3按钮Jquery on/off和on/off

Question

我有3个按钮:按钮1，按钮2和按钮3。按钮1显示/隐藏左侧。按钮2做同样的，但做右边。按钮3两边都是。怎样才是正确的方法？

https://jsfiddle.net/anthony0perez/m9h2n1vk/2/

<div id="list-map-button" class="btn map-btn">LIST</div>
<div id="map-map-button" class="btn map-btn">MAP</div>
<div id="both-map-button" class="btn map-btn map-btn-active">BOTH</div>
<br style="clear:both;"/>
<div class="container">
    <div id="left_side">
        map side
        <div  id="block-3">Block 3</div>
        <div id="block-4" class="not-active-block">Block 4</div>
    </div>
    <div id="right_side">
        list side
        <div id="block-5">Block 5</div>
        <div id="block-6" class="not-active-block">Block 6</div>
    </div>
    <br style="clear:both;"/>
</div>

    $(document).ready(function(){    
    $( "#both-map-button" ).click(function() {
        //alert('clicked both button');
        $("#block-4, #block-3, #block-5, #block-6").hide();
        $("#block-4, #block-3, #block-5, #block-6").addClass("acvite-btn");

        $("#block-4, #block-3").show();
        $("#block-5, #block-6").show();

    });
    $( "#map-map-button" ).click(function() {
        //alert('clicked map button');
        $("#block-5").toggle();
        $("#block-6").toggle();
    });
    $( "#list-map-button" ).click(function() {
         //alert('clicked list button');
        $("#block-4").toggle();
        $("#block-3").toggle();
    });
 });

Answer 1

您可以简单地使用一个变量来保持状态。注意，我为这个例子硬编码了循环，因为我们总是处理三个按钮。

​

inputs = document.getElementsByTagName('input');
state = 0;
for (var i = 0; i < 3; i++) {
  (function(i) {
    inputs[i].addEventListener('click', function() {
      state ^= (i + 1);
      for (var j = 1; j < 3; j++) {
        inputs[j - 1].classList.toggle('on', (j & state));
      }
    });
  }(i));
}

input {
  background: lightgray
}
input.side {
  background: red
}
input.on {
  background: limegreen
}

<input type="button" class="side" value="toggle left" />
<input type="button" class="side" value="toggle right" />
<input type="button" value="toggle both" />

​

如果您希望打开两个开关，如果其中一个已经打开，则只需添加一个单独的检查即可。

if (i & 2) {
    state = ~~(state > 0) * 3;
}