jQuery找到某个类的next/prev元素,但不一定是兄弟姐妹

Question

next,prev,nextAll和prevAll方法非常有用,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click按下具有id的span时,我想将下一个元素与类匹配find,在这种情况下,该类不是被点击元素的兄弟,所以next()或nextAll()不起作用.

Answer 1

试试这个.它将标记您的元素,创建一组与您的选择器匹配的元素,并从您的元素后面的集合中收集所有元素.

$.fn.findNext = function ( selector ) {
    var set = $( [] ), found = false;
    $( this ).attr( "findNext" , "true" );
    $( selector ).each( function( i , element ) {
        element = $( element );
        if ( found == true ) set = set.add( element )
        if ( element.attr("findNext") == "true" ) found = true;
    })
    $( this ).removeAttr( "findNext" )
    return set
}

编辑

使用jquerys索引方法更简单的解决方案.您调用方法的元素需要由同一个选择器选择

$.fn.findNext = function( selector ){
    var set = $( selector );
    return set.eq( set.index( this, ) + 1 )
}

为了解决这个障碍的功能,我们可以使用浏览器自己的compareDocumentposition

$.fn.findNext = function ( selector ) {
  // if the stack is empty, return the first found element
  if ( this.length < 1 ) return $(s).first();
  var found,
      that = this.get(0);
  $( selector )
    .each( function () {
       var pos = that.compareDocumentPosition( this );
       if ( pos === 4 || pos === 12 || pos === 20 ){
       // pos === 2 || 10 || 18 for previous elements 
         found = element; 
         return false;
       }    
    })
  // using pushStack, one can now go back to the previous elements like this
  // $("#someid").findNext("div").remove().end().attr("id")
  // will now return "someid" 
  return this.pushStack( [ found ] );
},  

编辑2 使用jQuery的$ .grep要容易得多.这是新代码

   $.fn.findNextAll = function( selector ){
      var that = this[ 0 ],
          selection = $( selector ).get();
      return this.pushStack(
         // if there are no elements in the original selection return everything
         !that && selection ||
         $.grep( selection, function( n ){
            return [4,12,20].indexOf( that.compareDocumentPosition( n ) ) > -1
         // if you are looking for previous elements it should be [2,10,18]
         })
      );
   }
   $.fn.findNext = function( selector ){
      return this.pushStack( this.findNextAll( selector ).first() );
   }

当压缩变量名称时,这变成仅仅两个班轮.

使用按位运算编辑3,此功能可能更快？

$.fn.findNextAll = function( selector ){
  var that = this[ 0 ],
    selection = $( selector ).get();
  return this.pushStack(
    !that && selection || $.grep( selection, function(n){
       return that.compareDocumentPosition(n) & (1<<2);
       // if you are looking for previous elements it should be & (1<<1);
    })
  );
}
$.fn.findNext = function( selector ){
  return this.pushStack( this.findNextAll( selector ).first() );
}

Answer 2

我今天自己正在研究这个问题,这就是我想出的:

/**
 * 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>

你现在可以使用找到并操纵'p'元素

$('#click').findNext('.find').html('testing 123');

我怀疑它在大型结构上表现不错,但在这里它是:)