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css - 不滑动我的切换滑动按钮确认否

问题描述

我已经通过 CSS 设置了我的切换滑动按钮。但我也有一个关于滑动的警报弹出窗口。当用户滑动时，会弹出警报并要求用户确认。如果用户选择“是”，它会正确滑动。但是如果用户选择“取消”，它仍然会滑动。如果用户选择在警报框中取消，我如何设置我的 CSS 不滑动？

这是我的 CSS 和 HTML 代码的样子

<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1">
<style>
.switch {
  top: 25px;
  right: 65px;
  position: relative;
  display: inline-block;
  width: 100px;
  height: 34px;
}

.switch input { 
  opacity: 0;
  width: 0;
  height: 0;
}

.slider {
  position: absolute;
  cursor: pointer;
  top: 0;
  left: 0;
  right: 0;
  bottom: 0;
  background-color: #ccc;
  -webkit-transition: .4s;
  transition: .4s;
}

.slider:before {
  position: absolute;
  content: "";
  height: 26px;
  width: 26px;
  left: 4px;
  bottom: 4px;
  background-color: white;
  -webkit-transition: .4s;
  transition: .4s;
}

input:checked + .slider {
  background-color: #2196F3;
}

input:focus + .slider {
  box-shadow: 0 0 1px #2196F3;
}

input:checked + .slider:before {
  -webkit-transform: translateX(66px);
  -ms-transform: translateX(66px);
  transform: translateX(66px);
}

.on
{
  display: none;
}

.on, .off
{
  color: white;
  position: absolute;
  transform: translate(-50%,-50%);
  top: 50%;
  left: 50%;
  font-size: 10px;
  font-family: Verdana, sans-serif;
}

input:checked+ .slider .on
{display: block;}

input:checked + .slider .off
{display: none;}

/* Rounded sliders */
.slider.round {
  border-radius: 34px;
}

.slider.round:before {
  border-radius: 50%;
}
</style>
</head>
<body>

<h2>Toggle Switch</h2>
<tr>
<td>
<button type="button" class="btn btn-primary">HELLO
</button>
<label class="switch">
  <input type="checkbox">
  <span class="slider"></span>
</label>
</td>
<td>
<label class="switch">
  <input type="checkbox" checked>
  <span class="slider"></span>
</label><br><br>
</td>
</tr>
<tr>
<td>
<label class="switch">
  <input type="checkbox">
  <span class="slider round">
    <span class="on">Enable</span> 
    <span class="off">Disable</span>
  </span>
</label>
</td>
<td>
<label class="switch" onclick="confirm('Yes');">
  <input type="checkbox" checked>
  <span class="slider round"></span>
</label>
</td>
</tr>
</body>
</html>

标签： cssbuttontoggle

解决方案

单靠 CSS 无法解决这个问题，因为您想反转用户操作，您必须使用 JavaScript/jQuery。检查以下代码段

function confirmClick() {
  var r = confirm("Press Cancel");
  if (r !== true) {
    document.getElementById('switchCheck').checked = !document.getElementById('switchCheck').checked
    
  }

}
.switch {
  top: 25px;
  right: 65px;
  position: relative;
  display: inline-block;
  width: 100px;
  height: 34px;
}

.switch input { 
  opacity: 0;
  width: 0;
  height: 0;
}

.slider {
  position: absolute;
  cursor: pointer;
  top: 0;
  left: 0;
  right: 0;
  bottom: 0;
  background-color: #ccc;
  -webkit-transition: .4s;
  transition: .4s;
}

.slider:before {
  position: absolute;
  content: "";
  height: 26px;
  width: 26px;
  left: 4px;
  bottom: 4px;
  background-color: white;
  -webkit-transition: .4s;
  transition: .4s;
}

input:checked + .slider {
  background-color: #2196F3;
}

input:focus + .slider {
  box-shadow: 0 0 1px #2196F3;
}

input:checked + .slider:before {
  -webkit-transform: translateX(66px);
  -ms-transform: translateX(66px);
  transform: translateX(66px);
}

.on
{
  display: none;
}

.on, .off
{
  color: white;
  position: absolute;
  transform: translate(-50%,-50%);
  top: 50%;
  left: 50%;
  font-size: 10px;
  font-family: Verdana, sans-serif;
}

input:checked+ .slider .on
{display: block;}

input:checked + .slider .off
{display: none;}

/* Rounded sliders */
.slider.round {
  border-radius: 34px;
}

.slider.round:before {
  border-radius: 50%;
}
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="width=device-width, initial-scale=1">
</head>
<body>

<h2>Toggle Switch</h2>
<tr>

<label class="switch" onchange="confirmClick();">
  <input id="switchCheck" type="checkbox" checked>
  <span class="slider round"></span>
</label>
</td>
</tr>
</body>
</html>

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