Preventing a dropdown from appearing behind other dropdowns

Question

Preventing a dropdown from appearing behind other dropdowns

88 Views Asked by vvye At 16 November 2025 at 10:05

I'm building a custom dropdown. The real thing is going to have all the necessary ARIA attributes of course, but here's the barebones version:

[...document.querySelectorAll('.select')].forEach(select => {
  select.addEventListener('click', function() {
    select.nextElementSibling.classList.toggle('visible');
  });
});

.dropdown {
  position: relative;
  width: 16rem;
  margin-bottom: 2rem;
}

.select {
  display: flex;
  align-items: center;
  height: 2rem;
  background-color: #ccc;
}

.popup {
  display: none;
  position: absolute;
  left: 0;
  right: 0;
  height: 10rem;
  background-color: #eee;
  box-shadow: 0 0 0.5rem red;
}

.popup.visible {
  display: block;
}

<!doctype html>
<html>

<body>

  <div class="dropdown">
    <div class="select">Button 1 ▼</div>
    <div class="popup">Popup 1</div>
  </div>

  <div class="dropdown">
    <div class="select">Button 2 ▼</div>
    <div class="popup">Popup 2</div>
  </div>

</body>

</html>

The obvious issue is that, when you open the first dropdown, popup 1 appears behind button 2. The obvious solution would be to give .popup a z-index, and make it an absurdly large value like 999 to make sure it appears above other elements on the page as well.

However, in my case, I would also like the popup to appear behind its corresponding button (in order to hide its box-shadow).

If I give the button a z-index greater than the popup's, the original problem returns: popup 1 appears behind button 2. If I instead give the z-index: 999 to the entire .dropdown and create a new stacking context, the same thing happens.

Is there any way I can meet my two requirements at the same time (popup behind its button, and only that one, but above everything else on the page)?

Original Q&A

There are 1 best solutions below

**GoodMan** · Accepted Answer

You could track the dropdown .open state. And use that to toggle the display property of its child .popup. However the .dropdown.open state will have a z-index:1, that way it will always show up on top of elements below it.

[...document.querySelectorAll('.select')].forEach(select => {
  select.addEventListener('click', function(event) {
    const {
      target: {
        parentElement: activeDropdown
      }
    } = event;
    activeDropdown.classList.toggle('open');
    const container = select.parentElement.parentElement;
    const dropdowns = container.querySelectorAll('.dropdown');
    Array.from(dropdowns).forEach(item => {
      if (item !== activeDropdown) {
        item.classList.remove('open')
      }
    })
  });
});

.dropdown {
  position: relative;
  width: 16rem;
  margin-bottom: 2rem;
}

.dropdown.open {
  z-index: 1;
}

.dropdown.open>.popup {
  display: block;
}

.select {
  display: flex;
  align-items: center;
  height: 2rem;
  background-color: #ccc;
}

.popup {
  display: none;
  position: absolute;
  left: 0;
  right: 0;
  height: 10rem;
  background-color: #fff;
  box-shadow: 0 0 0.5rem red;
}

<!doctype html>
<html>

<body>

  <div class="dropdown">
    <div class="select">Button 1 ▼</div>
    <div class="popup">Popup 1</div>
  </div>

  <div class="dropdown">
    <div class="select">Button 2 ▼</div>
    <div class="popup">Popup 2</div>
  </div>

</body>

</html>

Preventing a dropdown from appearing behind other dropdowns

There are 1 best solutions below

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