显示/隐藏Div也是固定的

Question

我一直来回来回，我确信有一个简单的答案，但这开始让我感到挫败。

我需要的是一个div框，它可以在点击一个按钮时切换。这是一个聊天脚本，将始终停留在底部角落。我希望按钮始终可见，然后如果您单击该按钮，聊天脚本将打开。然后，您可以通过再次点击按钮来尽可能减少它。我通过多次玩这么近，但我需要2个div（因为第一个div将包含始终可见的按钮，而第二个div将在点击时出现）。

更新：上面的当前脚本打开并关闭div没有问题。当滚动时，我不能同时遵循页面，尽管它们是“位置：固定的”。

$(function() { 
 
    $('.button1').on('click', function() { 
 
    $('.initiallyHidden').toggle(); 
 
    }); 
 
});

#calendar { 
 
    position: fixed; 
 
    bottom: 0; 
 
    right: 0; 
 
    float: right; 
 
} 
 

 

 
} 
 
#initiallyHidden { 
 
    position: fixed; 
 
    bottom: 0; 
 
    right: 0; 
 
    float: right; 
 
    width: 100px; 
 
    height: 100px; 
 
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 

 
<div class="calendar"> 
 
    <button class="button1">Quick Chat</button> 
 
</div> 
 

 
<div class="initiallyHidden" style="display: none;"> 
 
    <iframe src="'.$url.'" marginheight="0" marginwidth="0" scrolling="yes" allowtransparency="yes" frameborder="0"> 
 
</div>

上述当前的脚本打开和关闭，没有问题的股利。当滚动时，我不能同时按这个页面，尽管他们是position:fixed;。

请帮帮我。

Answer 1

$("#button1").on("click",()=>{ 
 
\t $("#initiallyHidden").toggle(); 
 
});

#chatbox { 
 
    position:fixed; 
 
    bottom:0; 
 
    right:0; 
 
    float:right; 
 
} 
 
#initiallyHidden { 
 
    bottom:10; 
 
    right:0; 
 
    float:right; 
 
    width:100px; 
 
    height:100px; 
 
    border-style: solid; 
 
    background-color:white; 
 
} 
 
#content { 
 
    width:100px; 
 
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> 
 
<div id="chatbox"> 
 
<div id="initiallyHidden" style="display: none;"> 
 
    Chat box 
 
</div> 
 
<button id="button1">Quick Chat</button> 
 
</div> 
 
<div id="content"> 
 
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</div>