我正在尝试使用JS创建一个菜单,当我点击第一个菜单项并且它打开时,我希望当我点击下一个项目时我希望第一个关闭,第二个打开和当我点击打开的菜单项时,我希望它关闭。当我点击任何地方,如果我可以让它关闭... 这里是我的代码 HTML
<div class="container-header">
<div class="navbar">
<ul>
<li>
<a onclick="dropdownToggle(event)"><span>Products</span></a>
<div class="dropdown">
<ul class="submenu-img">
<li><a><img src="2015_images/img.png">title</a></li>
<li><a><img src="2015_images/img.png">title</a></li>
<li><a><img src="2015_images/img.png">title</a></li>
<li><a><img src="2015_images/img.png">title</a></li>
<li><a><img src="2015_images/img.png">title</a></li>
</ul>
</div>
</li>
<li><a onclick="dropdownToggle(event)"><span>Services</span></a>
<div class="dropdown">
<ul class="submenu">
<li><a>item</a></li>
<li><a>item</a></li>
<li><a>item</a></li>
<li><a>item</a></li>
</ul>
</div>
</li>
<li><a onclick="dropdownToggle(event)"> <span>Softwares</span></a>
<div class="dropdown">
<ul class="submenu">
<li><a>item</a></li>
<li><a>item</a></li>
<li><a>item</a></li>
<li><a>item</a></li>
</ul>
</div>
</li>
</ul>
</div>
CSS
* {
box-sizing: border-box;
}
ul {
list-style: none;
}
.navbar {
width: 100%;
text-align: center;
position: absolute;
top: -500px;
background-color: #585859;
transition: top 0.3s;
z-index: 1;
}
@media screen and (min-width: 768px) {
.navbar {
line-height: 65px;
width: auto;
position: static;
}
}
.navbar ul {
margin: 0;
padding: 0;
}
.navbar li {
vertical-align: middle;
color: #fff;
padding: 10px;
border-bottom: 5px solid #fff;
}
@media screen and (min-width: 768px) {
.navbar li {
display: inline-block;
margin-right: 10px;
border: none;
padding: 0;
}
}
.dropdown {
display: none;
}
.dropdown-open {
display: block;
position: absolute;
top: 70px;
left: 0;
background: rgba(88, 88, 89, 0.7);
width: 100%;
}
.submenu {
width: 100%;
display: flex;
align-items: center;
flex-flow: row;
justify-content: center;
padding: 10px 0 !important;
}
.submenu li {
margin: 0 2px;
flex: 0 0 12%;
line-height: normal;
}
和JS
function dropdownToggle(event) {
var dropdownItem = event.target.parentElement.parentElement.getElementsByClassName("dropdown")[0];
if (dropdownItem.classList.contains("dropdown-open")) {
var dropdowns = document.querySelectorAll(".dropdown");
var arrayLength = dropdowns.length;
for (var i = 0; i < arrayLength; i++) {
dropdowns[i].classList.remove("dropdown-open");
}
}
else {
dropdownItem.classList.toggle("dropdown-open");
}
}
我尝试使用JSFiddle,但它没有工作,所以这里也是一个codepen .. http://codepen.io/nnns/pen/rLvdgE
答案 0 :(得分:0)
最好的方法是放
var dropdowns = document.querySelectorAll(".dropdown");
var arrayLength = dropdowns.length;
for (var i = 0; i < arrayLength; i++) {
dropdowns[i].classList.remove("dropdown-open");
}
在一个函数中,并在function dropdownToggle(event)
的开头调用它,所以每当你调用函数时,你都会关闭所有内容并打开你想要的东西。
修改强>
这样做:
function dropdownToggle(event) {
var dropdownItem =event.target.parentElement.parentElement.getElementsByClassName("dropdown")[0];
var dropdowns = document.querySelectorAll(".dropdown");
var arrayLength = dropdowns.length;
for (var i = 0; i < arrayLength; i++) {
dropdowns[i].classList.remove("dropdown-open");
}
dropdownItem.classList.add("dropdown-open");
}
答案 1 :(得分:0)
git merge feature/xxxxxx
这将删除所有下拉列表打开类并切换在a的父(li)中找到的下拉列表。 要关闭身体,请点击:
function dropdownToggle(event) {
var dropdownItem = event.target.parentElement.getElementsByClassName("dropdown")[0];
dropdownItem.classList.remove("dropdown");
var dropdowns = document.querySelectorAll(".dropdown");
var arrayLength = dropdowns.length;
for (var i = 0; i < arrayLength; i++) {
dropdowns[i].classList.remove("dropdown-open");
}
dropdownItem.classList.toggle("dropdown-open");
dropdownItem.classList.add("dropdown");
}