:

function loadMenus($activeMenu){
   GLOBAL $linkNamesArray, $linkURLsArray;

   $numMenus = count($linkNamesArray);
   for($idx=0; $idx<$numMenus; $idx++){
      if($idx==$activeMenu){
         echo("<a href='" . $linkURLsArray[$idx] . "' class='active'>");
         echo($linkNamesArray[$idx]);
         echo("</a><br>");
      }
      else{
         echo("<a href='" . $linkURLsArray[$idx] . "' class='normal'>");
         echo($linkNamesArray[$idx]);
         echo("</a><br>");
      }
   }
}

then on each page, you'd have something like:

$thisPage = 2;
loadMenus($thisPage);

Hope that helps,
Slick

Tomz

Yes, this it is!

I started experimenting with arrays on this myself already. However, it's always good to have someone else's fresh view on a problem and see it's really not a problem at all 🙂

slick

I've used this method many times on various sites to great success. It's really quick, reduces coding, is easy to debug, and easy to expand if you need to add more pages or items in the menu.

However, it does have its limitations - if you have a large number of menu items and you need to change the order of them, it can be a hassle updating all the page numbers. Also, sub-menus would be tricky to add.

But, if you know that your menus are going to increase by much over time, I've found that this method is the easiest and quickest to deploy.

Good luck!
Slick