javascript - 以编程方式更改翻转开关值

Question

我试过摆弄一些 HTML/JS/CSS但是卡在了如何将翻转开关动态更改为所需状态上。

CSS 来自这里 https://proto.io/freebies/onoff/

function on() {
  document.getElementById("innerswitchid").className = document.getElementById("innerswitchid").className + "onoffswitch-inner:before";
}

function off() {
  document.getElementById("innerswitchid").className = document.getElementById("innerswitchid").className + "onoffswitch-inner:after";
}

.onoffswitch {
  position: relative;
  width: 90px;
  -webkit-user-select: none;
  -moz-user-select: none;
  -ms-user-select: none;
}

.onoffswitch-checkbox {
  display: none;
}

.onoffswitch-label {
  display: block;
  overflow: hidden;
  cursor: pointer;
  border: 2px solid #999999;
  border-radius: 20px;
}

.onoffswitch-inner {
  display: block;
  width: 200%;
  margin-left: -100%;
  transition: margin 0.3s ease-in 0s;
}

.onoffswitch-inner:before,
.onoffswitch-inner:after {
  display: block;
  float: left;
  width: 50%;
  height: 30px;
  padding: 0;
  line-height: 30px;
  font-size: 14px;
  color: white;
  font-family: Trebuchet, Arial, sans-serif;
  font-weight: bold;
  box-sizing: border-box;
}

.onoffswitch-inner:before {
  content: "ON";
  padding-left: 10px;
  background-color: #34A7C1;
  color: #FFFFFF;
}

.onoffswitch-inner:after {
  content: "OFF";
  padding-right: 10px;
  background-color: #EEEEEE;
  color: #999999;
  text-align: right;
}

.onoffswitch-switch {
  display: block;
  width: 18px;
  margin: 6px;
  background: #FFFFFF;
  position: absolute;
  top: 0;
  bottom: 0;
  right: 56px;
  border: 2px solid #999999;
  border-radius: 20px;
  transition: all 0.3s ease-in 0s;
}

.onoffswitch-checkbox:checked+.onoffswitch-label .onoffswitch-inner {
  margin-left: 0;
}

.onoffswitch-checkbox:checked+.onoffswitch-label .onoffswitch-switch {
  right: 0px;
}

<div class="onoffswitch">
  <input type="checkbox" name="onoffswitch" class="onoffswitch-checkbox" id="myonoffswitch" checked>
  <label class="onoffswitch-label" for="myonoffswitch">
        <span id="innerswitchid" class="onoffswitch-inner"></span>
        <span class="onoffswitch-switch"></span>
      </label>
</div>
<button onclick="on()">ON</button>
<button onclick="off()">OFF</button>

然而，这只会让 flipswich 崩溃。如何使用 js 更改组件的 css 将 flipswich 动态更改为所需状态？

Answer 1

将您的代码更新为以下内容:

HTML

<button id="js-btn--on">
  ON
</button>
<button id="js-btn--off">
  OFF
</button>

JS

   document.getElementById("js-btn--on").onclick = on;
   document.getElementById("js-btn--off").onclick = off;

   function on() {
     document.getElementById('myonoffswitch').checked = true;

   }

   function off() {
     document.getElementById('myonoffswitch').checked = false;
   }

您可以切换复选框，让 CSS 完成它想要完成的工作，而不是混合使用 JS 和 CSS。