jQuery: Finding duplicate ID's and removing all but the first

问题

    $('[id]').each(function () {

        var ids = $('[id="' + this.id + '"]');

        // remove duplicate IDs
        if (ids.length > 1 && ids[0] == this) $('#' + this.id).remove();

    });

The above will remove the first duplicate ID, however I want to remove the last. I've tried $('#'+ this.id + ':last') but to no avail.

Fiddle

In the fiddle the input with the value 'sample' should be kept when the append action takes place.

回答1:

Use jquery filter :gt(0) to exclude first element.

$('[id]').each(function () {
    $('[id="' + this.id + '"]:gt(0)').remove();
});

Or select all the available elements, then exclude the first element using .slice(1).

$('[id]').each(function (i) {
    $('[id="' + this.id + '"]').slice(1).remove();
});

回答2:

Try:

 $('[id="' + this.id + '"]:not(#" + this.id + ":first)').remove();

回答3:

you can try

$("#ID").nextAll().remove();

回答4:

$('[id]').each(function() {
   var $ids = $('[id=' + this.id + ']');
   if ($ids.length > 1) {
     if(this.id === your_id)//which is duplicating
         $ids.not(':first').remove();
     }
});

回答5:

This code is longer than some of the others, but the double-nested loop should make its operation obvious.

The advantage of this approach is that it only has to scan the DOM to generate the list of elements with an id attribute once, and then uses the same list to find (and remove) the duplicates.

Elements that were already removed will have parentNode === null so can be skipped while iterating over the array.

var $elems = $('[id]');
var n = $elems.length;

for (var i = 0; i < n; ++i) {
    var el = $elems[i];
    if (el.parentNode) {  // ignore elements that aren't in the DOM any more
        var id = el.id;
        for (var j = i + 1; j < n; ++j) {
            var cmp = $elems[j];
            if (cmp.parentNode && (cmp.id === id)) {
                $(cmp).remove();  // use jQuery to ensure data/events are unbound
            }
        }
    }
}

回答6:

try this

var duplicated = {};

$('[id]').each(function () {   

    var ids = $('[id="' + this.id + '"]');

    if ( ids.length <= 1 ) return  ;

    if ( !duplicated[ this.id ] ){
         duplicated[ this.id ] = [];   
    }       

    duplicated[ this.id ].push( this );

});

// remove duplicate last ID, for elems > 1 
for ( var i in duplicated){

    if ( duplicated.hasOwnProperty(i) ){  

             $( duplicated[i].pop() ).remove();            
    }
}

and jsfiddle is http://jsfiddle.net/z4VYw/5/

回答7:

I'd use not() to remove the current element from ids and remove the rest:

(Fiddle)

$('body').on("click", ".placeholder", function() {

    data = '<section id="e1"><input name="e1name" /></section><section id="two"><input name="two" value="two section" /></section>';

    $('form').append(data);

        // ideally I'd like to run this function after append but after googling I find that's not possible.
        // for each ID ...
        $('[id]').each(function () {

            var ids = $('[id="' + this.id + '"]');

            if (ids.length>1) {                    
                ids.not(this).remove();            
            }

            return;

        });

});

回答8:

Try Using The below function Works fine for me:

 $('#favourities-ajax ul li').each(function(i) {
      $('[id="' + this.id + '"]').slice(1).remove();
      });

来源：https://stackoverflow.com/questions/15759882/jquery-finding-duplicate-ids-and-removing-all-but-the-first

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