CODEWITHSUNDEEP
jquery
Adam
Adam

Reputation: 1971

JQUERY load an array of elements into a DIV

I have an array of DIV elements in an array called 'noactiveMsg'. I want to load these DIV elements into a DIV called $('#chatCenterMembers').

I've tried:

noactiveMsg.appendTo($('#chatCenterMembers'));

But this doesn't work. Can someone advice on the syntax to load an array of elements/objects into a div?

code to made the array:

var noactiveMsg=[];
 $('div.chatmember').each(function() {                        
 if($(this).children().children('div').hasClass('chatactivestatus_offline')) {
 alert('offline');
 noactiveMsg.push($(this));
 }              
 });

Upvotes: 1

Views: 4081

Answers (3)

Joseph Silber
Joseph Silber

Reputation: 220156

You'll want to merge them all into one jQuery object:

var noactiveMsg = [$('<a>1</a>'), $('<a>1</a>')],
    collection;

$.each(noactiveMsg, function(i, e) {
    collection ? collection = collection.add(e) : collection = e;
});

EDIT:

Others have suggested adding every element in the array separately into the DOM. I would strongly advise you against it. Interacting with the DOM is an expensive operation, and should be avoided whenever possible.

Upvotes: 0

maček
maček

Reputation: 77826

If each element of your array is a jQuery object, you can try this:

Some sample html: 

<div>
    <p>Some stuff below!</p>
</div>

The jQuery stuff: 

$(function(){
    var foo = [];
    foo.push( $('<p>hello</p>') );
    foo.push( $('<p>world</p>') );
    foo.push( $('<p>Adam</p>') );

    // this will fail because each element is a 
    // separate jQuery object
    // $('div').append( foo );

    // instead try this
    $.each(foo, function(){
        $('div').append(this);
    })
});

see it work on jsFiddle

I also answered a similar question just recently

Upvotes: 1

BrainSpills
BrainSpills

Reputation: 1

I think this is the code you need. Assuming noactiveMsg is an array containing jQuery objects.

cnt = noactiveMsg.length();

for(i=0; i<cnt; i++) {

$('#chatCenterMembers').append(noactiveMsg[i].html());

}

Upvotes: 0

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