Javascript 获取jQuery集中最宽的元素

假设我有一堆具有不同文本内容的

元素

如何获得最宽的

jQuery很好。我只关心确定跨度，而不关心宽度本身的值

类似的问题也在讨论中。但它们只得到宽度的值

我是这样开始的：

$('span').each(function(id, value){
  if ($(this).width() > w) {
    largestSpan = id;
  }
});

---

使用jQuery识别div中最宽的跨度：

var oSpan;             // Container object for loop item
var oWidest;           // Container object for current widest in loop
var nWidth = 0;        // Int to store current span width
var nWidest = 0;       // Int to contain widest items width
var aWidest = [];      // Array to contain multiple matching spans

// Call each function on spans within div
$('#divContainer span').each(function(nIndex) 
{
    // Set reference to current span to avoid multiple calls per iteration 
    oSpan = $(this);
    // Set current width to avoid multiple calls per iteration
    nSpanWidth = oSpan.width();

    // Compare current width to widest width
    if (nSpanWidth == nWidest)
    {
        // If exact,add to array and set current widest items
        aWidest.push(oSpan);
        oWidest = oSpan;
        nWidest = nSpanWidth;
    }
    else if( nSpanWidth >= nWidest ) 
    {
        // If wider, reset array and set current widest item
        oWidest = oSpan;
        nWidest = nSpanWidth;
        aWidest = [].push(oWidest)
    }
    else { } // Move along short stuff.
});

---

几乎完全相同的答案之间间隔一分钟。仅此一项就投了赞成票。这一项是两项中最优雅、最紧凑的一项这里唯一的抱怨是每次执行

$（this）

时都在创建一个新的jQuery对象。在循环中，效率确实很重要，所以最好只存储一次

$element=$（此）

，然后每次引用

$element

而不是

$（此）

。@AlexKinnee是的，我同意此反馈。更新。再调整一次，您可以运行.width（）一次并存储它，而不是运行两次。

var oSpan;             // Container object for loop item
var oWidest;           // Container object for current widest in loop
var nWidth = 0;        // Int to store current span width
var nWidest = 0;       // Int to contain widest items width
var aWidest = [];      // Array to contain multiple matching spans

// Call each function on spans within div
$('#divContainer span').each(function(nIndex) 
{
    // Set reference to current span to avoid multiple calls per iteration 
    oSpan = $(this);
    // Set current width to avoid multiple calls per iteration
    nSpanWidth = oSpan.width();

    // Compare current width to widest width
    if (nSpanWidth == nWidest)
    {
        // If exact,add to array and set current widest items
        aWidest.push(oSpan);
        oWidest = oSpan;
        nWidest = nSpanWidth;
    }
    else if( nSpanWidth >= nWidest ) 
    {
        // If wider, reset array and set current widest item
        oWidest = oSpan;
        nWidest = nSpanWidth;
        aWidest = [].push(oWidest)
    }
    else { } // Move along short stuff.
});