在下拉菜单中切换相同 class 的各个元素

Question

I am building a dropdown menu. As is, when I open one submenu, it stays open, even if I open a second submenu. When I open the second submenu, it should close any other open menus.

 const submenuLinks = document.querySelectorAll('.has-submenu > a'); submenuLinks.forEach(element => element.addEventListener('click', function(e) { var submenu = element.nextElementSibling; if (submenu.classList.contains('menu-hide')) { submenu.classList.add('menu-show'); submenu.classList.remove('menu-hide'); } else { submenu.classList.add('menu-hide'); submenu.classList.remove('menu-show'); } }));

 .menu-hide { display: none; }.menu-show { display: block; }

 <nav> <ul class="menu"> <li> <a href="#">Menu Item 1</a> </li> <li class="has-submenu"> <a href="#">Menu Item 2 &raquo;</a> <ul class="menu-sub menu-hide"> <li><a href="#">Submenu Item 1</a></li> <li><a href="#">Submenu Item 2</a></li> <li><a href="#">Submenu Item 3</a></li> </ul> </li> <li class="has-submenu"> <a href="#">Menu Item 3 &raquo;</a> <ul class="menu-sub menu-hide"> <li><a href="#">Submenu Item 1</a></li> <li><a href="#">Submenu Item 2</a></li> <li><a href="#">Submenu Item 3</a></li> </ul> </li> </ul> </nav>

Answer 1

before your logic that hides the current submenu, you can hide every submenu (ul with class 'menu-sub'). In this way, only the current submenu will be expanded. Here's the edited javascript code:

const submenuLinks = document.querySelectorAll('.has-submenu > a');

    submenuLinks.forEach((element) =>
      element.addEventListener('click', function (e) {
        var allSubmenus = document.querySelectorAll('.menu-sub');
        allSubmenus.forEach((submenu) => {
          submenu.classList.remove(...submenu.classList);
          submenu.classList.add('menu-sub');
          submenu.classList.add('menu-hide');
        });
        var currentSubmenu = element.nextElementSibling;
        if (currentSubmenu.classList.contains('menu-hide')) {
          currentSubmenu.classList.add('menu-show');
          currentSubmenu.classList.remove('menu-hide');
        } else {
          currentSubmenu.classList.add('menu-hide');
          currentSubmenu.classList.remove('menu-show');
        }
      })
    );

What I did there extra was to remove all the classes from the submenus, add back the 'menu-sub' and 'menu-hide' classes.

Answer 2

You must loop over all ul's to check the classes. I commented what i add to you js code.

  const uls = document.querySelectorAll('.menu-sub');
  uls.forEach(a => {
      if (a.classList.contains('menu-show')) {        
        a.classList.remove('menu-show')  
        a.classList.add('menu-hide')  
      }      
    })

 const submenuLinks = document.querySelectorAll('.has-submenu > a'); submenuLinks.forEach(element => element.addEventListener('click', function(e) { /* start */ const uls = document.querySelectorAll('.menu-sub'); uls.forEach(a => { if (a.classList.contains('menu-show')) { a.classList.remove('menu-show') a.classList.add('menu-hide') } }) /* end */ var submenu = element.nextElementSibling; if (submenu.classList.contains('menu-hide')) { submenu.classList.add('menu-show'); submenu.classList.remove('menu-hide'); } else { submenu.classList.add('menu-hide'); submenu.classList.remove('menu-show'); } }));

 .menu-hide { display: none; }.menu-show { display: block; }

 <nav> <ul class="menu"> <li> <a href="#">Menu Item 1</a> </li> <li class="has-submenu"> <a href="#">Menu Item 2 &raquo;</a> <ul class="menu-sub menu-hide"> <li><a href="#">Submenu Item 1</a></li> <li><a href="#">Submenu Item 2</a></li> <li><a href="#">Submenu Item 3</a></li> </ul> </li> <li class="has-submenu"> <a href="#">Menu Item 3 &raquo;</a> <ul class="menu-sub menu-hide"> <li><a href="#">Submenu Item 1</a></li> <li><a href="#">Submenu Item 2</a></li> <li><a href="#">Submenu Item 3</a></li> </ul> </li> </ul> </nav>

Answer 3

• There is no need to use two different classes to show and hide the submenus.
• You can hide all submenus by default with CSS and only show the one with .menu-show

Try this

 const submenuLinks = document.querySelectorAll('.has-submenu > a'); submenuLinks.forEach(element => element.addEventListener('click', function(e) { var submenu = element.nextElementSibling; if (submenu.classList.contains('menu-show')) { submenu.classList.remove('menu-show'); } else { submenuLinks.forEach(a => a.nextElementSibling.classList.remove('menu-show')); submenu.classList.add('menu-show'); } }));

 .menu-sub { display: none; }.menu-show { display: block; }

 <nav> <ul class="menu"> <li> <a href="#">Menu Item 1</a> </li> <li class="has-submenu"> <a href="#">Menu Item 2 &raquo;</a> <ul class="menu-sub"> <li><a href="#">Submenu Item 1</a></li> <li><a href="#">Submenu Item 2</a></li> <li><a href="#">Submenu Item 3</a></li> </ul> </li> <li class="has-submenu"> <a href="#">Menu Item 3 &raquo;</a> <ul class="menu-sub"> <li><a href="#">Submenu Item 1</a></li> <li><a href="#">Submenu Item 2</a></li> <li><a href="#">Submenu Item 3</a></li> </ul> </li> </ul> </nav>