接下来的prev，nextAll和prevAll方法是非常有用的，但如果您尝试查找的元素不在同一个父元素中，则不会。我想做的是这样的：
<div>
<span id="click">Hello</span>
</div>
<div>
<p class="find">World></p>
</div>
当按住id点击的跨度时，我想将下一个元素与类find匹配，在这种情况下，它不是点击的元素的兄弟，所以next()或nextAll()将不起作用。
解决方法
我今天自己在处理这个问题，这是我想出来的：
/**
* Find the next element matching a certain selector. Differs from next() in
* that it searches outside the current element's parent.
*
* @param selector The selector to search for
* @param steps (optional) The number of steps to search,the default is 1
* @param scope (optional) The scope to search in,the default is document wide
*/
$.fn.findNext = function(selector,steps,scope)
{
// Steps given? Then parse to int
if (steps)
{
steps = Math.floor(steps);
}
else if (steps === 0)
{
// Stupid case :)
return this;
}
else
{
// Else,try the easy way
var next = this.next(selector);
if (next.length)
return next;
// Easy way Failed,try the hard way :)
steps = 1;
}
// Set scope to document or user-defined
scope = (scope) ? $(scope) : $(document);
// Find kids that match selector: used as exclusion filter
var kids = this.find(selector);
// Find in parent(s)
hay = $(this);
while(hay[0] != scope[0])
{
// Move up one level
hay = hay.parent();
// Select all kids of parent
// - excluding kids of current element (next != inside),// - add current element (will be added in document order)
var rs = hay.find(selector).not(kids).add($(this));
// Move the desired number of steps
var id = rs.index(this) + steps;
// Result found? then return
if (id > -1 && id < rs.length)
return $(rs[id]);
}
// Return empty result
return $([]);
}
所以在你的例子中
<div><span id="click">hello</span></div> <div><p class="find">world></p></div>
您现在可以使用“p”元素查找和操作
$('#click').findNext('.find').html('testing 123');
我怀疑它会在大型结构上表现很好，但这里是:)
