Question

<div class="div-0"></div>
<div class="div-1">
   <button-class="test-button">test1</button>
   <button-class="test-button">test2</button>
   <button-class="test-button">test3</button>

</div>
<div class="div-2"> 
   <button-class="test-button">test4</button>
   <button-class="test-button">test5</button>
   <button-class="test-button">test6</button>

</div>
<div class="div-3"></div>

我知道如何计算两个div之间的距离。

var disnace_new=$('.div-1').offset().top - $('.div-2').offset().top;

对于某些事件，我将增加margin-top属性.example

$(".test-button").on("click",function(){

     $(".div-2").css("margin-top","new-value");
       continue.......

});

在这里，我无法获得新的价值，也无法获得新的价值。在这种情况下，我想使两个div之间的距离为-10

怎么可能。

即$('.div-1').offset().top - $('.div-2').offset().top设置为 -10

$('.div-1').offset().top - $('.div-1').offset().top =-10;

这可能吗？

Answer 1

您可以尝试这样的事情：

＆＃13;

var distance_new = $('.div-1').offset().top - $('.div-2').offset().top;
console.log(distance_new);

$(".test-button").on("click",function(){

     $(".div-2").css("margin-top",distance_new+$('.div-1').outerHeight()+"px");
     console.log($('.div-1').offset().top - $('.div-2').offset().top);
})

＆＃13;

* {
box-sizing:border-box;
}

.div-1 {
  padding: 10px;
  margin: 20px 0;
}

.div-2 {
  padding: 10px;
  margin: 30px 0;
}

.space {
  height: 10px;
}

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="div-0"></div>
<div class="div-1">
  <button class="test-button">test1</button>
  <button class="test-button">test2</button>
  <button class="test-button">test3</button>
</div>
<div class="space"></div>
<div class="div-2">
  <button class="test-button">test4</button>
  <button class="test-button">test5</button>
  <button class="test-button">test6</button>
</div>
<div class="div-3"></div>

＆＃13;

设置两个div（两个类）相同

1 个答案: