Question

我希望有一个滑动（向下）切换，但希望面板整齐地沿着按钮的左边缘排列。我不想硬编码边距，位置，px等，但希望下滑面板始终与（和相对）（左下方）按钮“相关”。

我错过了什么？

<!doctype html>
<head>
    <script src='http://code.jquery.com/jquery-1.7.2.min.js'></script>

<style type="text/css">
    .slideToggleBox{
        float:left;
        padding:8px;
        margin:16px;
        border:1px solid red;
        width:200px;
        height:50px;
        background-color:blue;
        color:white;
        position:fixed;
        left:-10.5px;
        top:2.3%;
    }
    .clear{
        clear:both;
    }
</style>
</head>

<body>

<div class="clear">
     <button id=slideToggle>slideToggle()</button>
    <br/>
    <div class="slideToggleBox">
        slideToggle()
    </div>
</div>

<script type="text/javascript">
$("#slideToggle").click(function () {
   $('.slideToggleBox').slideToggle();
});
</script>
</body>
</html>

Answer 1

试试这段代码。添加了demo here

<强> HTML

<div class="wrapper">
  <button class=slideToggle>slideToggle()</button>
  <div class="slideToggleBox">slideToggle()</div>
</div>
<div class="wrapper">
  <button class=slideToggle>slideToggle()</button>
  <div class="slideToggleBox">slideToggle()</div>
</div>

<强> CSS

.wrapper{
  float:left;
  width:200px;
}
.slideToggleBox {
  float:left;
  border:1px solid red;
  width:200px;
  height:50px;
  background-color:blue;
  color:white;
  margin-top: -1px;
  margin-left:1px;
}

<强> JS

$(".slideToggle").click(function () {
  $(this).next().slideToggle();
});

按钮上的Jquery slideDown单击

1 个答案: