1
我在帶滾動和積極的菜單代碼中的問題,... 這裏是我的代碼和小提琴鏈接https://jsfiddle.net/ajm7Ljwu/如何根據滾動選擇菜單項?
只是檢查頁面3和第4的菜單,而滾動向下和向上.. 請幫我...謝謝(請不要給有關jQuery的任何解決方案)
CSS
<style type="text/css">
a{
position: fixed;
}
div{
float:left;
width:100px;
}
p{
height:500px;
}
</style>
HTML
<div>
<a href="" class="ele elemMe1">1</a><br>
<a href="" class="ele elemMe2">2</a><br>
<a href="" class="ele elemMe3">3</a><br>
<a href="" class="ele elemMe4">4</a><br>
</div>
<div>
<p class="elemMe" id="elemMe1">1</p>
<p class="elemMe" id="elemMe2">2</p>
<p class="elemMe" id="elemMe3">3</p>
<p class="elemMe" id="elemMe4">4</p>
</div>
的JavaScript
<script type="text/javascript">
var currentPagePosition = pageYOffset;
window.onscroll = function(e){
var allDiv = document.getElementsByClassName('elemMe');
console.log(allDiv);
for(var i = 0; i < allDiv.length; i++){
if(pageYOffset > allDiv[i].offsetTop-allDiv[i].offsetHeight-100){
var p_id = allDiv[i].attributes['id'].value;
var currentMenu = document.getElementsByClassName(p_id);
currentMenu[0].style.color = "red";
}else if(pageYOffset < allDiv[i].offsetTop-allDiv[i].offsetHeight){
currentMenu[0].style.color = 'black';
}
}
};
</script>