比較offset（）。top與對象屬性

Question

我正在嘗試構建一個'scrollSpy'類型的函數。我不知道如何將一個參數與一個對象中的某些值進行比較，並返回（數值上）最高值的名稱。

我有一些標記：

<ul class="nav"> 
    <li><a href="a"></a></li> 
    <li><a href="b"></a></li> 
    <li><a href="c"></a></li> 
    <li><a href="d"></a></li> 
    <li><a href="e"></a></li> 
    <li><a href="f"></a></li> 
</ul> 

<section id="a" class="scrollspy"></section> 
<section id="b" class="scrollspy"></section> 
<section id="c" class="scrollspy"></section> 
<section id="d" class="scrollspy"></section> 
<section id="e" class="scrollspy"></section> 
<section id="f" class="scrollspy"></section> 

還有一些腳本，創建一個對象包括各section，其距離的像素從頂部：

var obj = { 
    sectionOffset: {}, 
    scrollSpy: function (scrolled) { 
     // This should look in the object with the offset values, and from all 
     // the values that (scrolled) is higher than, take the largest, and 
     // return its name. 
    } 
} 


$(function() { 

    $(window).scroll(function() { 
     obj.scrollSpy($(document).scrollTop()); 
    }); 

    // Create an object consisting of all elements I want to 'spy' on. 
    // Names and values are made of element ID and top offset respectively. 
    $('.scrollspy').each(function() { 
     obj.sectionOffset[$(this).attr('id')] = (parseInt($(this).offset().top)); 
    }); 

}); 

後，我依次通過我想要的元素它會產生如下對象：

{ 
    d: 5195, 
    b: 3245, 
    a: 1319, 
    f: 5682, 
    c: 2139, 
    e: 3343 
} 

要明確，如果th e用戶在頁面上滾動3000px，函數應該返回c。

Answer 1

scrollSpy: function (scrolled) { 

    var highest = 0, highestName = null; 

    // iterate through the offsets 
    $.each(obj.sectionOffset, function(name, val) { 
     // check that scrolled is higher than the value, and that it's the highest found so far 
     if (scrolled > val && val > highest) { 
      highest = val; 
      highestName = name; 
     } 
    }); 

    return highestName; 
} 

Answer 2

$('.scrollspy').each(function() 
{ 
    if(($(this).offset().top + $(this).height()) > 0) 
    { 
     //this is the one you are looking for. 

     return false; //break the loop. 
    } 
}); 

這會像scrollSpy fiddle