Question

Any suggestions would be good. It is very simple but I cannot figure it out.

I am using a simple jQuery accordion script. Here is the HTML:

<ul id="menu">
    <li>
        <a>Weblog Tools</a> <!-- THIS IS MY LINK BUTTON -->
            <ul>
                <li><a href="http://www.pivotx.net/">PivotX</a></li>
                <li><a href="http://www.wordpress.org/">WordPress</a></li>
                <li><a href="http://www.textpattern.com/">Textpattern</a></li>
                <li><a href="http://typosphere.org/">Typo</a></li>
            </ul>
    </li>
    <li>
        <a>Programming Languages</a> <!-- THIS IS MY LINK BUTTON -->
            <ul>
                <li><a href="http://www.php.net/">PHP</a></li>
                <li><a href="http://www.ruby-lang.org/en/">Ruby</a></li>
                <li><a href="http://www.python.org/">Python</a></li>
                <li><a href="http://www.perl.org/">PERL</a></li>
                <li><a href="http://java.sun.com/">Java</a></li>
                <li><a href="http://en.wikipedia.org/wiki/C_Sharp">C#</a></li>
            </ul>
     </li>
</ul>

And here is the script:

function initMenu() {
    $('#menu ul').hide();
    $('#menu ul:first').show();

    $('a.collapse').click(

function() {
    var checkElement = $(this).next();
        if((checkElement.is('ul')) && (checkElement.is(':visible'))) {
    return false;
        }
        if((checkElement.is('ul')) && (!checkElement.is(':visible'))) {

    $('#menu ul:visible').slideUp('normal');


        checkElement.slideDown('normal');
    return false;
            }
        });
    }

    $(document).ready(function() {initMenu();});

Nice simple script, nice and light.

This is my question, any suggestions would be amazing!

I need to get the <a>Link Button</a> in the current visible menu to be display: none and alternate between open menus.

So the idea is, when the current visible menu is open, the link that opens it has a the style display: none added to it. But all the other <a>Link Buttons</a> should be visible and not have the style display: none applied. This is so the menu can still be functional.

Only the visible open menu should have display: none.

This would be so awesome if you can help. Thanks in advance.

Answer 1

Modify menu.js from

function initMenu() {
  $('#menu ul').hide();
  $('#menu ul:first').show();
  $('#menu li a').click(
    function() {
      var checkElement = $(this).next();
      if((checkElement.is('ul')) && (checkElement.is(':visible'))) {
        return false;
        }
      if((checkElement.is('ul')) && (!checkElement.is(':visible'))) {
        $('#menu ul:visible').slideUp('normal');
        checkElement.slideDown('normal');
        return false;
        }
      }
    );
  }
$(document).ready(function() {initMenu();});

To this:

function initMenu() {
  $('#menu ul').hide();
  $('#menu li a').click(
    function() {
      var checkElement = $(this).next();
      if((checkElement.is('ul')) && (checkElement.is(':visible'))) {
        return false;
        }
      if((checkElement.is('ul')) && (!checkElement.is(':visible'))) {
        $('#menu li a').show();
        $(this).hide();
        $('#menu ul:visible').slideUp('normal');
        checkElement.slideDown('normal');
        return false;
        }
      }
  );
}
$(document).ready(function() {initMenu();});

I modified the first part so it won’t show the first sub menus.
Then I added this lines so that it will show the other links, but hide the active link.

$('#menu li a').show();
$(this).hide();

EDIT:

You can also check a working implementation at http://jsfiddle.net/3cmPz/