CODEWITHSUNDEEP
jqueryhtmlclone
Avinash
Avinash

Reputation: 2005

Jquery clone and remove div

I have implemented jquery clone and remove. When ADD button is clicked a DIV A with certain form elements are cloned and inserted into another DIV B. In DIV A there is a hidden form button REMOVE. Now I require to enable the REMOVE button only in the clones when ADD button is clicked. i.e; I want to keep the form element in DIV A always hidden.

This is my code.

<div class="rule" id="rule">                        
            <div class="fm-req">
                <select name="rulefield" id="rulefield">
                    <option value="">select</option>
                </select>
            </div>
            <div class="fm-opt" >
                <input type="button" class='remove' value="-" style="display:none;"/>
            </div>
        </div>                  
    <div class="fm-rulebutton">
        <input type="button" id="addButton "class='add' value="+"/>
    </div>

        <div id='TextBoxesGroup' class="group">

here Div 'rule' is cloned into Div 'TextBoxesGroup' 

$(document).ready(function() {
    var id = 0;
    $('.add').click(function(){
            id++;
            $('.fm-opt').children('.remove').show();
            var prot = $(document).find(".rule").clone();
            prot.attr("class", 'rule'+id)
            prot.find(".id").attr("value", id);

            $(document).find("div .group").append(prot);
    });        


    $('.remove').live("click", function(){
            $(this).closest("#rule").remove();
    });
});

Upvotes: 2

Views: 9490

Answers (3)

redDevil
redDevil

Reputation: 1919

$('.add').click(function(){
            id++;
            $('.fm-opt').children('.remove').show();
            var prot = $(document).find(".rule").clone();
            prot.attr("class", 'rule'+id)
            prot.find(".id").attr("value", id);

            $(document).find("div .group").append(prot);
    });

should be changed as,

$('.add').click(function(){
            id++;
            var prot = $(document).find(".rule").clone();
            prot.attr("class", 'rule'+id)
            prot.children('.remove').show();
            prot.find(".id").attr("value", id);
            $(document).find("div .group").append(prot);
    });

You should make visible only those buttons which have been cloned.

Upvotes: 3

musefan
musefan

Reputation: 48435

The problem is you are calling .show() for all remove buttons. You need to limit it to only the new clone element. Like so:

$('.add').click(function(){
    id++;
    var prot = $(document).find(".rule").clone();
    prot.attr("class", 'rule'+id)
    prot.find(".id").attr("value", id);
    prot.find('input.remove').show();//<-- this is the important part

    $(document).find("div .group").append(prot);
});

This code will now only call .show() on the remove button that is found within the newly cloned element

Upvotes: 3

iPhoneDeveloper
iPhoneDeveloper

Reputation: 179

While cloning add unique Id="" value for cloned element.

In remove method, you can easily access the unique element and remove it.

Upvotes: 0

Related Questions