menu link selector in JQuery

I have this html menu:

<div id="menu">
    <ul>
          <li>
                <a href="1.html"> 1 </a>
          </li>
          <li>
                <a href="2.html"> 2 </a>
          </li>
          <li class="submenu">
                <a href="#"> 3 list </a>
                <ol>
                     <li>
                           <a href="31.html"> 31 </a>
                     </li>
                     <li>
                           <a href="32.html"> 32 </a>
                     </li>
                     <li>
                           <a href="33.html"> 33 </a>
                     </li>
                </ol>
          </li>
          <li class="submenu">
                <a href="#"> 4 list </a>
                <ol>
                     <li>
                           <a href="41.html"> 41 </a>
                     </li>
                     <li>
                           <a href="42.html"> 42 </a>
                     </li>
                </ol>
          </li>
          <li>
                <a href="5.html"> 5 </a>
          </li>
    </ul>
 </div>

I'm using JQuery and I wanted to get the link elements (a) that are not part of the submenu (1,2 and 5), so i tried:

$('#menu ul li a').each(function(){
   //do something with the link element
});

But i realized that this selector takes all the link elements(1,2,31,32,33,41,42 and 5).

I checked the JQuery manual and i could not find any way to get the link elements that I want in only one step, so I finally did this:

$('#menu ul li a').each(function(){
   if($(this).parent().parent().is('ul')){
      //do something with the link element
   }
});

That just takes all the link elements and in the conditional filters the ones that have an ul grandfather (1,2 and 5).

I would like to know if the开发者_如何学编程re is a better and more clean way to get the not submenu link elements.

Thanks!

---

Try this

$('#menu ul li a').not("li.submenu").each(function(){ });

Actually, that may return the inside li(s) still.

So maybe another not?

$('#menu ul li a').not(".submenu").not(".submenu *").each(function(){ });

---

Hopefully I'm not mistakenly on understanding your question..

You want to get first level li elements which don't have sub list? Then it means li element 1, 2 & 5? If so, try this:

$('#menu ul > li:not(:has(li))').each(...);

It only check for first level element after #menu by using > sign. Then filter again for each li that doesn't has li inside.

Hope this helps..