How can I select the “shallowest” matching descend

问题:

How can I select the first "shallowest" input?

My current selection will be the div marked "selected".
I won't know how many levels down it will be.

<div class="selected"> <!-- already have this -->
  <div class="unknown-number-of-wrapper-panels">
    ...
    <div class="collection">    
      <div class="child">
        <input type="text" value="2" /> <!-- don't want this -->
      </div>
    </div>
    <input type="text" value="2" /> <!-- need this -->
    <input type="text" value="2" />
    ...
  </div>
</div>

It seems like find().first() gives me the deepest one.

Edited for clarity. I need to find it based on the fact that it is shallower, not based on other unique attributes.

This might be like a reverse of closest() ?

回答1:

If I understand your issue, you need to recursively check the child nodes for elements with that class.

function findShallowest( root, sel ) {
    var children = root.children();
    if( children.length ) {
        var matching = children.filter( sel );
        if( matching.length ) {
            return matching.first();
        } else {
            return findShallowest( children, sel );
        }
    } else {
        return null;
    }
}

var selected = $('.selected');

findShallowest( selected, ':text' );

Example: http://jsfiddle.net/Qf2GM/

---

EDIT: Had forgotten a return statement, and had an ID selector instead of a class selector for the initial .selected.

---

Or make it into your own custom plugin:

Example: http://jsfiddle.net/qX94u/

(function($) {
    $.fn.findShallowest = function( sel) {
        return findShallowest( this, sel );
    };
    function findShallowest(root, sel) {
        var children = root.children();
        if (children.length) {
            var matching = children.filter(sel);
            if (matching.length) {
                return matching.first();
            } else {
                return findShallowest(children, sel);
            }
        } else {
            return $();
        }
    }
})(jQuery);

var result = $('.selected').findShallowest( ':text' );


alert( result.val() );

回答2:

You are after a breadth-first search rather than the depth-first search (which jQuery's find() uses). A quick google has found: http://plugins.jquery.com/project/closestChild

This could be used like this:

$(...).closestChild('input')

回答3:

Just to golf this "plugin" a bit - Uses @user113716's technique, just reduced code size.

$.fn.findShallowest = function( selector ) {
    var children = this.children(),
        matching = children.filter( selector );
    // return an empty set if there are no more children
    if ( !children.length ) {
        return children;
    }
    // if anything matches, return the first.
    if ( matching.length ) {
        return matching.first();
    }
    // check grand-children
    return children.findShallowest( selector );
};

Try on jsFiddle

回答4:

This is another approach. The idea is that you get the matching element with the least number of ancestors:

(function($) {
    $.fn.nearest = function(selector) {
        var $result = $();
        this.each(function() {
            var min = null,
                mins = {};
            $(this).find(selector).each(function() {
                var n_parents = $(this).parents().length,                
                if(!mins[n_parents]) {
                     mins[n_parents] = this;
                     min = (min === null || n_parents < min) ? n_parents : min;
                }              
             });
             $result = $result.add(mins[min]);
       });

       return $result;
    };
}(jQuery));

Usage:

$('selected').nearest('input');

DEMO

findShallowest, as @patrick has it, might be a better method name ;)

回答5:

If you don't know the class name of the bottom level element you can always use something like

$('.unknown-number-of-wrapper-panels').children().last();

回答6:

Well given your markup, would the following work?

$('.selected div > input:first')